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I'm having a little trouble calculating the double series given by the proposition below:

$$\int_0^1\frac{\left[\text{Li}_2(x)\right]^2}{x}\mathrm dx=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{1}{n^2k^2(n+k)}=2\zeta(2)\zeta(3)-3\zeta(5).\tag{1}$$

In fact, this integral has already been evaluated here and here, but i have no intention of evaluating it by integration or by using harmonic series. My interest is to attack the double sum in $(1)$ only using the manipulation of sums. Having clarified my interest in this topic, i naively proceeded as follows: $$\begin{align*}\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{1}{n^2k^2(n+k)}&=\sum_{n,k\in\mathbb{N}_1}\frac{1}{n^2k^2(n+k)}\\ &=\sum_{\substack{{n,k\in\mathbb{N}_1}\\{n=k}}}\frac{1}{n^2k^2(n+k)}+\sum_{\substack{{n,k\in\mathbb{N}_1}\\{n>k}}}\frac{1}{n^2k^2(n+k)}+\sum_{\substack{{n,k\in\mathbb{N}_1}\\{n<k}}}\frac{1}{n^2k^2(n+k)}\\ &=\frac{\zeta(5)}{2}+\sum_{\substack{{n,k\in\mathbb{N}_1}\\{n>k}}}\frac{1}{n^2k^2(n+k)}+\sum_{\substack{{n,k\in\mathbb{N}_1}\\{n<k}}}\frac{1}{n^2k^2(n+k)}\tag{a}\end{align*}$$

Here's my difficulty with that approach! How can i solve the last two double sums in $(\text a)$ with the indexes $ n <k $ and $ n> k $ ?. Studying other posts (links at the end of the question) related to my question, i found here another possible approach using the fact that: $ \frac{1}{n^2k^2}=\frac{1}{(n+k)^2n^2}+\frac{1}{(n+k)^2k^2}+\frac{2}{(n+k)^3k}+\frac{2}{(n+k)^3n}$, then multiplying by $\frac{1}{n+k}$ on both sides and taking the double sum, we get: $$\sum_{n,k\in\mathbb{N}_1}\frac{1}{n^2k^2(n+k)}=2\sum_{n,k\in\mathbb{N}_1}\frac{1}{(n+k)^3n^2}+4\sum_{n,k\in\mathbb{N}_1}\frac{1}{(n+k)^4n}.\tag{b} $$

According to this article, $(\text b)$ is equal to $T(2,2,1)=2T(2,0,3)+4T(1,0,4)$, where $T(a,b,c)$ is Tornheim’s double series defined by $T(a,b,c)=\sum_{n,m=1}\frac{1}{n^am^b(n+m)^c}$. In this second approach, the expression in $(\text b)$ seems to me a little less aggressive, but if i apply the concepts of the mentioned article, i would feel a little difficult to understand, since it starts with already established formulas. In short, two questions hang in my mind: How to work $(\text a)$ and $(\text b)$ without using integrals or harmonic numbers?

Related links:

A. $\displaystyle \sum_{m,n=1\, m\neq n}^\infty{\frac{m^2+n^2}{mn(m^2-n^2)^2}}$

B. $\displaystyle \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^2k^2(n+k)^2}$

C. $\displaystyle \sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2$

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One way of doing the sum is as follows

\begin{align} \sum\limits_{n,k = 1}^\infty \frac{1}{n^2k^2(n+k)} &= \sum\limits_{n,k = 1}^\infty \frac{k(n+k) - k^2}{n^3k^3(n+k)} \\&= \sum\limits_{n,k = 1}^\infty \frac{1}{n^3k^2} - \sum\limits_{n,k = 1}^\infty \frac{1}{n^3k(n+k)} \\&= \zeta(2)\zeta(3) - \sum\limits_{n=1}^\infty \frac{1}{n^3} \sum\limits_{k=1}^\infty \frac{1}{k(n+k)} \\&= \zeta(2)\zeta(3) - \sum\limits_{n=1}^\infty \frac{H_n}{n^4} \end{align}

where, we used the identity $\displaystyle \sum\limits_{k=1}^\infty \frac{1}{k(n+k)} = \frac{H_n}{n}$ in the second last line. The last sum is the well known Euler sum, which itself can be derived from partial fraction decomposition of $\displaystyle \frac{1}{n^3k(n+k)}$. Hope this helps! :)

As to the last Euler sum you may see Rob's answer here Generalized Euler sum $\sum_{n=1}^\infty \frac{H_n}{n^q}$.

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\begin{align*} \sum_{n=1}^\infty\sum_{k=1}^\infty\frac{1}{n^2k^2(n+k)}&=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{1}{n^2k^2}\int_0^1 x^{n+k-1}\,\text{d}x\\ &=\int_0^1 \left(\sum_{n=1}^\infty\frac{x^{n}}{n^2}\right)\left(\sum_{n=1}^\infty\frac{x^{k-1}}{k^2}\right)\,\text{d}x\\ &=\int_0^1 \left(\text{Li}_2(x)\right)\left(\sum_{n=1}^\infty\frac{x^{k-1}}{k^2}\right)\,\text{d}x\\ &=\sum_{k=1}^\infty\frac1{k^2}\int_0^1 x^{k-1}\text{Li}_2(x)\,\text{d}x\\ &\overset{\text{IBP}}{=}\sum_{k=1}^\infty\frac1{k^2}\left(\frac{\zeta(2)}{k}-\frac{H_k}{k^2}\right)\\ &=\zeta(2)\zeta(3)-\sum_{k=1}^\infty\frac{H_k}{k^4}\\ &=\zeta(2)\zeta(3)-[3\zeta(5)-\zeta(2)\zeta(3)]\\ &=2\zeta(2)\zeta(3)-3\zeta(5).\\ \end{align*}

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