How to prove the following problem $\displaystyle\sum_{n=1}^{\infty}na_{n}$ converges. Prove that:
$$
n\sum_{k=n}^{\infty}a_{k}\to0
$$
Wish someone can provide me with some clue about this problem.
 A: Nonnegativity is not required.
Suppose the series $\sum k a_k$ converges and
$$\lim_{n \to \infty} S_n = \lim_{n \to \infty} \sum_{k=1}^n k a_k  = S$$
Summing by parts we get
$$n \sum_{k=n}^m a_k = n \sum_{k=n}^m ka_k \frac{1}{k} = n \left[\frac{S_m}{m} - \frac{S_{n-1}}{n} + \sum_{k=n}^mS_k\left(\frac{1}{k} - \frac{1}{k+1} \right)\right]$$
The last sum on the RHS converges as $m \to \infty$ since the summand is $\mathcal{O}(k^{-2})$.  Taking the limit of both sides as $m \to \infty$ and noting that $S_m/m \to 0$ we get
$$n \sum_{k=n}^\infty a_k = - S_{n-1} + n\sum_{k=n}^\infty S_k\left(\frac{1}{k} - \frac{1}{k+1} \right)$$
For any $\epsilon> 0$ and all sufficiently large $n$ we have for $k \geqslant n$, 
$$S- \epsilon \leqslant  S_k \leqslant S + \epsilon,$$
and 
$$-S_{n-1} +n (S- \epsilon)\sum_{k=n}^\infty \left(\frac{1}{k} - \frac{1}{k+1} \right) \leqslant n\sum_{k=n}^\infty a_k \leqslant -S_{n-1} +n (S+ \epsilon)\sum_{k=n}^\infty \left(\frac{1}{k} - \frac{1}{k+1} \right) $$
Noting the telescoping sum converging to $1/n$, it follows that
$$-S_{n-1} + S - \epsilon\ \leqslant n\sum_{k=n}^\infty a_k \leqslant -S_{n-1} + S+ \epsilon,$$
whence,
$$-\epsilon \leqslant \liminf_{n \to \infty}\,\, n\sum_{k=n}^\infty a_k\leqslant \limsup_{n \to \infty}\,\,n\sum_{k=n}^\infty a_k \leqslant \epsilon$$
Since $\epsilon$ can be arbitrarily close to $0$ we must have
$$\lim_{n \to \infty} n \sum_{k=n}^\infty a_k = 0$$
