4

The problem I was considering about is the evaluation of the following series:

\begin{align*} \sum_{n=1}^{\infty}\frac{(H_{n})^2}{n(n+1)} \end{align*}

The attempt I could make was to change $(H_{n})^2$ to $H_{n}(H_{n+1}-\frac{1}{n+1})$ to see if the denominator and the numerator could match each other. Then the given series becomes

\begin{align*} \sum_{n=1}^{\infty}\left(\frac{H_{n}H_{n+1}}{n(n+1)}-\frac{H_n}{n(n+1)^2}\right) \end{align*}

The second term in the above series calculates to $\zeta{(3)}-\zeta{(2)}$. I have no idea how to evaluate the first term. Wolfram alpha gives 3$\zeta(3)$ for the value of the desired series. So the first term must be somehow evaluated to $2\zeta(3)+\zeta(2)$.

Other ways to calculate the desired series are also appreciated, particularly if integrals were used.

Did
  • 279,727
민찬홍
  • 154

5 Answers5

3

In terms of Stirling numbers of the first kind we have $$ \log^2(1-x)=\sum_{n\geq 1}\frac{2H_{n-1}}{n}\,x^n,\qquad -\log^3(1-x) = \sum_{n\geq 1}\frac{3\left(H_{n-1}^2-H_{n-1}^{(2)}\right)}{n}\,x^n$$ hence $$ \sum_{n\geq 1}\frac{3\left(H_{n-1}^2-H_{n-1}^{(2)}\right)}{n(n+1)}=-\int_{0}^{1}\log(1-x)^3\,dx = 6\tag{A}$$

$$\begin{eqnarray*} \sum_{n\geq 1}\frac{H_n^2}{n(n+1)}&=&\sum_{n\geq 1}\frac{H_{n-1}^2+\frac{2H_{n-1}}{n}+\frac{1}{n^2}}{n(n+1)}\\&=&\int_{0}^{1}\text{Li}_3(x)\,dx+\int_{0}^{1}\frac{(1-x)\log^2(1-x)}{x}\,dx+2+\sum_{n\geq 1}\frac{H_{n-1}^{(2)}}{n(n+1)}\\&\stackrel{\text{SBP}}{=}&\left(1-\zeta(2)+\zeta(3)\right)+2(\zeta(3)-1)+2+\sum_{n\geq 1}\left(1-\frac{1}{n+1}\right)\frac{1}{n^2}\\&=&1-\zeta(2)+3\,\zeta(3)+\zeta(2)-1\stackrel{\color{green}{\checkmark}}{=}\color{blue}{3\,\zeta(3)}.\tag{B} \end{eqnarray*}$$

$\text{SBP}$ stands for summation by parts.
This might not be the most efficient approach, but its logic is pretty simple to explain:

  1. $H_n^2-H_n^{(2)}$ is a nicer weight than $H_n^2$, due to the Taylor series of $\log^3(1-x)$;
  2. Both the integrals $\int_{0}^{1}\frac{x^m \log^{h}(x)}{1-x}\,dx$ and $\int_{0}^{1}\text{Li}_k(x)\,dx$ are elementary;
  3. Every series of the $\sum_{n\geq 1}\frac{H_n^{(k)}}{n(n+1)}$ kind can be simply computed by $\text{SBP}$.

By $\text{SBP}$ directly, $$\begin{eqnarray*} \sum_{n=1}^{N}\frac{H_n^2}{n(n+1)}&=&\left(1-\frac{1}{N+1}\right)H_N^2-\sum_{n=1}^{N-1}\left(1-\frac{1}{n+1}\right)\frac{H_n+H_{n+1}}{n+1}\\&=&H_N^2-\frac{H_N^2}{N+1}-\sum_{n=1}^{N-1}\frac{2H_n}{n+1}+\sum_{n=1}^{N-1}\frac{2H_n}{(n+1)^2}-H_N^{(2)}+H_N^{(3)}\\ &=&-\frac{H_N^2}{N+1}+\sum_{n=1}^{N-1}\frac{2H_n}{(n+1)^2}+H_N^{(3)}\end{eqnarray*}$$ hence $$\begin{eqnarray*} \sum_{n\geq 1}\frac{H_n^2}{n(n+1)}=\zeta(3)+2\sum_{n\geq 1}\frac{H_n}{(n+1)^2}&=&\zeta(3)+2\int_{0}^{1}\frac{\log(1-x)\log(x)}{1-x}\,dx \\&\stackrel{\text{IBP}}{=}&\zeta(3)+\int_{0}^{1}\frac{\log^2(1-x)}{x}\,dx\\&=&\zeta(3)+\int_{0}^{1}\frac{\log^2(x)\,dx}{1-x}\\&=&\zeta(3)+\sum_{n\geq 0}\int_{0}^{1}x^n\log^2(x)\,dx\\&=&\zeta(3)+2\,\zeta(3).\end{eqnarray*}$$

Jack D'Aurizio
  • 353,855
1

using the identity $$ \sum_{n=1}^{\infty}x^n \left(H_n^2-H_n^{(2)}\right)=\frac{\ln^2(1-x)}{1-x}$$

divide both sides by $x$ then integrate w.r.t $x$ from $x=0$ to $y$ , we get

$$ \sum_{n=1}^{\infty}\frac{y^n}{n}\left(H_n^2-H_n^{(2)}\right)=\int_0^y\frac{\ln^2(1-x)}{1-x}\ dx-\frac13 {\ln^3(1-y)}$$

integrate both sides w.r.t $y$ from $y=0$ to $1$ , we get

\begin{align*} \sum_{n=1}^{\infty}\frac{H_n^2-H_n^{(2)}}{n(n+1)}&=\int_0^1\int_0^y\frac{\ln^2(1-x)}{x}\ dxdy-\frac13\int_0^1{\ln^3(1-y)}\ dy\\ &=\int_0^1\frac{\ln^2(1-x)}{x}\left(\int_x^1dy\right)dx-\frac13\int_0^1{\ln^3(y)}\ dy\\ &=\int_0^1\frac{\ln^2(1-x)}{x}\left(1-x\right)dx-\frac13(-6)\\ &=\int_0^1\frac{\ln^2(1-x)}{x}dx-\int_0^1\ln^2(1-x)\ dx+2\\ &=\int_0^1\frac{\ln^2x}{1-x}dx-(2)+2\\ &=2\zeta(3)\tag{1} \end{align*}

\begin{align*} \sum_{n=1}^{N}\frac{H_n^{(2)}}{n(n+1)}&=\sum_{n=1}^{N}\frac{H_n^{(2)}}{n}-\sum_{n=0}^{N}\frac{H_n^{(2)}}{n+1}\\ &=\sum_{n=1}^{N}\frac{H_n^{(2)}}{n}-\sum_{n=1}^{N+1}\frac{H_{n-1}^{(2)}}{n}\\ &=\sum_{n=1}^{N}\frac{H_n^{(2)}}{n}-\sum_{n=1}^{N+1}\frac{H_{n}^{(2)}}{n}+\sum_{n=1}^{N+1}\frac1{n^3}\\ &=\sum_{n=1}^{N}\frac{H_n^{(2)}}{n}-\sum_{n=1}^{N}\frac{H_{n}^{(2)}}{n}-\frac{H_{N+1}^{(2)}}{N+1}+\sum_{n=1}^{N+1}\frac1{n^3}\\ &=-\frac{H_{N+1}^{(2)}}{N+1}+\sum_{n=1}^{N+1}\frac1{n^3} \end{align*} letting $N$ approach $\infty$ yields $$\sum_{n=1}^{\infty}\frac{H_n^{(2)}}{n(n+1)}=0+\sum_{n=1}^\infty\frac1{n^3}=\zeta(3)\tag{2}$$

plugging $(2)$ in $(1)$ we have

$$\sum_{n=1}^{\infty}\frac{H_n^2}{n(n+1)}=3\zeta(3)$$

Ali Shadhar
  • 25,498
1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ With $\ds{N \in \mathbb{N}_{\ \geq\ 1}\,}$: \begin{align} &\bbox[10px,#ffd]{\sum_{n = 1}^{N} {H_{n}^{2} \over n\pars{n + 1}}} = \sum_{n = 1}^{N}{H_{n}^{2} \over n} - \sum_{n = 1}^{N}{H_{n}^{2} \over n + 1} \\[5mm] = &\ \sum_{n = 1}^{N}{H_{n}^{2} \over n} - \sum_{n = 2}^{N + 1}{H_{n - 1}^{2} \over n} \\[5mm] = &\ \sum_{n = 1}^{N}{H_{n}^{2} \over n} - \sum_{n = 1}^{N}{H_{n}^{2} - 2H_{n}/n + 1/n^{2} \over n} - {H_{N}^{2} \over N + 1} \\[5mm] \stackrel{\mrm{as}\ N\ \to\ \infty}{\Huge \to}\,\,\, &\ 2\sum_{n = 1}^{\infty}H_{n}\ \overbrace{\bracks{-\int_{0}^{\infty}\ln\pars{x}x^{n - 1}\,\dd x}} ^{\qquad\ds{1 \over n^{2}}}\ -\ \overbrace{\sum_{n = 1}^{\infty}{1 \over n^{3}}}^{\ds{\zeta\pars{3}}} \\[5mm] = &\ -2\int_{0}^{1}\ln\pars{x}\ \overbrace{\sum_{n = 1}^{\infty}H_{n}\ x^{n}}^{\ds{-\,{\ln\pars{1 - x} \over 1 - x}}}\,{\dd x \over x} - \zeta\pars{3} \\[5mm] = &\ 2\int_{0}^{1}{\ln\pars{x}\ln\pars{1 - x} \over 1 - x}\,\dd x \\[2mm] + &\ 2\int_{0}^{1}{\ln\pars{x}\ln\pars{1 - x} \over x}\,\dd x - \zeta\pars{3} \\[5mm] = &\ 4\int_{0}^{1}{\ln\pars{x}\ln\pars{1 - x} \over x}\,\dd x - \zeta\pars{3} \\[5mm] = &\ -4\int_{0}^{1}\mrm{Li}_{2}'\pars{x}\ln\pars{x}\,\dd x - \zeta\pars{3} \\[5mm] = &\ 4\int_{0}^{1}\ \overbrace{\mrm{Li}_{2}\pars{x} \over x} ^{\ds{\mrm{Li}_{3}'\pars{x}}}\ \,\dd x - \zeta\pars{3} = \bbx{\large 3\zeta\pars{3}} \\ & \end{align}

Felix Marin
  • 89,464
1

Using the generating function

$$\sum_{n=1}^\infty\frac{H_{n}^2}{n}x^{n}=\operatorname{Li}_3(x)-\ln(1-x)\operatorname{Li}_2(x)-\frac13\ln^3(1-x)$$

Integrate both sides from $x=0$ to $1$ we get

$$\sum_{n=1}^\infty\frac{H_{n}^2}{n(n+1)}=\int_0^1\operatorname{Li}_3(x)dx-\int_0^1\ln(1-x)\operatorname{Li}_2(x)dx-\frac13\int_0^1\ln^3(1-x)dx$$

$$=(\zeta(3)-\zeta(2)+1)-(-2\zeta(3)-\zeta(2)+3)-\frac13(-6)=\boxed{3\zeta(3)}$$

Ali Shadhar
  • 25,498
1

We have

$$\sum_{n=1}^\infty\frac{H_n}{n+1}x^{n+1}=\frac12\ln^2(1-x)$$

multiply both sides by $-\frac{\ln(1-x)}{x^2}$ then $\int_0^1$ and use that $-\int_0^1 x^{n-1}\ln(1-x)dx=\frac{H_n}{n}$ we get

$$\sum_{n=1}^\infty\frac{H_n^2}{n(n+1)}=-\frac12\int_0^1\frac{\ln^3(1-x)}{x^2}dx=-\frac12\int_0^1\frac{\ln^3(x)}{(1-x)^2}dx$$

$$=-\frac12\sum_{n=1}^\infty n\int_0^1 x^{n-1}\ln^3(x)dx=3\sum_{n=1}^\infty\frac{1}{n^3}=3\zeta(3)$$

Ali Shadhar
  • 25,498