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\begin{align} \sum_{m=1}^{\infty} \left(\zeta(2m) - \zeta(2m+1)\right) &= \sum_{m=1}^{\infty} \sum_{n=2}^{\infty} \left( \frac{1}{n^{2m}} - \frac{1}{n^{2m+1}} \right) \\ &= \sum_{n=2}^{\infty} \sum_{m=1}^{\infty} \left( \frac{1}{n^{2m}} - \frac{1}{n^{2m+1}} \right) \tag{*}\\ &= \sum_{n=2}^{\infty} \left( \sum_{m=1}^{\infty} \frac{1}{n^{2m}} - \sum_{m=1}^{\infty} \frac{1}{n^{2m+1}} \right) \\ &= \sum_{n=2}^{\infty} \left( \frac{1}{n^2 - 1} - \frac{1}{n(n^2 - 1)} \right) \\ &= \sum_{n=2}^{\infty} \frac{n - 1}{n(n^2 - 1)}\\ &= \sum_{n=2}^{\infty} \frac{n - 1}{n(n-1)(n+1)} \\ &= \sum_{n=2}^{\infty}\frac{1}{n(n+1)} \\ &= \sum_{n=2}^{\infty}\left(\frac{1}{n} - \frac{1}{n+1}\right) \\ &= \left( \frac{1}{2} - \frac{1}{3} \right ) + \left( \frac{1}{3} - \frac{1}{4} \right ) + \left( \frac{1}{4} - \frac{1}{5} \right ) + \cdots \\ &= \frac{1}{2} \end{align}

Is interchanging the order of summation at $(*)$ valid?

I couldn't find any proofs of this identity online, although WolframAlpha verifies the result.

Reference

Closed form for $\sum_{k=1}^{\infty} \zeta(2k)-\zeta(2k+1)$

vengy
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  • Sure looks right. According to Wolfram Alpha, the sum of the first 20 terms is about 1/2-1e-12 – marty cohen Dec 19 '24 at 19:39
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    You can also do it this way, although you would now have to justify the interchange of the sum and integral (absolute convergence/uniform convergence/Fubini's theorem): $$\sum_{m=1}^{\infty} \left( \frac{1}{n^{2m}} - \frac{1}{n^{2m+1}} \right)=-\log (n)\sum_{m=1}^\infty\int_0^1\frac{1}{n^{2m+x}}dx=-\log(n)\int_0^1\frac{n^{-x}}{n^2-1}dx.$$ Integrating, $$-\frac{1}{n^2-1}\left[n^{-x}\right]_0^1=-\frac{1}{n^2-1}\left(\frac{1}{n}-1\right)=\frac{n-1}{n(n-1)(n+1)}=\frac{1}{n(n+1)}.$$ – pshmath0 Dec 19 '24 at 20:30
  • Interestingly, $\sum_{m=1}^{\infty} \left(\zeta(2m) - \zeta(2m+2)\right) = \zeta(2) - 1$. – vengy Dec 19 '24 at 21:21
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    @vengy that's a telescoping sum, e.g. $$\zeta(2)+(-\zeta(4)+\zeta(4))+(-\zeta(6)+\zeta(6))+\cdots=\zeta(2)+0+0+\cdots$$ – pshmath0 Dec 19 '24 at 21:52
  • @pshmath0 I've submitted this question, $$\sum_{m=1}^\infty \left( \zeta(2m) - \zeta(2m + 2k) \right) = \sum_{m=1}^k \zeta(2m) - k$$ so for $k=1$, the result is $\zeta(2)-1$. – vengy Dec 20 '24 at 02:49
  • @vengy Yes, the $-1$ comes from the last term which is $-\zeta(2m+2)$ which escapes the telescoping, and since $\displaystyle\lim_{n\to\infty}\zeta(n)=1$ you get the $-1$ on the end ! – pshmath0 Dec 20 '24 at 10:40

2 Answers2

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You should change the $n=1$ endpoint to $n=2$ before doing (*), but other than that, yes, it's valid. One justification for interchanging the summations is that the double series (starting at $n=2$) is absolutely convergent; the proof of that absolute convergence is pretty much the same calculation but with the negative terms turned positive.

Greg Martin
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This is an application of Tonelli's theorem. Since $f(m,n)=\frac{1}{n^{2m}}-\frac{1}{n^{2m+1}}>0$ for all $m\geq1$ and $n\geq2$ with $m,n\in\mathbb{N}$, we have a positive measurable function over the entire product measure space (i.e., $\sum_{m=1}^{\infty}\sum_{n=2}^{\infty}f(m,n)$ is positive with no singularities or discontinuities).

To see why getting rid of the $n=1$ term is necessary before swapping the order of summation, jump to the telescoping sum for a quick proof by contradiction: the telescoping would give an answer of $1$ after swapping the order if it is left in, but the original series levels off quickly after only $10$ terms to $0.4999998\dots.$ If the $n=1$ term were needed, then these two values would be equal. Since they aren't, we say that $\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}f(m,n)$ is not measurable at $n=1$ due to the discrepancy, and therefore Tonelli's theorem does not apply without the term's removal. One can do so by observing that $\sum_{m=1}^{\infty}0=0$.

However, when $m\geq1$ and $n\geq2$, we no longer have any discontinuities. By Tonelli's theorem, these conditions are sufficient to freely swap the order of summation with no impact on the final evaluation. Therefore, your step (*) is valid.

teadawg1337
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