Rate of convergence of summable sequence Suppose $a_n$ is a nonnegative real sequence such that 
\begin{equation}
\sum_n a_n <\infty.
\end{equation}
What do we know about $a_n$? We know $a_n\to 0$. We know $$\lim\inf n a_n = 0.$$ But can we say $$\lim n a_n = 0?$$ If not, why not? That is, can you construct a sequence that is summable but this is not the case?
 A: If $n$ is a perfect square, let $a_n=\frac{1}{n}$. If $n$ is not a perfect square, let $a_n=\frac{1}{2^n}$. 
The series $\sum a_n$ converges, but $na_n=1$ whenever $n$ is a perfect square. So it is not true that $\lim_{n\to\infty} na_n=0$.
The idea is that if we make $a_n=\frac{1}{n}$ (kind of big) on a sparse enough set, with the rest of the $a_n$ "small," we will have convergence. 
A: The answer is no.  More strongly the following is true.
Let $b_n$ be any sequence whose limit is 0.  Then there is a sequence $a_n$ which is summable, and is equal to $b_n$ for infinitely many $n$.
Proof.  Let $a_n = b_n$ the first time that $|b_n| < 1$, the first time after that that $|b_n| < \frac{1}{2}$, the first time after that that $|b_n| < \frac{1}{4}$ and so on.  At all other points let $a_n$ be 0.  By construction it is equal to $b_n$ infinitely often, yet the sum of the absolute value of its terms is bounded above by $1+\frac{1}{2} + \frac{1}{4} + ... = 2$.
Now apply this result to $b_n = \frac{1}{n}$.  Or, more pathologically, $\frac{1}{\sqrt{n}}$.  Or any other o(1) sequence that you want.
A: As the other anwsers have stated, this is not true in the general case.
However, if you suppose $(a_n)$ non increasing, then you have $\lim\limits_{n \to \infty} n \cdot a_n = 0 $.
To see why, consider $\sum\limits_{k=n+1}^{2n} a_k$.
In one hand, $\sum\limits_{k=n+1}^{2n} a_k \ge \sum\limits_{k=n+1}^{2n} a_{2n} = n \cdot a_{2n} $.
On the other hand, $\sum\limits_{k=n+1}^{2n} a_k = \sum\limits_{k=0}^{\infty} a_k - \sum\limits_{k=0}^{n} a_k- \sum\limits_{k=2n+1}^{\infty} a_k  \to   0 $ when $ n \to \infty$.
Therefore $\lim\limits_{n \to \infty} 2n \cdot a_{2n} = 0$.
And finally $(2n+1) \cdot a_{2n+1} \le (2n+1) \cdot a_{2n} =  2n \cdot a_{2n}  + a_{2n} \to   0 $ when $ n \to \infty$.
