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 $0$ 


It is true that $a_n \cdot n \rightarrow 0$ under this condition.
To prove this, observe that if the series is convergent, the sequence of partial sums is a Cauchy sequence.  In particular, for every $\varepsilon > 0$, there is some $N_0 = N_0(\varepsilon)$ such that whenever $n > m > N_0$, $\left| \sum_{i=1}^n a_i - \sum_{i=1}^m a_i \right| < \varepsilon$.
Now consider taking $m = N_0 + 1$ and $n \ge 2m$.  Then we have
$$ \sum_{i=1}^n a_i - \sum_{i=1}^m a_i = \sum_{i=m+1}^n a_i = a_n \cdot (n-m) + \sum_{i=m+1}^n (a_i - a_n) \ge a_n \cdot (n-m),$$
by the non-increasing property of the sequence $(a_n)_n$.
Hence $a_n \cdot (n-m) < \varepsilon$ whenever $n \ge 2(N_0 + 1)$.  However, since $n \ge 2m$, $n-m \ge \frac12 n$, and so $a_n \cdot n \le 2 a_n \cdot (n-m) < 2 \varepsilon$.
Since $\varepsilon$ can be taken to be arbitrarily small, this, coupled with the non-negativity of $a_n$, proves $a_n \cdot n \rightarrow 0$.
