6

Evaluate:

$$S_n=1+\left(\frac{1+\frac12}{2}\right)^2+\left(\frac{1+\frac12+\frac13}{3}\right)^2+\left(\frac{1+\frac12+\frac13+\frac14}{4}\right)^2+...$$

a_n are the individual terms to be summed.

My Try : \begin{align} &a_1=1\\ &a_2=\left(\frac{3}{4}\right)^2=\frac{9}{16}\\ &a_3=\left(\frac{11}{18}\right)^2\\ &a_4=\left(\frac{25}{48}\right)^2 \end{align} now :?

unseen_rider
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Almot1960
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2 Answers2

5

By setting $H_n = \sum_{k=1}^{n}\frac{1}{k}$ we have to compute $\sum_{n\geq 1}\left(\frac{H_n}{n}\right)^2$. We may notice that

$$ \sum_{n=1}^{N}\frac{H_n}{n}=\sum_{1\leq m\leq n\leq N}\frac{1}{mn}=\frac{H_N^2+H_N^{(2)}}{2}\tag{1}$$ and for the same reason $$ \sum_{n=1}^{N}\frac{H_n^{(2)}}{n^2} = \frac{1}{2}\left[\left(\sum_{n=1}^{N}\frac{1}{n^2}\right)^2+\sum_{n=1}^{N}\frac{1}{n^4}\right]\stackrel{N\to +\infty}{\longrightarrow}\frac{\zeta(2)^2+\zeta(4)}{2}=\frac{7\pi^4}{360} \tag{2}$$ Since $-\log(1-x)=\sum_{n\geq 1}\frac{1}{n}\,x^n$, by multiplying both sides by $\frac{1}{1-x}$ and applying termwise integration we have $$ \sum_{n\geq 1}\frac{H_{n}}{n}\,x^{n} = \text{Li}_2(x)+\frac{1}{2}\log^2(1-x) \tag{3}$$ hence by $(1)$ it follows that: $$ \sum_{N\geq 1}\frac{H_N^2+H_N^{(2)}}{2}x^{N} = \frac{\text{Li}_2(x)}{1-x}+\frac{1}{2}\cdot\frac{\log^2(1-x)}{1-x}\tag{4} $$ and by multiplying both sides of $(4)$ by $-\frac{2\log x}{x}$ and performing termwise integration over $(0,1)$:

$$ \sum_{N\geq 1}\frac{H_N^2+H_N^{(2)}}{N^2} = -\int_{0}^{1}\left[\frac{2\text{Li}_2(x)\log(x)}{x(1-x)}+\frac{\log^2(1-x)\log(x)}{x(1-x)}\right]\,dx.\tag{5} $$ The integral $-\int_{0}^{1}\frac{\log^2(1-x)\log(x)}{x(1-x)}\,dx$ can be computed by differentiating Euler's beta function, and it equals $\frac{\pi^4}{36}$. Since $\int\frac{\log(x)}{x(1-x)}\,dx=\frac{1}{2}\log^2(x)+\text{Li}_2(1-x)$ and $\frac{d}{dx}\text{Li}_2(x)=-\frac{\log(1-x)}{x}$, by integration by parts the whole problem boils down to computing: $$ I = \int_{0}^{1}\frac{\text{Li}_2(x)\log(x)}{1-x}\,dx \tag{6}$$ but we have already done that in $(2)$, since $\frac{\text{Li}_2(x)}{1-x}=\sum_{n\geq 1}H_n^{(2)}x^n.$ Collecting pieces,

$$ \sum_{n\geq 1}\left(\frac{H_n}{n}\right)^2 = \color{red}{\frac{17\pi^4}{360}}.$$

Jack D'Aurizio
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  • Hi Jack, I'm having hard time on the last sentence you said "collecting pieces". $\sum_{n\ge1}\left(\frac{H_n}n\right)^2$ never appeared since the 1st line and it just appeared on the last line. If you can elaborate a little bit, it would be appreciated. – Kay K. Mar 02 '17 at 15:51
  • @KayK.: I disagree. $\sum_{n\geq 1}\frac{H_n^2}{n^2}$ appears in the LHS of $(5)$, together with $\sum_{n\geq 1}\frac{H_n^{(2)}}{n^2}$, previously computed. It follows that by giving a closed form to the RHS of $(5)$ we also find the value of the wanted series. – Jack D'Aurizio Mar 02 '17 at 15:58
  • Thanks. Sorry I somehow couldn't find it... – Kay K. Mar 03 '17 at 00:56
4

Recall that the multiple zeta values are defined by the series $$ \zeta(s_1,\ldots,s_k):=\sum_{n_1>\ldots>n_k\geq 1}\frac{1}{n_1^{s_1}\ldots n_k^{s_k}}. $$ The sum $S$ can be expressed as a linear combination of multiple zeta values. We have $$ \begin{align*} S&=\sum_{n=1}^\infty \sum_{k_1,k_2=1}^n\frac{1}{n^2k_1k_2}\\ &=\left(2\sum_{n>k_1>k_2}+\sum_{n>k_1=k_2}+2\sum_{n=k_1>k_2}+\sum_{n=k_1=k_2}\right)\frac{1}{n^2k_1k_2}\\ &=2\zeta(2,1,1)+\zeta(2,2)+2\zeta(3,1)+\zeta(4). \end{align*} $$ Each of these multiple zeta values is a rational multiple of $\pi^4$. The expressions have been tabulated for instance on the MZV data mine: $$ \begin{align*} \zeta(2,1,1)&=\frac{\pi^4}{90},\\ \zeta(2,2)&=\frac{\pi^4}{120},\\ \zeta(3,1)&=\frac{\pi^4}{360},\\ \zeta(4)&=\frac{\pi^4}{90}. \end{align*} $$ So we get $$ S=\frac{17\pi^4}{360} $$

Julian Rosen
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