A convergence problem in real non-negative sequence, $\sum_{n=1}^\infty a_n$. We now have $a_n\geq 0$, $\forall n=1,2,...,$ and $\sum_{n=1}^\infty a_n <\infty$. Then I guess  that $\lim_{n\to\infty} a_n \cdot n = 0$. But I realized that it is wrong. Since if we let $a_n = 1/n $ if $n = 2^i$ for some $i=1,2,...$ and $a_n = 0$ for the rest of the $n$.
Then we have that $\sum_{n=1}^\infty a_n = 1/2 + 1/4 + 1/8 + \cdots < \infty$, and $a_n\cdot n$ does not converges to $0$. 
Now I add another condition that $a_n$ is non-increasing. Does this result hold this time. i.e. the formal question is as follows:

$a_n\geq 0$, $\forall n=1,2,...,$ and $a_n$ is non-increasing, and  $\sum_{n=1}^\infty a_n <\infty$. Then prove $a_n \cdot n \to 0$, or give a counterexample that $a_n\cdot n$ does not necessarily converge to $0$ 


Define $ b_n:= n$
$\frac{1}{b_n} \sum_{i = 1}^n b_i a_i  = \frac{1}{n} \sum_{i = 1}^n i a_i \geq^{(1)} \frac{1}{n} \sum_{i = 1}^n i a_n \\\
=  a_n \frac{1}{n} \sum_{i = 1}^n i  =   a_n \frac{1}{n} \frac{(n+1)n}2 =  a_n \frac{1}{n} \frac{(n+1)n}2 = a_n \frac{n+1}{2}  \geq 0$
By Kronecker's lemma, $\lim_{n\to \infty}\frac{1}{b_n} \sum_{i = 1}^n b_i a_i = 0$
By squeezee theorem, $ 0 = \lim_{n\to \infty} \frac{1}{b_n} \sum_{i = 1}^n b_i a_i \geq \lim_{n\to \infty} a_n \frac{n+1}{2} \geq 0$
Hence, $\lim_{n\to \infty} a_n (n+1) = 0$
By the convegence of summation of $a_n$, $\lim_{n \to \infty} -a_n = 0$.
Therefore, $\lim_{n\to \infty} a_n n = \lim_{n\to \infty} a_n (n+1) - a_n = 0 $
Where the (1) comes from the decreasing of the sequence.
