Ramanujan, the "Euler" of 20th century

      The first time I heard of the name Ramanujan was in 2017. At that time, I was wondering on the internet, searching information about infinite series connected to the work of the titan Leonard Euler. I stumbled across a page of Wolfram and I was totally hooked by the great exposition of its author. I was highly intrigued by many infinite series identities established by Ramanujan in his first letter to Hardy. Since then, I have held the deepest respect for Ramanujan for his unmatched ability of working with infinite processes (infinite series, infinite nested radicals, infinite products, infinite continued fractions). His mental power rivals that of Euler, yet his result is still highly distinctive from the master. On his blog, arguably one of the greatest mathematicians in our generation, Terence Tao, said like this about Ramanujan:

"I of course only know of Ramanujan through his work and through secondary sources, but I understand that he performed a prodigious amount of numerical computation and experimentation in the course of his research, which may have pointed him the way to some of the amazing identities and other mathematical results that he discovered, and would have given him a rather different intuition and “box of tools” than what a more mainstream, theory-based approach to mathematics would give. Still, his talent was exceptionally unusual, and he is one of the few successful mathematicians of the modern era that I would see as a plausible contender for the title of “genius” as the term is popularly used. But the bulk of mathematical progress nowadays comes from more prosaic individual and collaborative mathematical effort – much more Hardy than Ramanujan, so to speak."

There are 2 main articles here that were written specifically on the life and work of Ramanujan:

Who was Ramanujan by Stephen Wolfram

Thoughts on Ramanujan by Paramanand Sigh

They have covered almost everything that needs to be known about the life of Ramanujan. In this article, I only wish to point out certain similarities and differences between Leonard Euler and Ramanujan.

Similarity

Both Euler and Ramanujan possessed powerful memories and massive mental power for calculations. Their skills in manipulating mathematical symbols remain unmatched. Euler was very fond of infinite series and perform all kinds of operations on them, a passion that was shared by Ramanujan. However, Euler was born in the 18th century, and at that time, the theory of convergence of numerical, power and trigonometric series were not very well understood, thoroughly investigated and tightly defined. In the case of Ramanujan, his poverty-stricken background kept him from knowing many latest advances in the theory of infinite series in particular, and in mathematics in general. As such, he obtained some results that would not make sense to the 20th and 21th century mathematicians. Nevertheless, they were both highly skillful in exploiting known boundaries to discover more beautiful and deep identities.

Both Euler and Ramanujan were deeply religious. Euler was a devout protestant who were not concerned much with other issues except for numbers and mathematics. Ramanujan, till the end of his life, was a deeply conservative Hindu, who observed strict diets even though he had fallen ill many times during his stay in Cambridge.

Difference

Euler

Although there was little doubt that Euler was born to be a magnificent mathematician, he was very lucky to be born in a well-connected family. At an early age, Euler was already taken under the wing of Johann Bernoulli, and was close to his sons Daniel and Nicolaus Bernoulli. They were all highly ranked, if not one of the best mathematicians of that century. Needless to say, Euler was tutored by the greatest minds of Europe. This wonderful background of education, coupling with his incredible talent for Mathematics was what would make Euler the master of us all.

Euler was, in many senses, also lucky to born in Europe. He had many chances to maintain contact with other prominent mathematicians of his time, like Joseph Lagrange, the Bernoullis, Fagnano, etc. This means he could keep tab on the latest development of Mathematics and Mathematical Physics. His extensive correspondence portraited him as a man who were knowledgeable of all scientific fields in his days.

Euler was also, due to his talent, accepted into and sheltered by 2 great Academies in Europe, one in Russia, Petropolis (now Saint Petersburg), and one in Berlin. Thus, one could assume that he lived a very comfortable life and did not have to worry about mundane financial matters. It was a very prestigious position, and many mathematicians like de Moivre spent their lives to achieve just that.

Finally, Euler lived a very long life, a long and glorious life that covered almost the whole century that bore his name whenever people talked about XVIII century mathematics. He published 5 textbooks in total: Arithmetic, Elements of Algebra, Introduction to analysis of the infinite, Foundations in differential calculus, Foundations in integral calculus. Apart from all these, his papers published in many journals were all available to general leaders. Historians of mathematics had a lot of materials to research the life and work of this giant.

Ramanujan

In contrast to Euler, Ramanujan was much more unfortunate. He was not born in a well-connected family, was not accepted into any prestigious academy that would provide him good pension, and was not educated by any great mathematicians like Euler. Yet his profound talent had dragged him from the dredge of poverty to the shining summit of learning, when Hardy found ways to let him travel to England. 

Ramanujan developed his mathematics in isolation, with little help from outsiders. This placed a cost on his discoveries: many of his results were already known in Europe. Yet, even after taking away these pearls, the rest of his works were still startling and opened new doors for later mathematicians like Bruce C. Berndt.

Ramanujan, unlike Euler, did not have enough materials to write down derivations and proofs for many of his results. What we were left with were only 4 notebooks and few articles. This had led many to falsely assume that Ramanujan did not know the proofs and could just come up with the results. This was unattainable given the fact that his notebooks showed a fragment of the massive number of calculations that he went through. Lacking in money, he wrote down his final results only. Yet, a few demonstrations he showed indicate a very clear and sophisticated method behind all his formulae.

Some interesting infinite series of Ramanujan presented in the letter he sent to Hardy:

$$\sum_{n=1}^{+\infty} \dfrac{\coth(n\pi)}{n^7}=\dfrac{\coth(\pi)}{1^7}+\dfrac{\coth(2\pi)}{2^7}+\dfrac{\coth(3\pi)}{3^7}...=\dfrac{19\pi^7}{56700}$$

This series is divergent. Since:

$$\coth(x)=\dfrac{e^x+e^{-x}}{2}$$

$$\dfrac{\coth(n\pi)}{n^7}=\dfrac{e^{n\pi}+e^{-n\pi}}{2n^7}$$

$\displaystyle\lim_{n\to\infty}\dfrac{e^{-n\pi}}{2n^7}=0$ and $\displaystyle\lim_{n\to\infty}\dfrac{e^{n\pi}}{2n^7}=+\infty$

Thus the series diverge and the sum is not meaningful in ordinary sense. 

$$\dfrac{1}{1^5\cosh\left(\dfrac{\pi}{2}\right)}+\dfrac{1}{3^5\cosh\left(\dfrac{3\pi}{2}\right)}+\dfrac{1}{5^5\cosh\left(\dfrac{5\pi}{2}\right)}...=\dfrac{\pi^5}{768}$$

From Littlewood's Miscellany, (page 95, 96):

"Ramanujan great gift is a "formal" one, he dealt in "formulae". To be quite clear what is meant, I give two examples (the second is at random, the first one is of supreme beauty).

$$p(4)+p(9)x+p(14)x^2+...=5\dfrac{[(1-x^5)(1-x^{10})(1-x^{15})...]^5}{[(1-x)^2(1-x^3)(1-x^5)...]^6}$$

where $p(n)$ is the number of partitions of n

$$\int_{0}^{\infty}\dfrac{\cos(\pi x)}{[\Gamma(\alpha+x)][\Gamma(\alpha-x)]^2}dx=\dfrac{1}{4\Gamma(2\alpha-1)[\Gamma(\alpha)]^{2}} \left(\alpha>\dfrac{1}{2}\right)$$

But the great day of formulae seems to be over. No-one, if we are again to take the highest standpoint, seems to be able to discover a radically new type, though Ramanujan comes near in his work of partition series; it is futile to multiply examples in the sphere of Cauchy's theorem and elliptic function theory,  and some general theory dominates, if in a less degree, every other field. A hundred years or so ago, his power would have had ample scope. Discoveries alter the general mathematical atmosphere and have very remote effects, and we are not prone to attach great weight to rediscoveries, however independent they seem. How much are we allow for this, how great a mathematician might Ramanujan have been 100 or 150 years ago; and what would have happened if he had come in touch with Euler at the right moment?...

Derving two general formulae of trigonometric series found in "Subsidium Calculi Sinuum" of Euler

 In this post, with the help of A-Student, I have proved the two trigonometrical formulae of Euler. In this thread, I will prove 2 more general formlae found in his paper "Subsidium Calculi Sinuum" (Contribution to the calculation of sine) by Euler.

First exploration of Fourier series Part 2

First exploration of Fourier series Part 1

In the last post, I have entered into the field of trigonometric series. Naturally, this introduces me to Fourier series.

Lagrange and the trigonometric series $\sum_{n=1}^\infty \cos(nx)$

 This series $\displaystyle\sum_{n=1}^\infty \cos(nx)=\cos(x)+\cos(2x)+\cos(3x)+...$ has attracted my curiosity because many prominent mathematicians in the XVIII century came up with contradictory results. Chiefly among these mathematicians are Euler and Lagrange. I have introduced this series in this thread, so please visit it to learn more on how Euler derived it.

Lagrange's approach

I have kept the links to many papers written by Euler, Lagrange, Daniel Bernoulli and d'Alembert on the famous problem of vibrating string. Among these papers, I am particularly interested in a paper of Lagrange entitled "Recherche sur la nature et la propagation du son" (Research on the nature of the propagation of wave) (see the entire work here). On page 110, he wrote:

"Supposons, pour simplifier le calcul, que la séries dont on veut prendre la somme soit généralement"

$$\cos(x)+\cos(2x)+\cos(3x)+\cos(4x)...$$

He rewrote the series in exponential forms, and formed a sum of two infinite terms:

$$=\dfrac{e^{ix}+e^{-ix}}{2}+\dfrac{e^{2ix}+e^{-2ix}}{2}+\dfrac{e^{3ix}+e^{-3ix}}{2}+\dfrac{e^{4ix}+e^{-4ix}}{2}...$$

$$=\dfrac{e^{ix}+e^{2ix}+e^{3ix}+e^{4ix}}{2}...+=\dfrac{e^{-ix}+e^{-2ix}+e^{-3ix}+e^{-4ix}}{2}...$$

$$=\dfrac{1}{2}\left[e^{ix}+e^{2ix}+e^{3ix}+e^{4ix}...\right]+\dfrac{1}{2}\left[e^{-ix}+e^{-2ix}+e^{-3ix}+e^{-4ix}...\right]$$

Here we have two infinite geometric series with the common ratio $e^{ix}$ and $e^{-ix}$. We can write it in the form $\dfrac{1}{1-e^{ix}}$, but we need to subtract $1$ because the geometric series has $1$ as its first term, while our series do not have this.

$$=\dfrac{1}{2}\left[\dfrac{1}{1-e^{ix}}-1\right]+\dfrac{1}{2}\left[\dfrac{1}{1-e^{-ix}}-1\right]$$

$$=\dfrac{1}{2}\left[\dfrac{1}{1-e^{ix}}-\dfrac{1-e^{ix}}{1-e^{ix}}\right]+\dfrac{1}{2}\left[\dfrac{1}{1-e^{-ix}}-\dfrac{1-e^{-ix}}{1-e^{-ix}}\right]$$

$$=\dfrac{e^{ix}}{2(1-e^{ix})}+\dfrac{e^{-ix}}{2(1-e^{-ix})}$$

Adding them together, we have:

$$\dfrac{e^{ix}(1-e^{-ix})+e^{-ix}(1-e^{ix})}{2(1-e^{ix})(1-e^{-ix})}$$

$$=\dfrac{e^{ix}-+e^{-ix}-2}{2(1-e^{ix}-e^{-ix}+1)}=\dfrac{e^{ix}+e^{-ix}}{2(-e^{ix}-e^{-ix}+2)}-\dfrac{2}{2(2-e^{ix}-e^{ix})}$$

$$=\dfrac{\cos(x)-1}{2-e^{ix}-e^{-ix}}==\dfrac{\cos(x)-1}{2-2\left(\dfrac{e^{ix}-e^{-ix}}{2}\right)}=\dfrac{\cos(x)-1}{2(1-\cos(x))}=-\dfrac{1}{2}$$

Thus, according to Lagrange, this series has the sum of $-\dfrac{1}{2}$.

However, we plug $x=0$ into $\cos(x)+\cos(2x)+\cos(3x)+...$, we will have a series of the form $1+1+1+1+1...=+\infty$

So this series admits two value. In our modern view, it is divergent. Yet Lagrange held on so tightly that he offered us an explanation that was very characteristic of mathematicians of his generation:

Mais, dira-t-on, comment peut-il se faire que la somme de la suite infinis $\cos(x)+\cos(2x)+\cos(3x)+...$ soit toujours égale à $-\dfrac{1}{2}$, puisque dans le cas $x=0$, elle devient nécessairement égale à une suite d'autant d'unités? Je ré ponds que cela provient des termes qui se détruisent naturellement dans tous les cas, excep té dans celui òu $x=0$. (page 111)

Lagrange then proceeds to examine the partial sum of this series. 

"Pour rendre la chose plus sensible, cherchons la somme de la suite:"

$$\cos(x)+\cos(2x)+\cos(3x)+\cos(4x)+...\cos(mx)$$

He used the formula for summing finite geometric series $\dfrac{1-r^{n+1}}{1-r}$. Rewrite the finite given series above as:

$$=\dfrac{e^{ix}+e^{2ix}+e^{3ix}+e^{4ix}+...e^{mix}}{2}+\dfrac{e^{-ix}+e^{-2ix}+e^{-3ix}+e^{-4ix}+...e^{-mix}}{2}$$

Apply the formula, substracting $1$ from the series as above, we have:

$$\dfrac{1-e^{(m+1)ix}}{1-e^{ix}}-\dfrac{1-e^{ix}}{1-e^{ix}}+\dfrac{1-e^{-(m+1)ix}}{1-e^{-ix}}-\dfrac{1-e^{-ix}}{1-e^{-ix}}=\dfrac{e^{ix}-e^{(m+1)ix}}{2(1-e^{ix})}+\dfrac{e^{-ix}-e^{-(m+1)ix}}{2(1-e^{ix})}$$

$$=\dfrac{(e^{ix}-e^{(m+1)ix})(2-2e^{-ix})+(2-2e^{-ix})(e^{-ix}-e^{-(m+1)ix})}{4(2-e^{ix}-e^{-ix})}$$

$$=\dfrac{\color{red}{2}(e^{ix}-1-e^{(m+1)ix}+e^{mix}+e^{-ix}-e^{-(m+1)ix}-1+e^{-mix})}{\color{red}{4}(2-e^{ix}-e^{-ix})}$$

$$=\dfrac{(e^{ix}-1-e^{(m+1)ix}+e^{mix}+e^{-ix}-e^{-(m+1)ix}-1+e^{-mix})}{2(2-e^{ix}-e^{-ix})}$$

$$=\dfrac{(e^{ix}+e^{-ix}-e^{mix}+e^{-mix}-(e^{(m+1)ix}+e^{-(m+1)ix})-2)}{2(2-e^{ix}-e^{-ix})}$$

$$=\dfrac{e^{ix}+e^{-ix}}{2(2-e^{ix}-e^{-ix})}+\dfrac{e^{mix}+e^{-mix}}{2(2-e^{ix}-e^{-ix})}-\dfrac{e^{(m+1)ix}+e^{-(m+1)ix}}{2(2-e^{ix}-e^{-ix})}-\dfrac{\color{red}{2}}{\color{red}{2}(2-e^{ix}-e^{-ix})}$$

$$=\dfrac{\cos(x)+\cos(mx)-\cos[(m+1)x]-1}{2-2\left(\dfrac{e^{ix}+e^{-ix}}{2}\right)}=\dfrac{\color{red}{\cos(x)}+\cos(mx)-\cos(m+1)\color{red}{-1}}{2\color{red}{(1-\left(\cos(x)\right)})}$$

$$=\dfrac{\cos(mx)-\cos[(m+1)x]}{2{(1-\left(\cos(x)\right)})}-\dfrac{1}{2}$$

He stated that in the case when $m$ is an infinite number, $1$ vanishes along with $m$, where the term $\cos(m+1)$ equals to $\cos(mx)$, the formula is equal to:

$$\dfrac{\cos(x)+\cos(mx)-\cos[(m+1)x]}{2{(1-\left(\cos(x)\right)})}=0$$

Differentiating the expression, we have:

$$\dfrac{(m+1)\sin[(m+1)x]-m\sin(mx)}{2\sin(x)}$$

Let $x=0$, the expression has the form $\frac{0}{0}$

Differentiating twice, we have:

$$\dfrac{(m+1)^2\sin[(m+1)x]-m^2\sin(mx)}{2\cos(x)}$$

And by making $x=0$

$$\dfrac{(m+1)^2-m^2}{2}=m+\dfrac{1}{2}$$

So the value is:

$$m+\dfrac{1}{2}-\dfrac{1}{2}=m$$

Proof that the concerned series is divergent for all $x$

In order to prove this, we need two theorems that are encountered in Calculus and Analysis:

Theorem 1: If a series is convergent, this implies that the limit of the sequence that makes up that series must tend to 0. 

Theorem 2 (this theorem is pointed out to me by Nguyễn Mạnh Linh, but the blog that contains this theorem along with its proof has been deleted):

Let $(x_{n})_{n=1}^{+\infty}$ be a sequence of real numbers, with $x$ being given. If all subsequences of $(x_{n})_{n=1}^{+\infty}$ contain a subsequence that converges to a limit, then $(x_{n})_{n=1}^{+\infty}$ also converges to that limit.

Now, using theorem 1, let assume that the series $\displaystyle\sum_{n=1}^\infty cos(nx)$ is convergent. This implies that the sequence $(cos(nx))_{n=1}^{+\infty}$ converges to zero as n tends to infinity. Applying theorem 2, this imples that all subsequences of $(cos(nx))_{n=1}^{+\infty}$ contain a subsequence that converges to 0. Now, for the sequence $(cos(nx))_{n=1}^{+\infty}=\cos(x), \cos(2x), \cos(3x), \cos(4x),...$, we can extract a subsequence $(cos(2nx))_{n=1}^{+\infty}=\cos(2x), \cos(4x),...$ that tends to zero as n approaches infinity. This means $\displaystyle\lim_{n\to\infty} \cos(2nx)=0$.

From trigonometry, we know that $\cos(2nx)=2\cos^2(nx)-1$. If $\displaystyle\lim_{n\to\infty} \cos(nx)=0$, this means $\displaystyle\lim_{n\to\infty} 2\cos^2(nx)-1=2\cdot 0-1=-1$. This proves that  $\displaystyle\lim_{n\to\infty} \cos(2nx)=-1$, hence  $\displaystyle\lim_{n\to\infty} \cos(nx)=0$ is impossible. Thus, the series is divergent.

I wish to thank Vũ Tuấn Hiền for providing me guidance to this problem. His direction is important for me to decisively prove that this series is divergent.

Historical commentary:

This series and the method of using to sum it is demonstrated by the famed French mathematician Henri Lebesgue in his small lecture note on trigonometric series entitled "Leçons sur les séries trigonométriques professées au Collège de France" (page 35, section 21). He wrote: "Que l'on donne à la methode l'une ou l'autre forme, elle n'est rigoureuse que si l'on a étudié pour $|z|=1$, la série qui joue le rôle de (Z). Dans ce cas particulier, elle conduit à un résulta exact, mais, dans d'autres cas, elle peut conduire à des résultas incorrects, c'est ainsi que Lagrange écrivait l'égailité

$$0=\dfrac{1}{2}+\cos(x)+\cos(2x)+\cos(3x)+...$$

alors que la série du second membre est divergent, comme on le voit en calculant la somme de ses n premiers termes."

The $C+iS$ method for "summing" infinite trigonometric series-Part 1

  I was introduced to the so-called $C+iS$ method when I tried to prove these 2 formulae from the work of Euler. I don't know how to call this method of technique. Upon searching it on youtube, I've come across several videos made by Indian teachers who call it $C+iS$ method. $C$ stands for cosine series, and $S$ stands for sine series. Essentially, it makes use of the Euler formula $e^{ix}=\cos(x)+i\sin(x)$. The method was described by A-Level student, who helped me to prove those 2 formulae, "pretty powerful". While I don't deny its versatility, I think it can be dangerous to over-rely on this method because the issue of convergence of trigonometric series is highly complicated and has a lot of nuances (I don't even have the basic understanding of their conditions of convergence). In order to fully grasp all kinds of convergence of trigonometric and Fourier series, one must have enough prerequisites to read and understand reference works such as Antoni Zygmund's (a famed Polish mathematician) "Trigonometric Series" or Nina Karlovna Bary's "A Treatise of trigonometric series". I have read the more elementary, serving as an introduction to the subject, "Fourier series" by Georgi P. Tolstov, but many nuances do not make their way to my head. It is a highly difficult subject, and I don't have a background in measure theory to fully appreciate it. This issue is what I think dangerous if one use the $C+iS$ method all the times to derive the "sum" of a given trigonometric series, no matter how meaningful that is.

Below, I give a rough sketch and some exercises I have done with this method.

Suppose you are given a sine or cosine series of the form:

$$a_0\sin(x)+a_1\sin(x+y)+a_2\sin(x+2y)+a_3\sin(x+3y)+...$$

$$a_0\cos(x)+a_1\cos(x+y)+a_2\cos(x+2y)+a_3\cos(x+3y)+...$$

One may note that the arguments of these 2 series form an arithemtic progression.

One put:

$$C=a_0\cos(x)+a_1\cos(x+y)+a_2\cos(x+2y)+a_3\cos(x+3y)+...$$

$$S=a_0\sin(x)+a_1\sin(x+y)+a_2\sin(x+2y)+a_3\sin(x+3y)+...$$

Then:

$$C+iS=a_0[\cos(x)+i\sin(x)]+a_1[\cos(x+y)+i\sin(x+y)]+a_2[\cos(x+2y)+i\sin(x+2y)]+...$$

Apply the Euler formula:

$$e^{ix}=\cos(x)+i\sin(x)$$

$$C+iS=a_0e^{ix}+a_1e^{i(x+y)}+a_2e^{i(x+2y)}+a_3e^{i(x+3y)}...$$

According to Sydney Luxton Loney's "A Treatise of Plane Trigonometry, part 2: Analytical Trigonometry" page 114, the $C+iS$ series fall into 4 categories:

1) Those depending on the summation on a geometrical progression

2)Those depending on the binomial series

3)Those depending on the exponential theorems, including sub-cases, the sine and cosine series

4)Those depending on the logarithmic and, as subcase, Gregory's series (the series for $\arctan(x)$ and $artanh(x)$

One can utilize these series to sum up the $C+iS$ series above. Please consult this table of Maclaurin series for reference.

Then, after summing up and doing some more works to refine it, one can express it in the form of $A+iB$. Equating this with $C+iS$, one can easily point out $C=A$ and $S=B$, thus one obtains the "sum" of these two trigonometric series.

$C+iS$ series depending on the geometrical series:

Problem 1: 

Find the sum of $S_1=\displaystyle\sum_{n=1}^{\infty} (-1)^{n-1}\sin(nx)$ and $C_1= \displaystyle\sum_{n=1}^{\infty} (-1)^{n-1}\cos(nx)$

Notes: This series can be derived easily using this formula:

$$S_1 = \sin(x)-\sin(2x)+\sin(3x)-\sin(4x)...=\displaystyle\sum_{n=1}^{\infty} (-1)^{n-1}\sin(nx)$$

$$C_1 = \cos(x)-\cos(2x)+\cos(3x)-\cos(4x)...= \displaystyle\sum_{n=1}^{\infty} (-1)^{n-1}\cos(nx)$$

$$-1+C_1 = -1+\cos(x)-\cos(2x)+\cos(3x)-\cos(4x)...$$

$$-1+C_1+iS_1=-1+[\cos(x)+\sin(x)]-[\cos(2x)+\sin(2x)+[\cos(3x)+\sin(3x)]...$$

$$-1+e^{ix}-(e^{ix})^2+(e^{ix})^3-(e^{ix})^4+...$$

This is the geometric series, and  its "sum" is $-\dfrac{1}{1+e^{ix}}$

So, we have:

$$-\frac{1}{1+e^{ix}}=-\frac{1+e^{-ix}}{(1+e^{ix})(1+e^{-ix})}$$

$$=-\frac{1+\cos(x)-i\sin(x)}{(1+e^{ix})(1+e^{-ix})}=-\frac{1+\cos(x)-i\sin(x)}{2\cos(x)+2}$$

Equating the real and imaginary part with $C+iS$, we have:

$$\boxed{-1+C_1=-\frac{1+\cos(x)}{2+2\cos(x)}\implies C=\frac{1}{2}}$$

$$\boxed{S_1=\frac{\sin(x)}{2\cos(x)+2}}$$

Using the same precedure, one can also derive the series of:

$$\boxed {S_2 =\displaystyle\sum_{n=1}^{\infty} \sin(nx) = \sin(x)+\sin(2x)+\sin(3x)+\sin(4x)...=\dfrac{\sin(x)}{2\cos(x)-2}}$$

$$ \boxed {C_2 = \displaystyle\sum_{n=1}^{\infty}\cos(nx)=\cos(x)-\cos(2x)+\cos(3x)-\cos(4x)...= -\dfrac{1}{2}}$$

Remark:

For the two cosine series $C_1$ and $C_2$ above, the "sums" obtaining by this purely algebraical method doesn't make sense since they are divergent series for all x. For a proof, please refer to this post. The sum can only made sense not in the sense of convergence. But this lies outside of my domain of knowledge right now. These series appear in "Subsidium Calculi Sinuum" of Euler, the last part of this paper.

For the two sine series $S_1$ and $S_2$ above, it is easy to notice that at $x=0$ and $x=k\pi$, with $k$ being any integers, the two series attain a finite limit, that is $0$. At $x=\pi/2$, for example, the $S_2$ series turn into $1+0-1+0+1...$, which is divergent. Hence the series only converge for $x=0$ and $x=k\pi$.

$C+iS$ series depending on the logarithmic series:

Problem 2

$$S_1 = a\sin(x)-\dfrac{1}{2}a^2\sin(2x)+\dfrac{1}{3}a^3\sin(3x)-\dfrac{1}{4}a^4\sin(4x)...=\displaystyle\sum_{n=1}^{\infty} (-1)^{n-1}\dfrac{1}{n}a^n\sin(nx)$$

$$C_1 = a\cos(x)-\dfrac{1}{2}a^2\cos(2x)+\dfrac{1}{3}a^3\cos(3x)-\dfrac{1}{4}a^4\cos(4x)...=\displaystyle\sum_{n=1}^{\infty} (-1)^{n-1}\dfrac{1}{n}a^n\cos(nx)$$

$$C_1+iS_1=a[\cos(x)+i\sin(x)]-\dfrac{1}{2}a^2[\cos(2x)+i\sin(2x)]+\dfrac{1}{3}a^3[\cos(3x)+i\sin(3x)]-...$$

$$C_1+iS_1=ae^{ix}-\dfrac{1}{2}(ae^{ix})^2+\dfrac{1}{3}(ae^{ix})^3-\dfrac{1}{4}(ae^{ix})^4...$$

Let $t=ae^{ix}$, we have

$$C_1+iS_1=t-\dfrac{t^2}{2}+\dfrac{t^3}{3}-\dfrac{t^4}{4}...=\log(1+t)=\log(1+ae^{ix})$$

$$C_1+iS_1=\log[\color{green}{1+a\cos(x)}+i\color{blue}{a\sin(x)]}$$

We apply the formula for complex logarithm: $\log(\color{green}{a}+i\color{blue}{b})=\frac{1}{2}\log[a^2+b^2]+i\arctan(\frac{b}{a})$

$$\log[1+a\cos(x)+ia\sin(x)]=\dfrac{1}{2}\log[(1+a\cos(x)]^2+a\sin^2(x)]+i\arctan\left(\dfrac{a\sin(x)}{1+a\cos(x)}\right)$$

$$\boxed{C_1=\dfrac{1}{2}\log[(1+a\cos(x)]^2+a\sin^2(x)]}$$

$$\boxed{S_2=\arctan\left(\dfrac{a\sin(x)}{1+a\cos(x)}\right)}$$

Problem 3

$$S_2 = \sin(x)-\dfrac{1}{2}sin(2x)+\dfrac{1}{3}\sin(3x)-\dfrac{1}{4}\sin(4x)...=\displaystyle\sum_{n=1}^{\infty} (-1)^{n-1}\dfrac{1}{n}\sin(nx)$$

$$C_2 = \cos(x)-\dfrac{1}{2}\cos(2x)+\dfrac{1}{3}\cos(3x)-\dfrac{1}{4}\cos(4x)...=\displaystyle\sum_{n=1}^{\infty} (-1)^{n-1}\dfrac{1}{n}\cos(nx)$$

This is just a special case of the problem above, where $a=1$, skipping all these steps, one arrive at:

$$C_2=\dfrac{1}{2}\log[(1+\cos(x)]^2+\sin^2(x)]$$

$$S_2=\arctan\left(\dfrac{\sin(x)}{1+\cos(x)}\right)$$

Use the identity $\tan\left(\dfrac{x}{2}\right)=\dfrac{\sin(x)}{1+\cos(x)}$, we can rewrite 

$$S_2=\arctan\left(\tan\left(\dfrac{x}{2}\right)\right)=\dfrac{x}{2}$$'

Remark: 

Again, here, we don't know what is the interval where this "sum" $\dfrac{x}{2}$ is valid using this algebraical method. In Fourier analysis, one can rigorously derive this series by stating that $S_2 = \displaystyle\sum_{n=1}^{\infty} (-1)^{n-1}\dfrac{1}{n}\sin(nx)$ is the Fourier series representation of the function $\dfrac{x}{2}$ on the interval $[\pi, \pi]$. Since the $\dfrac{x}{2}$ is an odd function, it only has the sine series as its representation.

We have the formula for the coefficients of the Fourier series:

$$a_0+\sum_{n=1}^{+\infty}(a_n\cos(nx)+b_n\sin(nx))$$

$$a_0=\dfrac{1}{2 \pi}\displaystyle\int_{-\pi}^{\pi} \dfrac{x}{2}dx=\dfrac{1}{2 \pi}\left[\dfrac{x^2}{4}\right]_{-\pi}^{\pi}=\dfrac{1}{2 \pi}\left(\dfrac{(\pi)^2}{2}-\dfrac{(-\pi)^2}{2}\right)=0$$

$$b_n=\dfrac{1}{\pi}\displaystyle\int_{-\pi}^{\pi} \dfrac{x}{2}\sin(nx)=\displaystyle\sum_{n=1}^{+\infty} (-1)^{n-1}\dfrac{\sin(nx)}{n}$$

Problem 4

Find the sum of the series $\displaystyle\sum_{n=1}^{+\infty}\dfrac{1}{n}\sin(nx)$

$$S_3 = \sin(x)+\dfrac{1}{2}sin(2x)+\dfrac{1}{3}\sin(3x)+\dfrac{1}{4}\sin(4x)...=\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n}\sin(nx)$$

$$C_3 = \cos(x)+\dfrac{1}{2}\cos(2x)+\dfrac{1}{3}\cos(3x)+\dfrac{1}{4}\cos(4x)...=\displaystyle\sum_{n=1}^{\infty} \dfrac{1}{n}\cos(nx)$$

$$C_3+iS_3=[\cos(x)+i\sin(x)]+\dfrac{1}{2}[\cos(2x)+i\sin(2x)]+\dfrac{1}{3}[\cos(3x)+i\sin(3x)]+...$$

$$C_3+iS_3=e^{ix}+\dfrac{1}{2}(e^{ix})^2+\dfrac{1}{3}(e^{ix})^3+\dfrac{1}{4}(e^{ix})^4...$$

Let $t=e^{ix}$, we have

$$C_3+iS_3=t+\dfrac{t^2}{2}+\dfrac{t^3}{3}+\dfrac{t^4}{4}...=-\log(1-t)=-\log(1-e^{ix})$$

$$C_3+iS_3=-\log[\color{green}{1+a\cos(x)}-i\color{blue}{a\sin(x)]}$$

We apply the formula for complex logarithm: $\log(\color{green}{a}+i\color{blue}{b})=\frac{1}{2}\log[a^2+b^2]+i\arctan(\frac{b}{a})$

$$-\log[1-\cos(x)-i\sin(x)]=-\dfrac{1}{2}\log[(1-\cos(x)]^2-\sin^2(x)]-i\arctan\left(-\dfrac{\sin(x)}{1-\cos(x)}\right)$$

$$\boxed{C_1=-\dfrac{1}{2}\log[(1-\cos(x)]^2-a\sin^2(x)]}$$

$$iS_3=-i\arctan\left(-\dfrac{\sin(x)}{1-\cos(x)}\right)$$

Using the elementary trigonometric identities $\dfrac{\sin(x)}{1-\cos(x)}=\cot\left(\dfrac{x}{2}\right)$ and $\tan\left(\dfrac{\pi-x}{2}\right)$

$$\boxed{S_3=-\arctan\left(-\tan\left(\dfrac{\pi-x}{2}\right)\right)=\arctan\left(\tan\left(\dfrac{\pi-x}{2}\right)\right)=\dfrac{\pi-x}{2}}$$

Remark

Like above, $C+iS$ doesn't indicate the interval in which this equation holds. It should be noted that since the series in problem 3 is valid for the interval $[-\pi, \pi]$, replacing $x$ by $\pi-x$ sends $[-\pi, pi]$ to $[0, 2\pi]$

Historical remark

This series could be found in Euler's "Institutiones calculi differentialis" (Foundations in differential calculus, published first time in 1755) (chapter 4 "Concerning the conversion into series" page 526, Ian Bruce's translation) (page 271, Russian edition entitled "дифференциальное исчислениие", 1949) (page 388, first Latin edition, 1755). He derived series after a long and complicated consideration for the series of $\arctan(x)$. He wrote the series as below and differentiate this sine series to obtain the series in problem 1:

$$S=\dfrac{\pi}{2}=\dfrac{u}{2}+\dfrac{1}{1}\sin(u)+\dfrac{1}{2}\sin(2u)+\dfrac{1}{3}\sin(3u)...$$

$$S'=C=0=\dfrac{1}{2}+\cos(u)+\cos(2u)+\cos(3u)+\cos(4u)+\cos(5u)+...$$

$C$ is obviously a divergent series for all x.

If we integrate this divergent $C$ series, we obtained a true identity (though the interval for validity must be indicated).

Mathematicians of Euler's time freely differentiated and integrated trigonometric series since they viewed them as purely algebraic formulae to obtain accurate numerical series. The accuracy of the numerical series was what that mattered. To Euler, Lagrange and other XVIII century mathematicians, algebraic rules that held for finite algebraic expression can be extended to infinite expressions. This is what Cauchy call the "generality of algebra". However, it is untrue, as claimed by many historians, that Euler and his contemporaries had virtually no understanding of convergence. Joseph Louis François Bertrand (1822-1900) wrote in his "Traité Calcul différentiel et intégral" (published in 1864), page 230, section 228, a scathing critic about mathematicians of XVIII century as followed "Les géomètres du XVIII siècle attache peu importance à la convergence des séries. Les plus illustres d'entre eux fait illusion sur le peu de rigeur des raisonments  où elle interviennent". This line of attack has its own merits, because convergence of functional series were still poorly understood. Another author, Martin Ohm, older brother of the famed German physicist Georg Ohm, wrote in his tract "The Spirit of Mathematical Analysis: And Its Relation to a Logical System" that:

"The theory of infinite series rests at the present upon a very bad foundation. All operations are applied to them as if they were finite, but is this permissible? I think not. Where is it demonstrated that the differential of the infinite series is found by taking the differential of each term? Nothing is easier than to give examples where this rule is not correct; for example,

$$\dfrac{x}{2}=\sin(x)-\dfrac{1}{2}\sin(2x)+\dfrac{1}{3}\sin(3x)-\dfrac{1}{4}\sin(4x)+...$$

by differentiating we obtain

$$\dfrac{1}{2}=\cos(x)-\cos(2x)+\cos(3x)-\cos(4x)+$$

an utterly false result, for this series is divergent." (page 2)

Euler was very well aware of the divergence of series, but he still found divergent series useful in obtaining accurate results. Louis Antoine de Bougainville, an explorer but had been a mathematician in his youth, a student of d'Alembert, called the "sum" of divergent series as "faux" (false), and convergent series as "vrais"(true) (see his Traité du Calcul Intégral pour servir de suite à l'analyse des infiniments petits de M. le Marquis de l'Hopital, volume 1):

Les series convergents sont celles dont les termes vont diminuant à l'infini. Nous prouverons plus-bas que ces series sont les seuls vraies.

Les series divergents sont celles dont les termes vont en augmentant à l'infini. (page 302)

The criteria for determining the convergence is rather simple and far from being sophisticated as what is presented in a Calculus course. Of all illustrous mathematicians of Euler's generation, it seems that d'Alembert was the most careful in assessing the convergence of series. Perhaps, for this reason, the d'Alembert criteria, which is found in modern textbook, bears his name. In the early 19th century, several great mathematicians such as Abel and Cauchy spent great efforts in establishing criteria for convergence of numerical and functional series. It took mathematicians of the later generation, from the first half of the 19th century till the 20th century, to figure out the issue of convergence and other issues related to trigonometric and Fourier series.

Yet divergent series have made a come back to mathematics. Euler's view was somewhat vindicated by the emergence of the field of mathematics that study asymptotic expansions (or series) of functions. But I shall stop my comment here as I have no knowledge of this field of mathematics.

The $C+iS$ method for "summing" infinite trigonometric series-Part 2

We will continue to explore this method with several other types of series.

How did Euler first derive the power series for $\sin(x)$ and $\cos(x)$ without using differentiation?

The binomial series and binomial theorem

Zeta function at even integers, deriving the formula for summing infinite series of the form $\sum_{n=1}^{+\infty}\frac{1}{n^{2m}}$

 My recent excursion to the general formula for the power series of trigonometric and hyperbolic $\cot(x)$ and $\coth(x)$ leads me further to one of the most beautiful and deep results in modern mathematics. The similarity between the closed form summing formula of the Zeta function at even integers and the general formula of $\cot(x)$ puzzles me, for I can see that there are some connections between the two. A long search on wikipedia and mathstackexchange proves my intuition. Yet, I am still a highly incompetent math student so I cannot, by myself, deduce the connection between $\cot(x)$ and the summing formula. The following derivation and results are owned not to me but to wikiproof and an answer of Jack D'Aurizio. Just compare between the general formula for $\cot(x)$ and the summing formula of Zeta function at even integers and you will see some connections:

$$\cot(x)=\sum_{n=0}^{+\infty}\dfrac{(-1)^nB_{2n}2^{2n}x^{2n-1}}{(2n)!}$$

$$\zeta(2n)=\dfrac{(-1)^{n-1}(2\pi)^{2n}}{2(2n)!}B_{2n}$$

The similarity is striking to me, suggesting that there ought to be a connection of both of them. It turns out that my intuition is right. The relevant function is a modification of $\cot(x)$, which is the $\pi x\cot(\pi x)$. My initial guess is that the function $cot(\pi x)$ may be involved, but hours spent on worthless manipulation didn't lead me to the correct result. So as I mentioned above, I resorted to searching on google and I come up with $\pi x\cot(\pi x)$. 

History of problems

The history of the problems of summing infinite series of the form $\displaystyle\sum_{n=1}^{+\infty} \dfrac{1}{n^{2m}}$ is just as fascinating as the result itself. Infinite series had been a central focus of study of mathematicians in the XVII and XVIII century. Summing series was indeed very in vogue during this time period, and many, not just leading mathematicians, involved in this business for recreational purposes. The polymath Pierre Louis Maupertuis (1698-1759) made this following remark about this phenomenon in the XVIII century England:

"Cette affaire des suites qui est tout ce qu'il ly a de plus desagreable dans les mathematiques n'est qu'un jeu pour les Anglais, ce livre et celui de Moivre en sont une preuve."

"This business of series, the most disagreeable thing mathematics, is no more than a game for the English, this book and that of M.de Moivre are the proof."

Pietro Mengoli (1626-1686), an Italian mathematician who had studied with Bonaventura Cavalieri, in his "Novae quadraturae arithmeticae"  was successful in summing these telescoping series:

$$\displaystyle\sum_{n=1}^{+\infty}\dfrac{1}{n(n+2)}=\dfrac{1}{1\cdot 3}+\dfrac{1}{2\cdot 4}+\dfrac{1}{3\cdot5}...=\dfrac{3}{4}$$

$$\displaystyle\sum_{n=1}^{+\infty}\dfrac{1}{n(n+3)}=\dfrac{1}{1\cdot 4}+\dfrac{1}{2\cdot 5}+\dfrac{1}{3\cdot6}...=\dfrac{11}{18}$$

$$\displaystyle\sum_{n=1}^{+\infty}\dfrac{1}{n(n+1)(n+2)}=\dfrac{1}{1\cdot 2\cdot 3}+\dfrac{1}{2\cdot 3\cdot 4}+\dfrac{1}{3\cdot 4\cdot 5}...=\dfrac{1}{4}$$

$\displaystyle\sum_{n=1}^{+\infty}\dfrac{1}{(2n+1)(2n+3)(2n+5)}=\dfrac{1}{3\cdot 5\cdot 7}+\dfrac{1}{5\cdot 7\cdot 9}+\dfrac{1}{7\cdot 9\cdot 11}...=\dfrac{1}{60}*$ *The book gives $\dfrac{1}{12}$, which is an error.

(See Ferraro, The rise and development of the theory of series up to the early 1820s, page 9).

He also successfully prove that the harmonic series is divergent $\sum_{n=1}^{+\infty}\dfrac{1}{n}$ (a result that was first proved by the late medieval French mathematician Nicole Oresme). More importantly, he shows that the alternating harmonic series converges to $\ln(2)$:

$$\displaystyle\sum_{n=1}^{+\infty}\frac{(-1)^n+1}{n}=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}...=\ln(2)$$

The now famous series:

$$\displaystyle\sum_{n=1}^{+\infty}\frac{1}{n^{2}}=1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}...=?$$

is beyond his power. He posed this problems to leading mathematicians of the time, yet it must be roughly 100 years later that Euler could finally provide various methods and proofs for the sum of this series, and similar series of higher even powers.

In his first textbook of teaching infinitesimal calculus to young students "Introductio a analysi infinitorum", he summed these series up to the 26th powers. The first 10 are included here:

$$1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+\dfrac{1}{5^2}+\dfrac{1}{6^2}...=\dfrac{\pi^2}{6}$$

$$1+\dfrac{1}{2^4}+\dfrac{1}{3^4}+\dfrac{1}{4^4}+\dfrac{1}{5^4}+\dfrac{1}{6^4}...=\dfrac{\pi^4}{90}$$

$$1+\dfrac{1}{2^6}+\dfrac{1}{3^6}+\dfrac{1}{4^6}+\dfrac{1}{5^6}+\dfrac{1}{6^6}...=\dfrac{\pi^6}{945}$$

$$1+\dfrac{1}{2^8}+\dfrac{1}{3^3}+\dfrac{1}{4^8}+\dfrac{1}{5^8}+\dfrac{1}{6^8}...=\dfrac{\pi^8}{9450}$$

$$1+\dfrac{1}{2^{10}}+\dfrac{1}{3^{10}}+\dfrac{1}{4^{10}}+\dfrac{1}{5^{10}}+\dfrac{1}{6^{10}}...=\dfrac{\pi^{10}}{93555}$$

My "failed" attempt at deriving the generating function for $\zeta(2n)$

My failed attempt at deriving the generating function for $-\dfrac{\pi x}{2}\cot(\pi x)$

When I look at the formula, I know that I only have adjust some bits of the general formula of the $\cot(x)$ function to prove the summing formula for the value of $\zeta(2n)$. But I forgot just one small bit to complete my derivation. This is because I was quite tired after tinkering with the formula.

We have 

$$\cot(x)=\sum_{n=0}^{+\infty}\dfrac{(-1)^nB_{2n}2^{2n}x^{2n-1}}{(2n)!}$$

$$\cot(\pi x)=\sum_{n=0}^{+\infty}\dfrac{(-1)^nB_{2n}2^{2n}(\pi x)^{2n-1}}{(2n)!}$$

$$-\cot(\pi x)=\sum_{n=0}^{+\infty}\dfrac{(-1)^{n+1}B_{2n}2^{2n}(\pi x)^{2n-1}}{(2n)!}$$

$$-x\cot(\pi x)=\sum_{n=0}^{+\infty}\dfrac{(-1)^{n+1}B_{2n}2^{2n}(\pi x)^\color{red}{{2n+1}}}{(2n)!}$$

$$-\dfrac{x}{2}\cot(\pi x)=\sum_{n=0}^{+\infty}\dfrac{(-1)^{n+1}B_{2n}2^{2n}(\pi x)^\color{red}{{2n+1}}}{2(2n)!}$$

I made the mistake red, it should be $2n$ and forgot to add another $\pi$ in to obtain the correct function. I thought I was wrong and running around looking for the right function. Wikiproof employs $\pi x\cot(\pi x)$ in its proof and I thought that my function was completely wrong. Only after seeing the generating function on wiki page of the Riemann zeta function that I know I was completely right, missing just one $\pi$.

So I have derived the formula all by myself, but  I was so unfortunate that I thought I was completely wrong. The correct function with its coefficients being $\zeta(2n)$ is:

$$-\dfrac{\pi x}{2}\cot(\pi x)=\sum_{n=0}^{+\infty}\dfrac{(-1)^{n+1}B_{2n}2^{2n}(\pi x)^{2n}}{(2n)!}=\sum_{n=0}^{+\infty}\zeta(2n)x^{2n}=\dfrac{1}{2}+\dfrac{\pi^2}{6}x^2+\dfrac{\pi^4}{90}x^4...$$

An interesting derivation of the formula of finding values of the Zeta function at even integers by Jack D'Aurizio

Jack D'Aurizio, an Italian mathematician on MathStackExchange provided this interesting derivation of the formula of finding values of the Zeta function at even integers. The starting point is not the $\cot(x)$ like my approach, but $\coth(x)$. I repost it here and add more intermediate steps to make it easier for readers to understand. The original answer is quite terse and omit middle steps: These middle steps are provided by NoName on Mathstack.

$$ \frac{t}{e^t-1} = \sum_{n\geq 0}\frac{B_n}{n!}t^n \tag{1}$$

$$ \coth z-\frac{1}{z} = \sum_{n\geq 1}\frac{4^n\,B_{2n}}{(2n)!}z^{2n-1}.\tag{2}$$

Jack expressed the function $\sinh(x)$ as the so-called Weierstrass product $\displaystyle\sinh z=z\prod_{n=1}^\infty\left(1+\frac{z^2}{n^2\pi^2}\right)$ which was derived first by Euler.

Since:

$$\coth(z)=\dfrac{d}{dz}\log(\sinh(z))\tag{3}$$

$$\coth z-\frac{1}{z} = \dfrac{d}{dz}\displaystyle\sinh z=\dfrac{d}{dz}\log\left( z\prod_{n=1}^\infty\left(1+\frac{z^2}{n^2\pi^2}\right)\right)$$

Using the law of log, we can rewrite

$$\log(z)+\log\left(\left(1+\dfrac{z^2}{\pi^2}\right)\left(1+\dfrac{z^2}{2\pi^2}\right)\left(1+\dfrac{z^2}{3\pi^2}\right)\right)...\tag{4}$$

$$\dfrac{1}{z}+\log\left(\left(1+\dfrac{z^2}{\pi^2}\right)\right)+\log\left(\left(1+\dfrac{z^2}{2\pi^2}\right)\right)+\log\left(\left(1+\dfrac{z^2}{3\pi^2}\right)\right)...$$

$$\coth z -\color{red}{\frac{1}{z}} = \color{red}{\dfrac{1}{z}} + \sum_{n\geq 1}\frac{d}{dz}\log\left(1+\frac{z^2}{n^2 \pi^2}\right) = \sum_{n\geq 1}\frac{2z}{\pi^2 n^2+z^2}\tag{5}$$

$$=\sum_{n\geq 1}\frac{2z}{\pi^2 n^2+z^2}=\dfrac{2z}{\pi^2 n^2}\cdot \color{green}{\dfrac{1}{1+\left(\dfrac{z^2}{(\pi n)^2}\right)}}$$

We notice that the green part is a geometric series, with its formula $\dfrac{1}{1+t}=\sum{n=1}^{\infty}(-1)^{m-t}t^{m-t}$

$$\sum_{n=1}^{\infty}\dfrac{2z}{\pi^2 n^2}\cdot\sum_{m=1}^{\infty}(-1)^{m-t}\cdot \dfrac{z^{2m-2}}{(\pi n)^{2m-1}}=\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\dfrac{2z^{2m-1}}{\pi^{2m}n^{2m}}=\sum_{m=1}^{\infty}\dfrac{2z^{2m-1}}{\pi^{2m}}\cdot\sum_{n=1}^{\infty}\dfrac{1}{n^{2m}}$$

And we know that $\sum_{n=1}^{\infty}\dfrac{1}{n^{2m}}=\zeta(2n)$

Thus we have:

$$\sum_{m=1}^{\infty}\dfrac{2z^{2m-1}\zeta(2n)}{\pi^{2m}}$$

Comparing the coefficients of $z^{2n-1}$

$$\frac{2^{2n}\,B_{2n}}{(2n)!} =\frac{2\,\zeta(2n)}{\pi^{2n}}(-1)^{n-1} $$

$$\zeta(2n)  =\frac{\,B_{2n} (-1)^{n-1}(2\pi)^{2n}}{2(2n)! }$$

An interesting function that approximates values of Zeta function for both odd and even intergers

The following function, found in this thread of Mathstack, approximates many values of Zeta function at both odd and even intergers

$$f(s) = \frac{\pi^s}{\left\lfloor((2^s - 1)\frac{\pi^s}{2^s}\right\rfloor-1} \approx\zeta(s),$$

First few values are here:

\begin{array}{|c|c|c|} s & f(s) & |\zeta(s) - f(s)| \\ \hline2 &  \frac{\pi^2}{6} & 0 \\ \hline3 &  \frac{\pi^3}{26} & 0.00950\ldots \\ \hline4 &  \frac{\pi^4}{90} & 0 \\ \hline5 &  \frac{\pi^5}{295} & 0.00042\ldots \\ \hline6 &  \frac{\pi^6}{945} & 0 \\ \hline7 &  \frac{\pi^7}{2995} & 0.00009\ldots \\ \hline8 & \frac{\pi^8}{9450} & 0 \\ \hline9 & \frac{\pi^9}{29749} & 0.000011\ldots \\ \hline\end{array}

According to user247327, the approximation of $\frac{\pi^9}{29749}$ is $1.0020202$, which is very far from the assigned value on the table.

Deriving the power series for $\tan(x)$ and $\tanh(x)$

Some times ago, we have been able to obtain the power series of $\tan(x)$ and $\tanh(x)$. But this method of long division fails to make visible the connection between the power series of $\tan(x)$ and $\tanh(x)$ and the Bernoulli numbers. It also comes quite unhandy. In the previous post, we have successfully derived the formula for $\cot(x)$ and $\coth(x)$. We are now ready to derive the general formulae of the power series of $\tan(x)$ and $\tanh(x)$

The crucial identities that we are going to use to derive the series are:

$$\tan(x)=\cot(x)-2\cot(2x)\tag{1}$$

$$\tanh(x)=2\coth(2x)-\coth(x)\tag{2}$$

The power series that we need to use are

$$\cot(x)=\displaystyle\sum_{n=0}^{+\infty}\dfrac{(-1)^{n}B_{2n}2^{2n}x^{2n-1}}{(2n)!}\tag{3}$$

$$\coth(x)=\displaystyle\sum_{n=0}^{+\infty}\dfrac{B_{2n}2^{2n}x^{2n-1}}{(2n)!}\tag{4}$$

$$2\cot(2x)=\displaystyle\sum_{n=0}^{+\infty}\dfrac{(-1)^{n}B_{2n}2^{2n+1}2^{2n-1}x^{2n-1}}{(2n)!}=\displaystyle\sum_{n=0}^{+\infty}\dfrac{(-1)^{n}B_{2n}2^{4n}x^{2n-1}}{(2n)!}\tag{5}$$

$$2\coth(2x)=\displaystyle\sum_{n=0}^{+\infty}\dfrac{B_{2n}2^{2n+1}2^{2n-1}x^{2n-1}}{(2n)!}=\displaystyle\sum_{n=0}^{+\infty}\dfrac{B_{2n}2^{4n}x^{2n-1}}{(2n)!}\tag{6}$$

Power series of $\tan(x)$

We apply the identity from $(1)$

$$\tan(x)=\displaystyle\sum_{n=0}^{+\infty}\dfrac{(-1)^{n}B_{2n}2^{2n}x^{2n-1}}{(2n)!}-\displaystyle\sum_{n=0}^{+\infty}\dfrac{(-1)^{n}B_{2n}2^{4n}x^{2n-1}}{(2n)!}$$

$$=\sum_{n=0}^{+\infty} \dfrac{(-1)^{n}B_{2n}2^{2n}x^{2n-1}-[(-1)^{n}B_{2n}2^{4n}x^{2n-1}]}{(2n)!}$$

$$=\sum_{n=0}^{+\infty} \dfrac{(-1)^{n}B_{2n}[2^{2n}x^{2n-1}-2^{4n}x^{2n-1}]}{(2n)!}$$

$$\boxed{=\sum_{n=0}^{+\infty} \dfrac{(-1)^{n}B_{2n}2^{2n}x^{2n-1}(1-2^{2n})}{(2n)!}}$$

or

$$\boxed{=\sum_{n=0}^{+\infty} \dfrac{(-1)^{n-1}B_{2n}2^{2n}x^{2n-1}(2^{2n}-1)}{(2n)!}}$$

Power series of $\tanh(x)$

$$\tanh(x)=\displaystyle\sum_{n=0}^{+\infty}\dfrac{B_{2n}2^{4n}x^{2n-1}}{(2n)!}-\displaystyle\sum_{n=0}^{+\infty}\dfrac{B_{2n}2^{2n}x^{2n-1}}{(2n)!}$$

$$=\sum_{n=0}^{+\infty} \dfrac{B_{2n}2^{4n}x^{2n-1}-B_{2n}2^{2n}x^{2n-1}}{(2n)!}$$

$$=\sum_{n=0}^{+\infty} \dfrac{B_{2n}[2^{4n}x^{2n-1}-2^{2n}x^{2n-1}]}{(2n)!}$$

$$\boxed{=\sum_{n=0}^{+\infty} \dfrac{B_{2n}2^{2n}x^{2n-1}(2^{2n}-1)}{(2n)!}}$$