<s>[INST] Simple series divergence problem I've got a problem here:
$$\sum_{n=1}^{\infty} \frac{5^n}{n(3^{n+1})}$$
I've used the ratio test and essentially did this:
$$\sum_{n=1}^{\infty} \left( \frac{5^{n + 1}}{n (3^{n+1+1})} / \frac{5^n}{n(3^{n+1})}\right) = \frac{5^n\,5}{9(n+1)3^n} \cdot \frac{n\,3^n}{5^n}$$
With a bunch of cancellations we get $\dfrac{5n}{9n+9}$, which means it converges, as $\frac{5}{9} < 1$. But the answer says it diverges! I even tried the root test and got the same result. Where am I going wrong?
 [/INST] $$\begin{align}
  a_n &= \frac{5^n}{n(3^{n+1})} \\[2em]
  \frac{a_{n+1}}{a_n}
  &= \frac{5^{n+1}}{(n+1)(3^{n+2})} / \frac{5^n}{n(3^{n+1})} \\[0.5em]
  &= \frac{5^{n+1} \cdot n \cdot 3^{n+1}}{5^n \cdot (n+1) \cdot 3^{n+2}} \\[0.5em]
  &= \frac{5}{3} \left( \frac{n}{n+1} \right) \\[0.5em]
  &= \frac{5}{3} \left( 1 - \frac{1}{n+1} \right) \\[0.5em]
  &\to \frac{5}{3}>1
\end{align}$$
Therefore, from the ratio test,we conclude that the series diverges.
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