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?

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- Feb 24th 2006, 04:10 PMCowPizzaRadius of convergence
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 PMThePerfectHackerQuote:

Originally Posted by**CowPizza**

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.