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!

Note that the desired conclusion is not true if the $a_n$ are not assumed to be nonincreasing:
We know a couple of facts: 


*

*If a sequence $(s_n)_{n\geq1}$ converges then we have that
$\displaystyle\qquad\liminf_{n\to\infty} s_n = \lim_{n\to\infty} s_n = \limsup_{n\to\infty} s_n$.

*There is a summable series $\sum_{n=1}^\infty a_n$ with non-negative $a_n$ such that $\limsup\limits_{n\to\infty} na_n > 0$.
Take these two together you get that the best you can hope to prove in this slightly more general situation is that $\liminf\limits_{n\to\infty} na_n = 0$, and that is indeed true.
For an example showing that 2 above is true, consider this: We start with the harmonic series and observe that
$\displaystyle\qquad \limsup_{n\to\infty} n\cdot \tfrac1n = 1$,
but it isn't summable. But we could replace infinitely many of the sequence elements by zeroes without changing the limit superior from being $1$. Define $S(n)$ to be $1$ when $n$ is a perfect square and $0$ otherwise. Then $a_n = \frac{S(n)}n$ has the desired property as
$\displaystyle\qquad \sum_{n=1}^\infty a_n = \sum_{n=1}^\infty \frac{S(n)}n = \sum_{k=1}^\infty \frac1{k^2} < \infty$.
