# Convergence of Sequence: (n^2)/(e^n)

1. Mar 28, 2012

### k_squared

NEVERMIND! IT IS 0! I SOMEHOW WAS STARING AT THE WRONG ANSWER SHEET FOR A LITTLE BIT! THANK YOU!

1. The problem statement, all variables and given/known data

Determinte whether the sequence converges or diverges:
(n^2)/(e^n)

2. Relevant equations

The book says that the solution is: e/(e-1).

However, the limit of the equation y=(n^2)/(e^n) as n goes to infinity is 0.

I don't know why I can't seem to apply the theorem that:

If the limit as x goes to infinity of f(x) = L and f(n)= an, then the limit of an as x approaches infinity is L.

3. The attempt at a solution

I used L'Hôpital's rule to prove the limit is 0. Wolfram alpha confirms this.

Last edited: Mar 28, 2012
2. Mar 28, 2012

### LCKurtz

The question is whether the sequence converges, so the answer would be yes or no, right? What question does your book say e/(e-1) is the answer to? You are correct that the sequence converges to 0. Have you stated the question fully?