Abstract Nonsense

Crushing one theorem at a time

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.

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December 27, 2010 - Posted by | Analysis, Computations, Fun Problems | , , , ,

1 Comment »

  1. […] begin now  by recalling that in our last post we derived the result […]

    Pingback by Another Interesting Series Involving the Harmonic Numbers « Abstract Nonsense | December 27, 2010 | Reply


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