# Radius and interval of convergence help

1. Nov 8, 2013

### xtrubambinoxpr

I have a problem:

the radius of convergence for the power series $\sum$(-1)^n $\frac{(x-3)^n}{(n+1)}$ is R=1. Determine the interval of convergence.

What does this mean? can anyone help me solve this please

2. Nov 8, 2013

### HallsofIvy

Staff Emeritus
Are you asking about the definition of "interval of convergence"? Surely if you are asked to find an interval of convergence, you should already have learned what it means!

To determine for what values of x the power series, $\sum a_n(x- x_0)^n$, converges, one can use the "ratio test":

"ratio test"- The numerical sum of positive numbers, $\sum x_n$, converges if the sequence of ratios, $x_{n+1}/x_n$, converges to a limit less than 1 and diverges if that sequence converges to a limit larger than 1. It may either converge or diverge if the limit is 1.

Of course, the numbers, for all x, are not necessarily positive so we have to talk about "absolute convergence", the convergence of $\sum |a_n x^n|$.

For example, the ratio of of the form $|a_{n+1}(x-x_0)^{n+1}|/|a_n x^n|$ reduces to $(a_{n+1}/a_n)|x- x_0|$. Saying the limit is less than 1 is exactly the same as saying that $|x-x_0|< \lim_{n\to\infty}|a_n/a_{n+1}$.

That is, if we let $r= \lim_{n\to\infty}|a_n/a_{n+1}|$, then we are saying that the series converges for $|x-x_0|< r$ which is, of course, the same as saying $-r< x- x_0< r$ or $x_0- r< x< x_0+ r$. That is precisely saying that the series converges for x in the open interval $(x_0- r, x_0+ r)$ and diverges outside the interval $x_0- r\le x\le x_0+ r$ (the endpoints, $x_0- r$ and $x_0+ r$, have to be checked separately). The interval $(x_0- r, x_0+ r)$, including one or both endpoints if the series converges there, is called the "interval of convergence" and "r" is called the "radius of convergence"

(When you work with power series or complex numbers, the "interval of convergence" in the "number line" becomes a disk in the "complex plane" and the "radius of convergence" really is a radius.)

Now, if we use the ratio test to determine the convergence of $\sum (-1)^n (x-3)^n/(n+1)$, we look at the ratio
$$\dfrac{|x- 3|^{n+1}}{n+2}\dfrac{n+3}{|x-3|^n}= \frac{n+2}{n+3}|x-3|$$
You should ge able to see that the limit, as n goes to infinity, of (n+2)/(n+3) is 1 so we know that this series converges absolutely for |x- 3|< 1. That is the same as saying that -1< x- 3< 1 or 3-1= 2< x< 3+ 1= 4. It diverges outside the interval. Whether or not it converges at x= 2 or x= 4 would have to be decided separately.