# Abstract Nonsense

## Another Interesting Series Involving the Harmonic Numbers

Point of post: In this post we calculate the value of

$\displaystyle \sum_{n=1}^{\infty}\frac{H_n}{n^2}$

and in doing so also calculate

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

where $H_n$ is the $n^{\text{th}}$ harmonic number.

Problem: Compute

$\displaystyle \sum_{n=1}^{\infty}\frac{H_n}{n^2}$

Proof:

Lemma: Let $\text{Re}(s)>1$. Then,

$\displaystyle \Gamma(s)\zeta(s)=\int_0^{\infty}\frac{x^{s-1}}{e^x-1}\text{ }dx$

where $\Gamma(s)$ is the Gamma Function and $\zeta(s)$ is the Riemann Zeta Function.

Proof: Recall by definition that

$\displaystyle \Gamma(s)=\int_0^{\infty} x^{s-1}e^{-x}\text{ }dx$

We note then that

\displaystyle \begin{aligned}\int_0^{\infty}x^{s-1}e^{-nx}\text{ }dx &= \frac{1}{n^s}\int_0^{\infty}u^{s-1}e^{-u}\text{ }du\\ &= \frac{\Gamma(s)}{n^s}\end{aligned}

where the second step was due to the substitution $nx=u$. It follows then that

\displaystyle \begin{aligned}\Gamma(s)\zeta(s) &= \Gamma(s)\sum_{n=1}^{\infty}\frac{1}{n^s}\\ &= \sum_{n=1}^{\infty}\frac{\Gamma(s)}{n^s}\\ &= \sum_{n=1}^{\infty}\int_0^{\infty}x^{s-1}e^{-nx}\text{ }dx\\ &= \int_0^{\infty}x^{s-1}\sum_{n=1}^{\infty}e^{-nx}\text{ }dx\\ &= \int_0^{\infty}x^{s-1}\frac{e^{-x}}{1-e^{-x}}\text{ }dx\\ &= \int_0^{\infty}\frac{x^{s-1}}{e^x-1}\text{ }dx\end{aligned}

where we’ve used tacit use of the uniform continuity a.e. for the summand and the geometric series. $\square$

We begin now  by recalling that in our last post we derived the result that

$\displaystyle \frac{-\log(1-x)}{1-x}=\sum_{n=1}^{\infty}H_n x^n$

for $x\in(-1,1)$. It follows then that for $x\in(-1,1)$ we have that

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

and thus for $x\in(-1,1)$ we have that

$\displaystyle \frac{1}{2}\int_0^ x\frac{\log^2(1-t)}{t}\text{ }dt=\sum_{n=1}^{\infty}\frac{H_n}{(n+1)^2}x^{n+1}$

Noting though that since this series on the right converges for $x\in[-1,1]$ since $H_n\sim \ln(n)+\gamma$ (where $\gamma$ is the Euler-Mascheroni constant) it follows that

$\displaystyle \frac{1}{2}\int_0^1\frac{\log^2(1-t)}{t}\text{ }dt=\sum_{n=1}^{\infty}\frac{H_n}{(n+1)^2}$

Note though that if one lets $\log(1-t)=w$ in this integral one arrives at

$\displaystyle \frac{1}{2}\int_0^{\infty}\frac{w^2}{e^w-1}\text{ }dw$

But, by our lemma this is equal to

$\frac{1}{2}\Gamma(3)\zeta(3)=\zeta(3)$

The problem follows since

\displaystyle \begin{aligned}\sum_{n=1}^{\infty}\frac{H_n}{n^2} &= 1+\sum_{n=2}^{\infty}\frac{H_n}{n^2}\\ &= 1+\sum_{n=2}^{\infty}\frac{H_{n-1}+\frac{1}{n}}{n^2}\\ &= 1+\sum_{n=1}^{\infty}\frac{H_{n-1}}{n^2}+\sum_{n=2}^{\infty}\frac{1}{n^3}\\ &= \sum_{n=2}^{\infty}\frac{H_{n-1}}{(n+1)^2}+\zeta(3)\\ &= \sum_{n=1}^{\infty}\frac{H_n}{(n+1)^2}+\zeta(3)\\ &= 2\zeta(3)\end{aligned}

References:

1. Apostol, Tom M. Mathematical Analysis. London: Addison Wesley Longman, 1974. Print.

2. Edwards, Harold M. Riemann’s Zeta Function. Mineola, NY: Dover Publications, 2001. Print.

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December 27, 2010 -

## 3 Comments »

1. […] was browsing thorough Alex Youcis’s blog Abstract Nonsense, when I stumbled upon a post where I saw the […]

Pingback by Zeta function and harmonic numbers « Hardy-Ramanujan Letters | March 14, 2011 | Reply

2. Hello Alex,

Interesting identity. Based on this idea, I was ale to derive a few more identities involving the harmonic number and the Riemann zeta function. I have mentioned about you in my post

http://hardyramanujan.wordpress.com/2011/03/14/zeta-function-and-harmonic-numbers/

Comment by Nilotpal Sinha | March 14, 2011 | Reply

• Dear Niloptal,

Thanks for visiting my blog! Out of all my posts, this is not one I would have thought someone would enjoy. Also, I have never seen your blog, but it is very, very impressive. It is definitely of a much different taste than what I am used to–all the better! Keep up the good work friend.

Best,
Alex Youcis

Comment by drexel28 | March 15, 2011 | Reply