1. ## 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=$\displaystyle \frac{5}{3}\leq x \leq \frac{7}{3}\right]$

2. ## 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.

3. ## Re: Radius and interval of convergence

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?

4. ## Re: Radius and interval of convergence

Hi again,
Back to basics: