Take the 2-minute tour ×

In the book I am reading they write that for Hurwitz zeta function, $\zeta(x,s)=\sum_{n=0}^{\infty} \frac{1}{(x+n)^s}$, the next sum in the RHS converges for $\Re(s)>-1$, and I don't see how exactly?!

$$\zeta(x,s)-(\zeta(s)-sx\zeta(s+1)= x^{-s} + \sum_{n=1}^{\infty} n^{-s}[(1+x/n)^{-s} - (1-x/n)]$$

I think the sum converges for $\Re (s+1) > 1$ which means $\Re s >0$ and not $>-1$.

The reference is Andrews' et al Special Functions red book, page 17. http://books.google.co.il/books?id=kGshpCa3eYwC&printsec=frontcover#v=onepage&q&f=false

share|improve this question

1 Answer 1

up vote 3 down vote accepted

Looks like there's a typo in your post. Andrews has $\left(1-\frac{sx}{n}\right)$ where you have $\left(1-\frac{x}{n}\right)$ in the rightmost term.

If you expand $\left(1+\frac{x}{n}\right)^{-s}$ as a power series in $\frac{x}{n}$, the series begins $1-s\frac{x}{n}+O(x^2/n^2)$. The first two terms are cancelled by the $-\left(1-\frac{sx}{n}\right)$, so the quantity between the square brackets is $O(n^{-2})$ as $n\to\infty$. This means the sum converges for Re$(s+2)>1$.

share|improve this answer

Your Answer


