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The infinite series I need to solve is $$\sum_{n=1,3,5...}^{\infty}\frac{1}{n^{2}}$$

and because the point of interest lies in the value of odd n,

the infinite series can be expressed as

$$\sum_{n=1}^{\infty}\frac{1}{(2n-1)^{2}}$$

This came up in a quantum mechanics problem involving the expectation value of the Hamiltonian.

Is there a good idea to verify the solution is indeed $$\frac{\pi^{2}}{8}$$ or is this something with which I must refer to a math table? Any good ideas would be helpful.

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marked as duplicate by Najib Idrissi, Alex M., jameselmore, Vlad, user98602 Dec 18 '15 at 15:51

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ you can depend on the $\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$ to prove it $\endgroup$ – E.H.E Dec 18 '15 at 10:24
  • $\begingroup$ You must use the Maclaurin expansion of the function sine (or cosine, I dont remember). $\endgroup$ – Masacroso Dec 18 '15 at 10:45
  • $\begingroup$ solve? what is the unknown variable to solve for? dont you mean evaluate in closed form? $\endgroup$ – Arjang Dec 18 '15 at 11:36
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    $\begingroup$ See also: math.stackexchange.com/questions/1454960/… and math.stackexchange.com/questions/389488/… $\endgroup$ – Martin Sleziak Dec 18 '15 at 14:33
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Consider $$\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\sum_{n=1}^{\infty}\frac{1}{(2n)^{2}}+\sum_{n=1}^{\infty}\frac{1}{(2n-1)^{2}}=\frac{1}4\sum_{n=1}^{\infty}\frac{1}{n^2}+\sum_{n=1}^{\infty}\frac{1}{(2n-1)^{2}}$$ So $$\sum_{n=1}^{\infty}\frac{1}{(2n-1)^{2}}=\frac{3}4 \sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{3}4 \times\frac{\pi ^2}{6}=\frac{\pi ^2}{8}$$

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We have $$ \sum_{n=1}^\infty \frac 1{n^2} = \frac{\pi^2}6 \tag+$$ Hence the even numbers sum to $$ \sum_{n=1}^\infty \frac 1{(2n)^2} = \frac 14 \sum_{n=1}^\infty \frac 1{n^2} = \frac{\pi^2}{24} $$ So, the difference is $$\sum_{n=1}^\infty \frac 1{(2n-1)^2} = \frac{\pi^2}6 - \frac{\pi^2}{24} = \frac{\pi^2}8 $$ The fact that $(+)$ holds, is "well-known", hence perhaps something you could refer to (looking it up in a table), or you compute $$ \int_0^1 \int_0^1 \frac 1{1- xy}\, dy\,dx $$ in two ways, expanding $\frac 1{1-xy} = \sum_{n=0}^\infty (xy)^n$ gives $$ \int_{[0,1]^2} \frac{1}{1-xy} \, d(x,y) = \sum_{n=1}^\infty \frac 1{n^2}$$ On the other hand, let $u = \frac 12(x+y)$, $v= \frac 12(y-x)$, then \begin{align*} \int_{[0,1]^2} \frac 1{1-xy}\, d(x,y) &= 4 \int_0^{1/2}\int_0^u \frac 1{1 - u^2+ v^2} \,dv \, du + 4 \int_{1/2}^1 \int_0^{1-u} \frac1{1 - u^2 + v^2}\, dv\, du\\ &= \frac{\pi^2}6 \end{align*}

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by using $$\frac{\pi^2}{6}=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...$$

$$\frac{\pi^2}{6}=1+\frac{1}{3^2}+\frac{1}{5^2}+..\frac{1}{2^2}(1+\frac{1}{2^2}+\frac{1}{3^2}+...)$$ $$\frac{\pi^2}{6}=1+\frac{1}{3^2}+\frac{1}{5^2}+..\frac{1}{2^2}(\frac{\pi^2}{6})$$ $$\frac{\pi^2}{6}-\frac{\pi^2}{24}=1+\frac{1}{3^2}+\frac{1}{5^2}+..$$ $$\frac{\pi^2}{8}=1+\frac{1}{3^2}+\frac{1}{5^2}+..$$

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We can use the residue theorem:

$$\sum_{n=1}^{\infty} \frac1{(2 n-1)^2} = \frac12 \sum_{n=-\infty}^{\infty} \frac1{(2 n-1)^2} = -\frac{\pi}{2} \operatorname*{Res}_{z=1/2} \frac{\cot{\pi z}}{(2 z-1)^2} = \frac{\pi^2}{8 \sin^2{(\pi/2)}} = \frac{\pi^2}{8}$$

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To use $\zeta(2)=\frac{\pi^2}{6}$ (have a look at this historical question) is a fast way to go, but plenty of approaches are possible. An Euler-style proof follows.

Step 1. The function $f(x)=\cos(\pi x)$ is an entire function having simple roots at $\mathbb{Z}+\frac{1}{2}$.

Step 2. The Mittag-Leffler theorem then gives the Weierstrass product: $$ \cos(\pi z) = \prod_{n\geq 0}\left(1-\frac{4z^2}{(2n+1)^2}\right) \tag{2a}$$ where the LHS can be written as a Taylor series: $$ \cos(\pi z) = 1-\frac{\pi^2 z^2}{2}+\frac{\pi^4 z^4}{24}-\ldots\tag{2b}$$

Step 3. By comparing the coefficient of $z^2$ in the RHS of $(2a)$ and in the RHS of $(2b)$ we get: $$ -\sum_{n\geq 0}\frac{4}{(2n+1)^2} = -\frac{\pi^2}{2}\tag{3} $$ and your claim immediately follows.

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