Show this series converges

[naslund19]

Hi, can anyone help me to show that this series converges:
\(\displaystyle \sum_{n=1}^{\infty} \frac{n5^n}{(n+1)3^{2n}}\)

I tried the ratio test but that gave me 1, and I would try the comparison test but I don't know what to compare.

Any help?

Thanks.

 

[akoaysigod]

I might have made a mistake but using the ratio test I got 5/9 which shows it converges.

in the end I got

(n+1)(n+1) 3^2n 5^(n+1)/n(n+2) 3^(2n+2) 5^n

the first part goes to 1 then 5^(n+1)/5^n is 5 since there is one more on the top and the 3^(2n)/3^(2n+2) = 9 since there are 2 more on the bottom then there are on the top. That's a poor mathematical way to explain it probably but its the only way I can make sense of it.

 

[BigGlenntheHeavy]

\(\displaystyle \sum_{n=1}^{\infty}\frac{n5^{n}}{(n+1)9^{n}}\)

\(\displaystyle Using \ the \ ratio \ test \ \implies \ \lim_{n\to\infty}\bigg|\frac{(n+1)5^{n+1}}{(n+2)9^{n+1}} \ * \ \frac{(n+1)9^{n}}{n5^{n}}\bigg|\)

\(\displaystyle = \ \frac{5}{9}\lim_{n\to\infty}\bigg|\frac{(n+1)^{2}}{n(n+1)}\bigg| \ = \ \frac{5}{9}(1) \ = \ \frac{5}{9} \ < \ 1, \ hence \ the \ series \ converges.\)