This is the series:
¥Ò n©ø / 3©ú
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:
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