Abstract Nonsense

Crushing one theorem at a time

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 - Posted by | Analysis, Computations, Fun Problems | , , , ,

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


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