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Recently I have been thinking a lot about variations of the Basel Problem, and methods to solve them. Here I found the following solution to the Basel Problem by Alfredo Z. (I include the entire answer due to its brevity)

Define the following series for $ x > 0 $

$$\frac{\sin x}{x} = 1 - \frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+\cdots\quad.$$

Now substitute $ x = \sqrt{y}\ $ to arrive at

$$\frac{\sin \sqrt{y}\ }{\sqrt{y}\ } = 1 - \frac{y}{3!}+\frac{y^2}{5!}-\frac{y^3}{7!}+\cdots\quad.$$

if we find the roots of $\frac{\sin \sqrt{y}\ }{\sqrt{y}\ } = 0 $ we find that

$ y = n^2\pi^2\ $ for $ n \neq 0 $ and $ n $ in the integers

With all of this in mind, recall that for a polynomial

$ P(x) = a_{n}x^n + a_{n-1}x^{n-1} +\cdots+a_{1}x + a_{0} $ with roots $ r_{1}, r_{2}, \cdots , r_{n} $

$$\frac{1}{r_{1}} + \frac{1}{r_{2}} + \cdots + \frac{1}{r_{n}} = -\frac{a_{1}}{a_{0}}$$

Treating the above series for $ \frac{\sin \sqrt{y}\ }{\sqrt{y}\ } $ as polynomial we see that

$$\frac{1}{1^2\pi^2} + \frac{1}{2^2\pi^2} + \frac{1}{3^2\pi^2} + \cdots = -\frac{-\frac{1}{3!}}{1}$$

then multiplying both sides by $ \pi^2 $ gives the desired series.

$$\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots = \frac{\pi^2}{6}$$

This solution fascinated me. Although it takes a few things for granted (such as that the Fundamental Theorem of Algebra applies to infinite polynomials) I nevertheless thought the proof was one of the most beautiful I've seen. In attempting to generalize this I stumbled across the following function and its associated power series: $$\cos\left(\frac{x}{\sqrt{2}}\right)\cosh\left(\frac{x}{\sqrt{2}}\right) = \sum_{k=0}^\infty \frac{x^{4k}}{(4k)!}(-1)^k$$
By a simple substitution we find that $$\cos\left(\frac{x^{1/4}}{\sqrt{2}}\right)\cosh\left(\frac{x^{1/4}}{\sqrt{2}}\right) = \sum_{k=0}^\infty \frac{x^{k}}{(4k)!}(-1)^k$$
Which is a polynomial in $x$; noting that $\cosh$ is nowhere zero (on the real line) we solve for the roots as such: $$\cos\left(\frac{x^{1/4}}{\sqrt{2}}\right) = 0 \implies \frac{x^{1/4}}{\sqrt{2}} = n\pi + \frac{\pi}{2} \implies x = \frac{\pi^4(2n+1)^4}{4}$$ Using the argument for Viete's Formula in the solution above, we find that
$$\frac{4}{\pi^4}\sum_{k=0}^\infty \frac{1}{(2k+1)^4} = -\frac{-\frac{1}{4!}}{1}$$ Upon manipulation we find that $$\color{red}{\sum_{k=0}^\infty \frac{1}{(2k+1)^4} = \frac{\pi^4}{96}}$$
From here we can bring this into a form more like that of the Basel Problem by noting that $$\sum_{k=0}^\infty \frac{1}{(2k+2)^4} = \frac{1}{16}\sum_{k=0}^\infty \frac{1}{(k+1)^4} = \frac{1}{16}\sum_{k=1}^\infty \frac{1}{k^4}$$
Noting that this computes the even terms of $\sum_{k=1}^\infty k^{-4}$ we find that the odd terms must equal $\frac{15}{16}\sum_{k=1}^\infty k^{-4}$ which is our series calculated above

Applying this, we find that, as desired,
$$\color{red}{\sum_{k=1}^\infty \frac{1}{k^4} = \frac{\pi^4}{90}}$$

However, I have been unable to generalize this approach further. Intuition tells me that the functions desired would have power series similar to those used, and would be a combination of trigonometric functions. Nevertheless, I have only been able to solve this for the case above (which required me to switch the power series from $\sin(x)$ to $\cos(x)$ from Alfredo's proof, as neglecting $\cosh(x)$ was desirable).

My question is thus: Can this proof format be applied to solve series of the form $\sum_{k=1}^\infty k^{-n}$ for any $n$ other than $2$ and $4$?

Note: I apologize for the length of this post, but I felt that a full presentation of the proof might assist in generalizing it. If anyone has any suggestions concerning this please let me know!

Note 2: In case anyone is troubled by this seemingly naive application of Viete's Formula to infinite polynomials, know that this is perfectly valid, and is known as the Root Linear Coefficient Theorem.

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  • $\begingroup$ See here for the part on roots: en.wikipedia.org/wiki/Weierstrass_factorization_theorem I'm so sure you can simply say "$\cosh$ is nowhere zero" since $\cosh(x)=\cos(ix)$. $\endgroup$ – Simply Beautiful Art Aug 12 '16 at 23:38
  • $\begingroup$ @SimpleArt Wonderful! Thank you for the reference. I suppose then that the Root Linear Coefficient Theorem along with this justifies applying "Viete's Formula and the Fundamental Theorem of Algebra to infinite polynomials" that plagued my mind. Glad to hear it is justified in general! Concerning $\cosh$, I was referring to the real line, but I ought to specify. Thank you for the catch! $\endgroup$ – Brevan Ellefsen Aug 12 '16 at 23:40
  • $\begingroup$ See here for more generalized method, following the same path: math.stackexchange.com/questions/1887144/… $\endgroup$ – Simply Beautiful Art Aug 12 '16 at 23:41
  • $\begingroup$ It is not justified in general, please read the link. $\endgroup$ – Simply Beautiful Art Aug 12 '16 at 23:42
  • $\begingroup$ @SimpleArt I apologize, I misused the term "general" to mean that it works in this case as a special case of something else. I definitely read the link, and it fascinated me. Thank you also for the reference to your post... I marked it as a favorite and meant to look back to see if it was answered but forgot to do so. I will read again through your question and the accepted answer and try to rework this into a solution of my own question. Thanks for your time! $\endgroup$ – Brevan Ellefsen Aug 12 '16 at 23:46
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To start, the "factoring" step is known as the Weierstrass factorization theorem, which asserts that you can express some functions as products of their factors.

From here, take note that you had

$$\frac{\sin(x)}x=\dots\left(1+\frac{x^2}{2^2\pi^2}\right)\left(1+\frac{x^2}{1^2\pi^2}\right)\left(1-\frac{x^2}{1^2\pi^2}\right)\left(1-\frac{x^2}{2^2\pi^2}\right)\dots$$

which is basically what you noted.

From here, I multiply similar terms to get

$$f(x):=\frac{\sin(x)}x=\left(1-\frac{x^2}{1^2\pi^2}\right)\left(1-\frac{x^2}{2^2\pi^2}\right)\left(1-\frac{x^2}{3^2\pi^2}\right)\dots$$

and define this as $f(x)$.

using $(1-ue^{\frac{2\pi i}n})(1-ue^{\frac{4\pi i}n})\dots(1-ue^{2\pi i})=1-u^n$, we can then get

$$f(xe^{\frac{2\pi i}n})f(xe^{\frac{4\pi i}n})\dots f(xe^{2\pi i})=\left(1-\frac{x^{2n}}{1^{2n}\pi^{2n}}\right)\left(1-\frac{x^{2n}}{2^{2n}\pi^{2n}}\right)\left(1-\frac{x^{2n}}{3^{2n}\pi^{2n}}\right)\dots$$

And since this is not generalizable to $n\notin\mathbb N$, we can only solve for

$$\sum_{k=1}^\infty\frac1{k^{2n}}$$

By using this method. As for the odd values, no closed form yet exists.

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  • $\begingroup$ along with the answer to your question linked to provide additional context, this perfectly answers my question. Thank you Simple Art! $\endgroup$ – Brevan Ellefsen Aug 12 '16 at 23:59
  • $\begingroup$ @BrevanEllefsen No problem. $\endgroup$ – Simply Beautiful Art Aug 13 '16 at 0:03

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