Determine whether the series converges or diverges. 3^(n+2) / 5^n

Question

johnt · Answer

$$a_n = \frac{ 3^{n+2} }{ 5^n } $$

johnt · Answer

I'm not sure where to start with this. WolframAlpha says the limit is 0, suggesting it converges, but I don't know how to prove that. It looks like you'd use L'Hospital's rule, but that doesn't do anything. You just get $$3^{n+2} \ln(3)$$ again, which doesn't help to integrate it.

If someone could help me understand where to begin, I'd appreciate it.

e.mccormick · Answer

Hmmm.... yah, that one is a little past me, but since nobody posted for an hour, let me link something that might cover it: http://tutorial.math.lamar.edu/Classes/CalcII/ConvergenceOfSeries.aspx

anonymous · Answer

Just because $\displaystyle\lim_{n\to\infty}a_n=0$ doesn't mean $\sum a_n$ converges. The converse is true, though: If $\sum a_n$ converges, then $\displaystyle\lim_{n\to\infty}a_n=0$.

For this series, you just have to do some rewriting:
$$\begin{align*}\sum_{n=0}^\infty \frac{3^{n+2}}{5^n}&=\sum_{n=0}^\infty \frac{3^{n}\cdot3^2}{5^n}\
&=9\sum_{n=0}^\infty \frac{3^n}{5^n}\
&=9\sum_{n=0}^\infty \left(\frac{3}{5}\right)^n\end{align*}$$
What do you know about the convergence/divergence of a geometric series?