# Abstract Nonsense

## 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.  $\blacksquare$

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.

December 28, 2010 -