Abstract Nonsense

Crushing one theorem at a time

An Interesting Sum Involving Riemann’s Zeta Function

Point of post: In this post we shoe precisely we compute the radius of convergence of the the power series

\displaystyle \sum_{m=2}^{\infty}\frac{\zeta(m)}{m}x^m

and find an explicit formula (in terms of known special functions and constants) for values where it converges.


Theorem: The power series

\displaystyle \sum_{m=2}^{\infty}\frac{\zeta(m)}{m}x^m

converges absolutely on (-1,1) and conditionally at -1; it diverges for every x\in\mathbb{R}-[-1,1). Moreover, for x\in[-1,1) we have that

\displaystyle \sum_{m=2}^{\infty}\frac{\zeta(m)}{m}x^m=\log\left(\Gamma(1-x)\right)-\gamma x

where \gamma is the Euler-Maschernoi constant.

Proof: We recall that the radius of convergence of our power series is

\displaystyle R=\left(\limsup_{n\to\infty}\sqrt[m]{\frac{\zeta(m)}{m}}\right)^{-1}

Recall though that for integer m>1 we have that

\displaystyle 1\leqslant \zeta(m)\leqslant\zeta(2)=\frac{\pi^2}{6}\quad\mathbf{(1)}

so that

\displaystyle \frac{1}{\sqrt[m]{m}}\leqslant\sqrt[m]{\frac{\zeta(m)}{m}}\leqslant \sqrt[m]{\frac{\pi^2}{6m}}

Thus, applying the squeeze theorem to both sides gives us that

\displaystyle \lim_{m\to\infty}\sqrt[m]{\frac{\zeta(m)}{m}}=1

and thus inputting this into the definition of radius of convergence gives us that R=1. Thus, it follows that

\displaystyle \sum_{m=2}^{\infty}\frac{\zeta(m)}{m}x^m

converges for |x|<1 and diverges for |x|>1. Thus, it remains to check the cases when x=\pm 1.  Using \mathbf{(1)} we readily attain that

\displaystyle \lim_{m\to\infty}\frac{\zeta(m)}{m}=0

from where

\displaystyle \sum_{m=2}^{\infty}\frac{(-1)^m\zeta(m)}{m}

converges by Dirichlet’s test. Just as easily from \mathbf{(1)} we obtain that

\displaystyle \frac{1}{m}\leqslant\frac{\zeta(m)}{m}

and thus

\displaystyle \sum_{m=2}^{\infty}\frac{\zeta(m)}{m}

diverges with comparison the harmonic series. Thus, we have the first part of our result, namely that

\displaystyle \sum_{m=2}^{\infty}\frac{\zeta(m)}{m}x^m

converges precisely when x\in[-1,1). Thus, we are now trying to prove that for such x we have that

\displaystyle \sum_{m=2}^{\infty}\frac{\zeta(m)}{m}x^m=\log\left(\Gamma(1-x)\right)-\gamma x

To do this we first recall that for x\in\mathbb{R}-\mathbb{N} we have

\displaystyle \Gamma(-x)=\frac{e^{\gamma x}}{-x}\prod_{n=1}^{\infty}\left(1-\frac{x}{n}\right)e^{\frac{-x}{n}}

Then, taking the logarithm of both sides (recalling the result of taking the logarithm of an infinite convergent series of non-negative terms)$ we get

\displaystyle \log\left(\Gamma(-x)\right)=\gamma x-\log(-x)+\sum_{n=1}^{\infty}\left(-\log\left(1-\frac{x}{n}\right)-\frac{x}{n}\right)

or, said differently

\displaystyle \log\left(\Gamma(1-x)\right)-\gamma x=\sum_{n=1}^{\infty}\left(-\log\left( 1-\frac{x}{n}\right)-\frac{x}{n}\right)

The problem is finished then by noticing that

\displaystyle \begin{aligned}\sum_{n=1}^{\infty}\left(-\ln\left(1-\frac{x}{n}\right)-\frac{x}{n}\right) &= \sum_{n=1}^{\infty}\left(\sum_{m=0}^{\infty}\frac{x^{m+1}}{n^{m+1}(m+1)}-\frac{x}{n}\right)\\ &= \sum_{n=1}^{\infty}\sum_{m=1}^{\infty}\frac{x^{m+1}}{n^{m+1}(m+1)}\\ &= \sum_{m=1}^{\infty}\frac{x^{m+1}}{m+1}\sum_{n=1}^{\infty}\frac{1}{n^{m+1}}\\ &= \sum_{m=1}^{\infty}\frac{\zeta(m+1)}{m+1}x^{m+1}\\ &= \sum_{m=2}^{\infty}\frac{\zeta(m)}{m}x^m\end{aligned}

where the interchanging of the sums was justified since for each x\in[-1,1) and each fixed n\in\mathbb{N} the inner sum converges absolutely.  \blacksquare



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.



December 28, 2010 - Posted by | Analysis, Computations, Fun Problems | , , , , ,

No comments yet.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: