• Feb 24th 2006, 04:10 PM
CowPizza
Hey, I'm supposed to find out the interval of convergence for this power series:

$\displaystyle \sum_{n = 0}^\infty\frac{(-1)^n \ast (x-5)^n}{n5^n}$

Any thoughts?
• Feb 25th 2006, 03:37 PM
ThePerfectHacker
Quote:

Originally Posted by CowPizza
Hey, I'm supposed to find out the interval of convergence for this power series:

$\displaystyle \sum_{n = 0}^\infty\frac{(-1)^n \ast (x-5)^n}{n5^n}$

Any thoughts?

First of all $\displaystyle n\not =0$ :eek: thus maybe you mean $\displaystyle \sum_{n = 1}^\infty\frac{(-1)^n \ast (x-5)^n}{n5^n}$
Well, I think it would be easier to consider,
$\displaystyle \sum^{\infty}_{k=1}\frac{(1-\frac{x}{5})^k}{k}$, which is equivalent to the one you gave.
Then let $\displaystyle u=1-\frac{x}{5}$, thus,
$\displaystyle \sum^{\infty}_{k=1}\frac{u^k}{k}$.
Notice that this is a power series, thus use the generalized ratio test, where $\displaystyle a_k=\frac{1}{k}$.
Thus,
$\displaystyle \lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|= \left|\frac{k}{k+1} \right|=1$, thus the radius of convergence is the reciprocal of this which is also 1.

Let us now find the interval of convergence, since $\displaystyle -1<u<1$ the series is absolutely convergent we have that $\displaystyle -1<1-\frac{x}{5}<1$, thus,
$\displaystyle 0<x<10$.
Notice when $\displaystyle x=0$ the series diverges because it forms a harmonic series and when $\displaystyle x=10$ the series converges conditionally because it form and alternating series. Thus,
$\displaystyle 0<x\leq 10$ is the interval of convergence.
Q.E.D.