# Abstract Nonsense

## Interesting Series Involving the Harmonic Numbers

Point of post: In this post we compute the value of the series

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

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

Problem: Compute

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

Proof: We begin by recalling that if $\displaystyle \sum_{n=0}^{\infty}a_n z^n$ and $\displaystyle \sum_{n=0}^{\infty}b_n z^n$ are two complex power series convergent on $B_R(0)$ then

$\displaystyle \sum_{n=0}^{\infty}\sum_{k=0}^{n}a_{k}b_{n-k}z^n$

converges on $B_R(0)$ and its value is

$\displaystyle \left(\sum_{n=0}^{\infty}a_n z^n\right)\left(\sum_{n=0}^{\infty}b_n z^n\right)$

(this is just a consequence of Merten’s Theorem). In particular, if the power series is real with

$\displaystyle f(x)=\sum_{n=1}^{\infty}a_n x^n$

on $(-1,1)$ then

\displaystyle \begin{aligned}\frac{f(x)}{1-x} &= \left(\sum_{n=0}^{\infty}a_n x^n\right)\left(\sum_{n=0}^{\infty}x^n\right)\\ &= \sum_{n=1}^{\infty}\sum_{k=0}^{n}a_k x^n\end{aligned}

Thus, noting that for $z\in(-1,1)$

$\displaystyle -\log(1-x)=\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}$

we may conclude that

\displaystyle \begin{aligned}\frac{-\log(1-x)}{1-x} &= \sum_{n=0}^{\infty}\sum_{k=0}^{n}\frac{1}{k+1}z^{n+1}\\ &= \sum_{n=0}^{\infty}\sum_{k=1}^{n+1}\frac{1}{k}z^{n+1}\\ &= \sum_{n=1}^{\infty}\sum_{k=1}^{n}\frac{1}{k} z^n\\ &= \sum_{n=1}^{\infty}H_n x^n\end{aligned}

Thus, it follows that

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

and thus

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

for $x\in(-1,1)$. In particular, it follows that

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

Thus, it suffices to compute

$\displaystyle \int_0^{\frac{1}{2}}\frac{-\log(1-x)}{x(1-x)}\text{ }dx$

But, letting $\displaystyle x=\frac{t}{t+1}$ we find that

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

Recall though that for $x\in(-1,1]$ we have that

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

and thus, in particular

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

But, this is just $\eta(2)$ where $\eta$ is the Dirichlet Eta Function. But, it’s a well-known fact that

$\displaystyle \eta(s)=\left(1-2^{1-s}\right)\zeta(s),\quad \text{Re}(s)>1$

where $\zeta$ is the Riemann Zeta Function. Thus, it follows that

$\displaystyle \sum_{n=1}^{\infty}\frac{H_n}{2^n n}=\frac{1}{2}\zeta(2)=\frac{\pi^2}{12}$

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.