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Not having a great series background, I turn to the wider world... :)

I need to find two sums:

$$\sum_{m=1}^{\infty}\frac{\sin^2(m\pi)}{(m^2-a^2)^2}$$

and

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

I've seen some related series on here, but not close enough that I can figure out how to get these sums from the ones posted. Any help would be much appreciated.

Michael Cloud
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  • What is $a$? If it is an integer then both series are undefined... – Servaes Nov 16 '18 at 23:55
  • a is not an integer. For my case $a=n[1+\frac{1}{4}\cos(n)]$ where n is an integer – Michael Cloud Nov 17 '18 at 00:43
  • In more detail the first sum should be $\sum_{m=1}^{\infty}\frac{\sin^2(\frac{m\pi}{\gamma_n}){(m^2-\gamma_n^2n^2)^2}$ where $\gamma_n=[1+\frac{1}{4}\cos(n)]$ and $n$ is a positive natural number – Michael Cloud Nov 17 '18 at 00:49
  • trying one more time to get the formatting right on my previous comment: \sum_{m=1}^{\infty}\frac{\sin^2(\frac{m\pi}{\gamma_n}){(m^2-\gamma_n^2n^2)^2} where \gamma_n=[1+\frac{1}{4}\cos(n)] and n is a positive natural number – Michael Cloud Nov 17 '18 at 00:56
  • You can add $ sign to format Mathjax. – Tianlalu Nov 17 '18 at 00:57
  • @MichaelCloud There should be an "edit" button near the bottom of your question. – epimorphic Nov 17 '18 at 04:17

2 Answers2

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The first one is $0$ because $\sin m\pi = 0$ for any $m\in \Bbb Z$.

For the second one, we split it into two sums, \begin{align*} S=\sum_{m=1}^\infty\frac{m^2}{(m^2-a^2)^2}&=\sum_{m=1}^\infty\frac{m^2-a^2+a^2}{(m^2-a^2)^2}\\ &=\underbrace{\sum_{m=1}^\infty\frac{1}{m^2-a^2}}_{S_1}+a^2\underbrace{\sum_{m=1}^\infty\frac{1}{(m^2-a^2)^2}}_{S_2}. \end{align*} From this answer, we know $$S_1=\frac1{2a^2}-\frac{\pi\cot\,\pi a}{2a}.\tag{*}$$ Differentiate $(*)$ with respect to $a$ (we can do that because $S_1$ uniformly converges), \begin{align*}S_1'=\frac d{da}\sum_{m=1}^\infty\frac{1}{m^2-a^2}&=2a\sum_{m=1}^\infty\frac{1}{(m^2-a^2)^2}=2aS_2\\ \implies S_2&=\frac{1}{2a}S_1'. \end{align*} $$\therefore S=S_1+\frac{a}{2}S_1'=\frac{\pi^2}{4}\csc^2\pi a-\frac{\pi\cot \pi a}{4a}.$$

Community
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Tianlalu
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  • Thanks @Tianlalu ! For the first one, I had overly pared down the real problem, should read: $sum_{m=1}{\infty} \frac{\sin^2(\frac{m\pi}{\gamma_n)}{(m^2-\gamma_n^2n^2)}$ where $\gamma_n=1+\frac{1}{4}\cos(n)$ with $n$ a positive natural – Michael Cloud Nov 17 '18 at 01:39
  • @MichaelCloud: I don't think there is a nice closed form for that. Even if we take some simple constants such as $\gamma_n=1/2$ and $n=1$, we will get a complicated answer $$\sum_{m=1}^\infty\frac{\sin(\frac{m\pi}{2})}{(m^2-1/4)^2}=G+\frac 18\zeta(2,\frac{5}{4})-2-\pi+\frac{5}{8}\pi^2,$$where $G$ is Catalan's constant and $\zeta(s,a)$ is the generalized zeta function. – Tianlalu Nov 17 '18 at 01:50
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I did find an answer to the second one:

$$\sum_{m=1}^{\infty} \frac{m^2}{(m^2-a^2)^2}= \frac{\pi[2\pi a-\sin(2\pi a)]}{8a\sin^2(\pi a)}$$

According to a Dodonov, Klimov, and Nikonov "Quantum Particle in a Box with Moving Walls" J. Math. Phys. 34 (8) August 1993

Michael Cloud
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