# How to show that $\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\frac{\pi^2}{6}$

Wolfram Alpha shows that $$\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\zeta(2)=\frac{\pi^2}{6}$$ I want to prove this.

Attempt:

I tried to treat this as a telescoping series: \begin{align} \sum_{n=1}^{\infty}\frac{H_n}{n^2+n}&=\sum_{n=1}^{\infty}H_n\left(\frac{1}{n}-\frac{1}{n+1}\right)\\ &=H_{1}\left(1-\frac{1}{2}\right)+H_{2}\left(\frac{1}{2}-\frac{1}{3}\right)+H_{3}\left(\frac{1}{3}-\frac{1}{4}\right)\\ &=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{6}+\frac{1}{3}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{9}-\frac{1}{12} \end{align} I think this method is not quite useful, so I tried another one: $$H_n=\int_{0}^{1}\frac{1-t^n}{1-t}dt$$ Then, $$\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\sum_{n=1}^{\infty}\frac{1}{n^2+n}\int_{0}^{1}\frac{1-t^n}{1-t}dt$$ At this point, I do not know how to proceed.

Beside telescoping, all you need is to reverse the order of the double sum: $$\sum_{n\ge 1}\frac{\sum_{k=1}^n k^{-1}}{n(n+1)}=\sum_{k\ge 1}k^{-1}\sum_{n\ge k}\frac{1}{n(n+1)}=\sum_{k\ge 1}k^{-2}=\frac{\pi^2}{6}.$$Edit to explain the reversal: we're summing all terms of the form $$\frac{k^{-1}}{n(n+1)}$$ with $$n,\,k$$ integers in the set $$\{(n,\,k)|n,\,k\ge 1,\,k\le n\}$$. But I could equivalently describe this set as $$\{(n,\,k)|k\ge 1,\,n\ge k\}$$.

• How do you reverse the order? I do not completely understand, especially how you manipulate the lower limit. – Larry Jan 6 '19 at 15:20
• @Larry See edit. – J.G. Jan 6 '19 at 15:23
• $\frac{H_n}{n^2(1+\frac{1}{n})}=\frac{1}{n^2}lim_{n→∞}\frac{H_n}{1+H_n}=\frac{1}{n^2}$ – sirous Jan 6 '19 at 15:24
• @J.G Thanks, I think I understand the reversal now. – Larry Jan 6 '19 at 15:25
• @sirous Your equality is not true: for example $$H_1/(1^2+1)\neq 1/1^2$$ – Nick Jan 6 '19 at 15:26

We have \begin{align} \sum_{n=1}^\infty\frac{H_n}{n^2+n} &=\sum_{n=1}^\infty H_n\left(\frac1n-\frac1{n+1}\right) \\&=\sum_{n=1}^\infty\sum_{m=1}^n\frac1m\left(\frac1n-\frac1{n+1}\right) \\&=\sum_{m=1}^\infty\frac1m\sum_{n=m}^\infty\left(\frac1n-\frac1{n+1}\right) \\&=\sum_{m=1}^\infty\frac1m\cdot\frac1m \\&=\frac{\pi^2}6. \end{align}

applying summation by parts \begin{align} \sum_{n=1}^N\frac{H_n}{n(n+1)}&=\sum_{n=1}^N\frac{H_n}{n}-\sum_{n=1}^N\frac{H_n}{n+1}\\ &=\sum_{n=1}^N\frac{H_n}{n}-\sum_{n=1}^{N+1}\frac{H_{n-1}}{n}\\ &=\sum_{n=1}^N\frac{H_n}{n}-\sum_{n=1}^{N+1}\frac{H_{n}}{n}+\sum_{n=1}^{N+1}\frac{1}{n^2}\\ &=\sum_{n=1}^N\frac{H_n}{n}-\sum_{n=1}^{N}\frac{H_{n}}{n}-\frac{H_{N+1}}{N+1}+\sum_{n=1}^{N+1}\frac{1}{n^2}\\ &=-\frac{H_{N+1}}{N+1}+\sum_{n=1}^{N+1}\frac{1}{n^2} \end{align} letting $$N$$ approach $$\infty$$, we get, $$\sum_{n=1}^\infty\frac{H_n}{n(n+1)}=0+\sum_{n=1}^{\infty}\frac{1}{n^2}=\zeta(2)$$

Using summation by parts we have $$\sum_{n\leq N}\frac{H_{n}}{n\left(n+1\right)}=H_{N}\sum_{n\leq N}\frac{1}{n\left(n+1\right)}-\sum_{n\leq N-1}\frac{1}{n+1}\left(\sum_{k\leq n}\frac{1}{k\left(k+1\right)}\right).$$ Clearly $$\sum_{n\leq N}\frac{1}{n\left(n+1\right)}=\left(1-\frac{1}{N+1}\right)$$ then $$\sum_{n\leq N}\frac{H_{n}}{n\left(n+1\right)}=H_{N}\left(1-\frac{1}{N+1}\right)-H_{N}+\sum_{n\leq N-1}\frac{1}{\left(n+1\right)^{2}}.$$ Now, since $$H_{N}\sim\log\left(N\right)$$ as $$N\rightarrow+\infty$$, the claim follows.

$$\newcommand{\Li}{\operatorname{Li}}$$Starting with a generating function for the Harmonic Numbers

$$-\frac{\log(1-x)}{1-x}~=~\sum_{n=1}^\infty H_nx^n\tag1$$

we divide both sides by $$x$$ and integrate afterwards to get

\small\begin{align*} \int-\frac{\log(1-x)}{x(1-x)}\mathrm dx&=\int\sum_{n=1}^\infty H_nx^{n-1}\mathrm dx\\ \Li_2(x)+\frac12\log^2(1-x)+c&=\sum_{n=1}^\infty \frac{H_n}nx^{n} \end{align*}

Plug in $$x=0$$ to determine the value of $$c$$ which turns out to be $$0$$ aswell. Integrating this equation yields to

\small\begin{align*} \int\Li_2(x)+\frac12\log^2(1-x)\mathrm dx&=\int\sum_{n=1}^\infty \frac{H_n}nx^{n}\mathrm dx\\ x\Li_2(x)-x+(x-1)\log(1-x)+\frac12(x-1)(\log^2(1-x)-2\log(1-x)+2)+c&=\sum_{n=1}^\infty\frac{H_n}{n(n+1)}x^{n} \end{align*}

The new constant again equals $$0$$ due the same argumentation as above. Now we can plug in $$x=1$$ to compute the value of your given series which overall is given by $$\Li_2(1)$$ and this is well-known to equal $$\zeta(2)$$.

$$\therefore~\sum_{n=1}^\infty\frac{H_n}{n(n+1)}~=~\frac{\pi^2}6\tag2$$

Here is another, slightly longer, approach. It will use a double integral in order to evaluate the sum.

Noting that $$\int_0^1 x^{n - 1} \, dx = \frac{1}{n} \quad \text{and} \quad \int_0^1 y^n \, dy = \frac{1}{n + 1},$$ the sum can be written as $$\sum_{n = 1}^\infty \frac{H_n}{n(n + 1)} = \int_0^1 \int_0^1 \frac{1}{x} \sum_{n = 1}^\infty H_n (xy)^n \, dx dy.\tag1$$ Here the order of the summation with the double integral has been changed and can be done due to the dominated convergence theorem.

Making use of the generating function for the Harmonic number, namely $$-\frac{\log(1-x)}{1 - x}~=~\sum_{n=1}^\infty H_n x^n,$$ the infinity sum appearing inside the integral in (1) can be written as $$\sum_{n = 1}^\infty \frac{H_n}{n(n + 1)} = -\int_0^1 \int_0^1 \frac{\ln (1 - xy)}{x(1 - xy)} \, dx dy.$$ Enforcing a substitution of $$x \mapsto x/y$$ gives \begin{align} \sum_{n = 1}^\infty \frac{H_n}{n(n + 1)} &= -\int_0^1 \int_0^y \frac{\ln (1 - x)}{x (1 - x)} \, dx dy\\ &= -\int_0^1 \int_0^y \left [\frac{\ln (1 - x)}{x} + \frac{\ln (1 - x)}{1 - x} \right ] \, dx \, dy\\ &= \int_0^1 \operatorname{Li}_2 (y) \, dy + \frac{1}{2} \int_0^1 \ln^2 (1 - y) \, dy\\ &= I_1 + \frac{1}{2} I_2. \end{align} Here $$\operatorname{Li}_2 (y)$$ is the dilogarithm function.

For the first of these integrals, writing the dilogarithm function in terms of its series representation, we have \begin{align} I_1 &= \int_0^1 \sum_{n = 1}^\infty \frac{y^n}{n^2} \, dy\\ &= \sum_{n = 1}^\infty \frac{1}{n^2} \int_0^1 y^n \, dy\\ &= \sum_{n = 1}^\infty \frac{1}{n^2(n + 1)}\\ &= \sum_{n = 1}^\infty \left [\frac{1}{n^2} + \frac{1}{n + 1} - \frac{1}{n} \right ]\\ &= \sum_{n = 1}^\infty \frac{1}{n^2} - \sum_{n = 1}^\infty \left (\frac{1}{n} - \frac{1}{n + 1} \right )\\ &= \frac{\pi^2}{6} - 1, \end{align} where the first of the series is the well-known Basel problem while the second telescopes.

For the second of the integrals, define $$I(a) = \int_0^1 (1 - y)^a \, dy, \quad a > - 1.$$ Observe that $$I''(0) = \int_0^1 \ln^2 ( 1 - y) \, dy,$$ where the derivative is with respect to $$a$$. Now, as $$I(a) = \operatorname{B} (1, a + 1) = \frac{\Gamma (a + 1)}{\Gamma (a + 2)} = \frac{1}{a + 1},$$ where $$\operatorname{B}(x,y)$$ is the Beta function, one readily has $$I''(a) = \frac{2}{(a + 1)^3},$$ yielding $$I''(0) = I_2 = 2$$.

Thus $$\sum_{n = 1}^\infty \frac{H_n}{n (n + 1)} = \frac{\pi^2}{6} - 1 + \frac{1}{2} \cdot 2 = \frac{\pi^2}{6}.$$