# Radius and interval of convergence

• April 19th 2014, 10:19 AM
May someone check my work?

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

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

To find the interval of convergence, C
$\left| x-2 \right|<\frac{1}{3} \Rightarrow \frac{5}{3}

For $x=\frac{5}{3}$:
$\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 $x=\frac{7}{3}$:
$\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= $\frac{5}{3}\leq x \leq \frac{7}{3}\right]$
• April 19th 2014, 04:46 PM
johng
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.
• April 19th 2014, 05:36 PM
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?
• April 20th 2014, 05:42 PM
johng
Re: Radius and interval of convergence
Hi again,
Back to basics:

http://i61.tinypic.com/1zg965v.png