Ratio Test Radius of Convergence

[checkmatechamp]

Homework Statement

∑ x²ⁿ / n!

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

Homework Equations

The Attempt at a Solution

So I take the limit as n approaches infinity of a_a+1 / a_n. 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?

 

[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?

 

[HallsofIvy]

So does that mean my interval of convergence is just the point x = 0?

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.