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In the evaluation of $$\sum_{k=1}^\infty \sum_{\ell=1}^{k-1}\sum_{m=1}^{\ell-1}\frac{\delta_{k, 2\ell-2m}}{m\left(\ell-m\right)\left(k-\ell\right)}.$$ Here $\delta_{k, 2\ell-2m}$ denotes the "Kronecker delta" (see here) and is given by $$\delta_{k, 2\ell-2m} = \begin{cases} 0 \qquad \mathrm{if} k \neq 2\ell-2m, \\ 1 \qquad \mathrm{if} k =2\ell-2m. \end{cases}$$ Thus we have the equality $$S = \sum_{\ell=1}^\infty\sum_{m=1}^{\ell-1} \frac{1}{m(\ell-m)(\ell-2m)}.$$ I was able to simplify this down to something along the lines of $$\sum_{n=1}^\infty \frac{H_{\left(n-3/2\right)}}{n^2}$$which I have no idea how to approach. Sorry for the little information given, I am just looking for an idea of how I can go about this. Thank you!

Varun Vejalla
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  • What does $\delta$ mean here? I haven't seen it yet. – Varun Vejalla May 31 '20 at 15:48
  • The Kronecker delta -- see here https://en.wikipedia.org/wiki/Kronecker_delta . In others words, we can easily collapse our triple sum to only the case where $k=2\ell-2m$ because everywhere else is zero. The function itself came from and integral with cosine of two different arguments, so we consider the case when the arguments are equal and the case where they are not equal. When they are not equal ($k \neq 2\ell-2m$), the summand goes to zero, but when they are equal ($k =2\ell-2m$), the sum does not go to zero; thus we only consider the latter case. I hope this helped! –  May 31 '20 at 16:39
  • In fact, I will put the equality in the post for more clarification; thanks for the concern! –  May 31 '20 at 16:40
  • Thanks for the explanation! I don't think you correctly obtained the second sum from the first. In the second sum, the summand will have a denominator of $0$ if $l = 2m$. In the first sum, it is possible for the denominator to be $0$. – Varun Vejalla May 31 '20 at 16:52
  • The displayed formula with $S$ on the LHS has a zero in the denominator on the RHS for each even value of $\ell$, when $m=\ell/2$. (Ah, what VVejalla just said!) – Barry Cipra May 31 '20 at 16:52
  • I find $\sum_{n=1}^\infty \frac{H_{\left(n-3/2\right)}}{n^2}= \frac{\pi^2}{3}-8 \log(2) -\frac{\pi^2}{3} \log(2) +\frac{7}{2} \zeta(3)\simeq -0.328473$ – Dr. Wolfgang Hintze May 31 '20 at 21:19

2 Answers2

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You did not reduce the sum properly. For a given even $k$, the set of $m$ that satisfies $k = 2\ell - 2m, 1 \le \ell \le k-1, 1 \le m \le \ell-1$ is the positive natural numbers less than or equal to $k-1$

This means that the sum should be $$\sum_{t = 1}^{\infty} \sum_{m=1}^{t-1} \frac{1}{m(\ell-m)(k-\ell)}$$

where $k = 2t$ and $\ell = \frac{k+2m}{2} = t+m$

Plugging those in and simplifying, I get $$\sum_{t = 1}^{\infty} \sum_{m=1}^{t-1} \frac{1}{mt(t-m)}$$

Changing the order of summation yields $$\sum_{m=1}^{\infty} \sum_{t=m+1}^{\infty} \frac{1}{mt(t-m)}$$

The inner sum can be simplified so that the double sum becomes $$\sum_{m=1}^{\infty}\frac{H_m}{m^2}$$

Using this answer to another question, the sum can be simplified to $$2\zeta(3) \approx 2.404$$

Varun Vejalla
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It is not a complete answer. Let \begin{eqnarray} f(x)&=&\sum_{m=1}^\infty \frac{H_{m-\frac32}}{m^2}x^m \end{eqnarray} and then \begin{eqnarray} f'(x)&=&\sum_{m=1}^\infty \frac{H_{m-\frac32}}{m}x^{m-1}\\ (xf'(x))'&=&\sum_{m=1}^\infty H_{m-\frac32}x^{m-1}=\sum_{m=1}^\infty x^{m-1}\int_0^1\frac{1-t^{m-\frac32}}{1-t}dt\\ &=&\int_0^1\sum_{m=1}^\infty \frac{1-t^{m-\frac32}}{1-t}x^{m-1}dt\\ &=&\int_0^1\bigg(\frac{1}{(1-t) (1-x)}-\frac{1}{(1-t)\sqrt t (1-t x)}\bigg)dt. \end{eqnarray} So \begin{eqnarray} xf'(x)&=&\int_0^x\int_0^1\bigg(\frac{1}{(1-t) (1-r)}-\frac{1}{(1-t) \sqrt t(1-t r)}\bigg)dtdr\\ &=&\int_0^1\int_0^x\bigg(\frac{1}{(1-t) (1-r)}-\frac{1}{(1-t)\sqrt t (1-t r)}\bigg)drdt\\ &=&\int_0^1\frac{-\log (1-x)+t^{-3/2} \log (1-t x)}{1-t}dt\\ f(1)&=&\int_0^1\frac1x\int_0^1\frac{-\log (1-x)+t^{-3/2} \log (1-t x)}{1-t}dtdx\\ &=&\int_0^1\int_0^1\frac1x\frac{-\log (1-x)+t^{-3/2} \log (1-t x)}{1-t}dxdt\\ &=&\int_0^1\frac{\pi ^2-6 t^{-3/2} \text{Li}_2(t)}{6-6 t}dt. \end{eqnarray} Now I have problem for the last step. But Wolfram Mathematica gives $$ \int_0^1\frac{\pi ^2-6 t^{-3/2} \text{Li}_2(t)}{6-6 t}dt=\frac{7 \zeta (3)}{2}-\frac{1}{3} \pi ^2 (\log (2)-1)-8 \log (2). $$

xpaul
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