This is the series:

¡Ä

¥Ò   n©ø / 3©ú

n=1

We can start with the fact that the infinite series x©ú = x / (1-x), and then differentiate a few times to end up with something of the form:

n©øx©ú

on the left hand side.  Then just substitute x = 1/3 to solve for the limit.  I've done it on paper and confirmed Charlie's solution.

Edit: Wow, the formatting appears to have gotten messed up.

I'll try again:

Start with the sum of the series x^n as n -> infinity.

Differentiate a few times until you end up with n^3 * x^n.

Plug in x = 1/3 to get the answer.

Edited on February 26, 2014, 1:13 pm

Posted by tomarken on 2014-02-26 13:11:36