Sequence - Converge or Diverge?

[kiddopop]

Consider the series inf E n=1 (e^n)/(3^(n-1). Does this series converge or diverge? If the series converges, find its sum. Give the exact value and an approximation rounded to three decimal places after the decimal point.

I've tried to figure out if the series converges or diverges, but I keep going around in circles not really getting an answer.

 

[BigGlenntheHeavy]

\(\displaystyle \sum_{n=1}^{\infty}\frac{e^{n}}{3^{n-1}}, \ use \ ratio \ test, \ to \ wit:\)

\(\displaystyle \lim_{n\to\infty}\bigg|\frac{a_{n+1}}{a_n}\bigg| \ = \ \lim_{n\to\infty}\bigg|\frac{e^{n+1}}{3^{n}}*\frac{3^{n-1}}{e^{n}}\bigg| \ = \ \lim_{n\to\infty}\frac{e}{3} \ = \ .90609 \ < \ 1, \ converges.\)

\(\displaystyle Now, \ \sum_{n=1}^{\infty}\frac{e^{n}}{3^{n-1}} \ = \ \sum_{n=0}^{\infty}\frac{e^{n+1}}{3^{n}} \ = \ e\sum_{n=0}^{\infty}\bigg(\frac{e}{3}\bigg)^{n}\)

\(\displaystyle Hence \ (geometric \ series), \ Sum \ = \ \frac{a}{1-r} \ = \ \frac{e}{1-\frac{e}{3}} \ = \ \frac{3e}{3-e} \ = \ 28.947\)