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I suppose my question could be restated as "why?" to the above statement. I'll rephrase my question if this breaks any rules asking it this way, but the expression is simply:

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

with the partial sum being: $$\sum_{m=2}^a \frac{1}{m^{2n}(m^2-1)} = ??$$

$m, n \in \mathbb N$

Are the above two somehow more insidious than they look?

user3108815
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    Since we don’t even know whether $\sum_{n=1}^{\infty}{n^{-5}}$ is irrational, it is not as surprising as you could think. – Aphelli Jul 01 '19 at 16:20
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    Given a nonnegative integer $n$ we should have an explicit result. If we do not give $n$, I guess that Mathematica considers $n$ as a complex argument. – Robert Z Jul 01 '19 at 16:29
  • @Mindlack The proposed series can be evaluated for any non-negative integer $n$. – Robert Z Jul 01 '19 at 16:52
  • Mathematica is too smart for its own good, WA even more so ;) – Yuriy S Jul 02 '19 at 05:29

2 Answers2

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Yes, your series is innocuous if $n$ is a non-negative integer: by partial fraction decomposition $$\frac{1}{m^{2n}(m^2-1)}=-\sum_{k=1}^n\frac{1}{m^{2k}}+\frac{1/2}{m-1}-\frac{1/2}{m+1}.$$ Hence $$\sum_{m=2}^{\infty} \frac{1}{m^{2n}(m^2-1)} = -\sum_{k=1}^n\zeta(2k)+n+\frac{3}{4}$$ where $\zeta(2k)=\frac{|B_{2k}|(2\pi)^{2k}}{2(2k)!}$ (see wiki).

However, I guess that wolframalpha considers $n$ as a complex argument (not as a non negative integer). On the other hand, given an explicit non negative integer $n$, wolframalpha will give you the result.

Robert Z
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$$ \begin{align} \sum_{m=2}^\infty\frac1{m^{2n}(m^2-1)} &=\sum_{m=2}^\infty\sum_{k=n+1}^\infty m^{2k}\\ &=\sum_{k=n+1}^\infty\sum_{m=2}^\infty m^{2k}\\ &=\sum_{k=n+1}^\infty(\zeta(2k)-1)\\ &=\bbox[5px,border:2px solid #C0A000]{\frac34-\sum_{k=1}^n(\zeta(2k)-1)}\tag1 \end{align} $$ since $$ \begin{align} \sum_{k=1}^\infty(\zeta(2k)-1) &=\sum_{k=1}^\infty\sum_{n=2}^\infty\frac1{n^{2k}}\\ &=\sum_{n=2}^\infty\sum_{k=1}^\infty\frac1{n^{2k}}\\ &=\sum_{n=2}^\infty\frac1{n^2-1}\\ &=\frac12\sum_{n=2}^\infty\left(\frac1{n-1}-\frac1{n+1}\right)\\[3pt] &=\frac34\tag2 \end{align} $$ In $(1)$, $\zeta(2k)$ is a rational multiple of $\pi^{2k}$, as shown in this answer.

Here are the first several sums: $$ \begin{array}{c|l} n&\frac34-\sum\limits_{k=1}^n(\zeta(2k)-1)\\\hline 1&\frac74-\frac{\pi^2}6\\ 2&\frac{11}4-\frac{\pi^2}6-\frac{\pi^4}{90}\\ 3&\frac{15}4-\frac{\pi^2}6-\frac{\pi^4}{90}-\frac{\pi^6}{945}\\ 4&\frac{19}4-\frac{\pi^2}6-\frac{\pi^4}{90}-\frac{\pi^6}{945}-\frac{\pi^8}{9450}\\ 5&\frac{23}4-\frac{\pi^2}6-\frac{\pi^4}{90}-\frac{\pi^6}{945}-\frac{\pi^8}{9450}-\frac{\pi^{10}}{93555}\\ 6&\frac{27}4-\frac{\pi^2}6-\frac{\pi^4}{90}-\frac{\pi^6}{945}-\frac{\pi^8}{9450}-\frac{\pi^{10}}{93555}-\frac{691\pi^{12}}{638512875} \end{array}\tag3 $$

robjohn
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  • I don't see where the closed form is expressed, it seems like replacing a series with an even harder to calculate series to me! – user3108815 Jul 01 '19 at 22:44
  • It is replacing an infinite series with a finite sum (note that the sum in $(1)$ is finite). $\zeta(2k)$ has a closed form as a rational multiple of $\pi^{2k}$ as shown in the cited answer. Both Robert Z and I got the same finite sum. I doubt there is a simpler form, though there could be. – robjohn Jul 01 '19 at 22:54