This is a problem adapted from Stewart Calculus, and I feel like the provided solution is too hand-wavy to be satisfactory.
Suppose $\{b_n\}$ is a sequence that converges to $\frac{1}{5}$. The goal is to determine whether or not $\sum \frac{b_n^n}{n}$ converges, using the ratio test.
If we let $a_n = \frac{b_n^n}{n}$, then $$a_{n+1} = \frac{b_{n+1}^{n+1}}{n+1}.$$ Applying the ratio test to evaluate the the limit of $| \frac{a_{n+1}}{a_n}|$ as $n$ goes to infinity, we get
$$b_{n+1} (\frac{b_{n+1}}{b_n})^n \cdot \frac{n}{n+1}.$$
Taking the limit, $b_{n+1}$ becomes $\frac{1}{5}$ and $\frac{n}{n+1}$ becomes $1$, but I don't know how to address the factor in the middle. It's a mistake to say that we can take the limit inside before taking the limit on the outside-- after all, $\lim_{x \to \infty} ({1 + 1/x})^x$ is $e$, not $1$.
What is the right way to do this problem?
You cannot use the ratio test for that, it is inconclusive. For instance, consider the sequence $(b_n)_n$ defined by
$$ b_{2n}=\frac{1}{5},\qquad b_{2n+1}=\frac{1}{5}\left(1+ \frac{1}{\sqrt{n}}\right) $$ for all $n\geq 0$. Then $$ \frac{a_{2n+1}}{a_{2n}} = \frac{1}{5}\cdot \frac{2n}{2n+1}\cdot \left(1+ \frac{1}{\sqrt{n}}\right)^{2n+1} \xrightarrow[n\to\infty]{} \infty $$ while $$ \frac{a_{2n+2}}{a_{2n+1}} \xrightarrow[n\to\infty]{} 0 $$ so the ratio test is inconclusive (the limit of the ratio fails to exist).
Other tests, like comparison to say $(9/10)^n/n$, will work, though, since $|b_n| \leq 9/10$ for all $n$ large enough.