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

\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!

Answer

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 $n a_n \to 0.$

Attribution
Source : Link , Question Author : dls , Answer Author : Ragib Zaman

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