Radius and interval of convergence

May someone check my work?

$\displaystyle \sum_{n}^{\infty}(n3^{-n})(x-2)^n$

To find R

$\displaystyle \lim_{n \rightarrow \infty} \left| \frac{n3^{-n}}{(n+1)3^{1-n}} \right| = \frac{1}{3}\lim_{n \rightarrow \infty} \left| \frac{n}{n+1} \right|=\frac{1}{3}$

To find the interval of convergence, C

$\displaystyle \left| x-2 \right|<\frac{1}{3} \Rightarrow \frac{5}{3}<x<\frac{7}{3}$

For $\displaystyle x=\frac{5}{3}$:

$\displaystyle \sum_{n}^{\infty}(n3^{-n})\left\(\frac{-1}{3}\right\)^n=\sum_{n}^{\infty}n(-1)^n\left\(\frac{1}{3}\right\)^{2n}$ which converges by alternating series test.

For $\displaystyle x=\frac{7}{3}$:

$\displaystyle \sum_{n}^{\infty}(n3^{-n})\left\(\frac{1}{3}\right\)^n=\sum_{n}^{\infty} \frac{n}{9^n}$ which converges by the root test.

So C=$\displaystyle \frac{5}{3}\leq x \leq \frac{7}{3}\right]$

Re: Radius and interval of convergence

Hi,

No, you made an algebraic error. Given $a_n=n3^{-n}$, then $a_{n+1}=(n+1)3^{-n-1}$ and so the quotient

$${a_n\over a_{n+1}}={n3^{-n}\over (n+1)3^{-n-1}}=3\,{n\over n+1}$$

So your interval of convergence is $|x-2|<3$ or $-1<x<5$.

Aside. When you find the interval of convergence of a power series, the ratio test or root test will always fail at the endpoints; the interval of convergence is found by one of these tests.

Re: Radius and interval of convergence

Quote:

Originally Posted by

**johng** When you find the interval of convergence of a power series, the ratio test or root test will always fail at the endpoints; the interval of convergence is found by one of these tests.

I'm not sure I understand what you're saying. If they always fail at the endpoints then why should we use them?

Re: Radius and interval of convergence