If $(a_n)\subset[0,\infty)$ is non-increasing and $\sum_{n=1}^\infty a_n<\infty$, then $\lim_{n\to\infty}{n a_n} = 0$ I'm studying for qualifying exams and ran into this problem.

Show that if $\{a_n\}$ is a nonincreasing sequence of positive real
numbers such that $\sum_n a_n$ converges, then $\lim_{n \rightarrow \infty} n a_n = 0$.

Using the definition of the limit, this is equivalent to showing
\begin{equation}
\forall \varepsilon > 0 \; \exists n_0 \text{ such that }
|n a_n| < \varepsilon \; \forall n > n_0
\end{equation}
or
\begin{equation}
\forall \varepsilon > 0 \; \exists n_0 \text{ such that }
a_n < \frac{\varepsilon}{n} \; \forall n > n_0
\end{equation}
Basically, the terms must be bounded by the harmonic series. Thanks, I'm really stuck on this seemingly simple problem!

By the Cauchy condensation test, $\displaystyle \sum 2^n a_{2^n} $ converges so $ 2^n a_{2^n} \to 0. $ For $ 2^n < k < 2^{n+1} $, 
$$ 2^n a_{2^{n+1}}    \leq k a_{k} \leq 2^{n+1} a_{2^n}$$
so $n a_n \to 0.$
