How Do You Simplify and Find the Interval of Convergence for a Power Series?

[tolove]

Edit: Nevermind, figured it out. Thank you for readingOriginal problem:
Find the interval of convergence
\sum^∞_n=1 xⁿ / n * √(n) * 3ⁿ

Ratio Test, right? a_n+1/a

I get to here and I can't figure out how to get rid of the ns:

lim n→∞ abs(x/3)* [n*√(n) / (n+1)*√(n+1)]

Solution,
They break apart evenly:
(n/(n+1)) * (n/(n+1)**(1/2)

(also, sorry this looks terrible. I'm not sure how to use the graphics options very well yet)

 

[Dick]

Original problem:
Find the interval of convergence
\sum^∞_n=1 xⁿ / n * √(n) * 3ⁿ

Ratio Test, right? a_n+1/a

I get to here and I can't figure out how to get rid of the ns:

lim n→∞ abs(x/3)* [n*√(n) / (n+1)*√(n+1)]

(also, sorry this looks terrible. I'm not sure how to use the graphics options very well yet)

Start with just the n/(n+1) part. Divide numerator and denominator by n and tell me what the limit of that is as n->infinity.

 

[tolove]

Start with just the n/(n+1) part. Divide numerator and denominator by n and tell me what the limit of that is as n->infinity.

Simplifies down to:
lim_n→∞ abs(x)/3 * (1/(1+1/n))**(3/2) = abs(x)/3

Thank you very much for your reply!