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!

Some hints:

If $S_{n} = \sum_{k=1}^{n} a_{k}$ 

then what is 

$\lim_{n \to \infty} S_{2n} - S_{n}$?

Now can you use the fact that $a_{n}$ is non-increasing to upper bound a certain term of the sequence $na_{n}$ with a multiple of $S_{2n} - S_{n}$?
