Radius/Interval of Convergence

**RockHard** · November 23rd 2009, 04:29 PM

We were being taught this subject today, however, the professor did not get to finish the lecture and mentioned something about testing the end points of the interval with another test of convergence, so they function doesn't diverge at the end points? How would I do this, as in checking the end points properly, just want to get a step ahead,

I was given the following example I reworked myself till I obtained the interval of convergence

$\sum_{n=1}^{\infty}\frac{(2x+3)^n}{n^2}$

I implemented a ratio test, here the end result

$|2x-3|\lim_{n\to\infty}(\frac{n}{n+1})^2$

$|2x-3|\lim_{n\to\infty}(1-\frac{1}{n+1})^2 = 1$

$= |2x+3|$

$-1\le2x+3\le1$

$-2\le x \le -1$

**Scott H** · November 24th 2009, 03:11 AM

Here is a hint: the series at $x=-2$ and $x=-1$ are

$\sum_{n=1}^{\infty}\frac{(-1)^n}{n^2}=-1+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{4^2}-\cdots$

and

$\sum_{n=1}^{\infty}\frac{1}{n^2}=1+\frac{1}{2^2}+\ frac{1}{3^2}+\frac{1}{4^2}+\cdots.$

Thread: Radius/Interval of Convergence

LinkBack

Thread Tools

Display

Radius/Interval of Convergence

Similar Math Help Forum Discussions

radius of convergence and interval of convergence of the series

radius and interval of convergence

Radius and interval of convergence?

Radius and Interval of Convergence

Radius and interval of convergence

Search Tags