# Ratio Test Radius of Convergence

1. Nov 10, 2014

### checkmatechamp

1. The problem statement, all variables and given/known data
∑ x2n / n!

The limits of the sum go from n = 0 to n = infinity

2. Relevant equations

3. The attempt at a solution

So I take the limit as n approaches infinity of aa+1 / an. So that gives me:

((x2n+2) * (n!)) / ((x2n) * (n + 1)!)

Canceling everything out gives me x2 / (n + 1)

The limit as n approaches infinity is x2 1 / (infinity + 1), which is x^2 * 0

So now where do I go from here? x^2 * 0 is 0. So does that mean my interval of convergence is just the point x = 0?

2. Nov 10, 2014

### TheMathSorcerer

You messed up, it's definitely worth fixing and understanding.

You should get $\lim n \to \infty \frac{x}{n + 1} = 0$. Convince yourself:)

Here $L = 0$. The ratio test says the series converges absolutely if $L < 1$. For what values of $x$ is this true?

3. Nov 10, 2014

### HallsofIvy

Staff Emeritus
Just the opposite. What you have shown is that the ratio is always less than 1 so it always converges. In general if the limit in the ratio test is 'A' then the radius of convergence is 1/A. If ratio is never less than 1, the radius of convergence is 0 (so the interval of convergence is just x= 0) and if the ratio is always less than 1, the radius of convergence is infinity.