# Radius of convergence

1. Mar 19, 2009

### incus

Having a hard time with this one: E 1/n^x , have tried too use n^-x=e^(-x ln n) which in turn e^(...) = lim n->OO (1-(x ln n)/n)^n and then go on with finding the centre, but I feel I'm far far off. How to get it like E an(x-c)^n and use the more straight foreward path.

2. Mar 19, 2009

### yyat

Hi incus!

The series

$$\sum_{n=0}^{\infty}\frac{1}{n^x}$$

is not a power series, so it does not have a radius convergence. It does however have a region of convergence (the x so that the series converges). Is x supposed to be a real or complex variable?

In the case where x is a real variable you can use the integral test to find the region of convergence.

The case where x is complex can be reduced to the real case by considering the real part of x and the absolute value of the terms in the series.

3. Mar 19, 2009

### Dick

You can't ask for a radius of convergence unless you say which point you are expanding around. I'm guessing the actual question is 'for what values of x does the series converge'. Is x complex? Hint: your series is a p-series. And your series defines part of the Riemann zeta function.

4. Mar 20, 2009

### incus

Thanks for steering me in the right direction yyat and Dick. Got blinded by the question.