Convergence of sequence and series

Question

Show that if $a_n$ is a convergent series of non-negative reals such that {$a_n$} is decreasing for all n, then the sequence $na_n$ converges to 0.

My attempt:

If series $a_n$ converges, then root test would imply that $\lim_{n\to\infty}a_n^{\frac{1}{n}}\leq1$.

Case 1:$\lim_{n\to\infty}a_n^{\frac{1}{n}}<1$

Applying root test on sequence $b_n=na_n$ would imply that $\lim_{n\to\infty}({na_n})^{\frac{1}{n}}<1$. Hence series $b_n$ converges. Therefore, $\lim_{n\to\infty}{na_n}=0$.

Case 2:$\lim_{n\to\infty}{a_n}^{\frac{1}{n}}=1$

I need help to proceed with this case.

Any help is appreciated.

Answer 1

hint

$\sum a_n $ converges $ \implies $

$$\lim S_n=\lim_{n\to\infty}(a_{n+1}+\cdots a_{2n})=0$$

but

$$0<na_{2n}<S_n$$ since $(a_n)$ is decreasing.

So $$\lim_{n\to+\infty}2na_{2n}=0$$

Do the same for $ (2n+1)a_{2n+1}$