# If $(a_n)\subset[0,\infty)$ is non-increasing and $\sum a_n<\infty$, then $\lim{n a_n} = 0$

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

$$\forall \varepsilon > 0 \; \exists n_0 \text{ such that } |n a_n| < \varepsilon \; \forall n > n_0$$

or

$$\forall \varepsilon > 0 \; \exists n_0 \text{ such that } a_n < \frac{\varepsilon}{n} \; \forall n > n_0$$

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 $$n a_n \to 0.$$