0
$\begingroup$

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

$\endgroup$

1 Answer 1

3
$\begingroup$

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$(s+2)>1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.