Divergent series. 
Prove that if the series $\sum a_n$ diverges then $\sum na_n$ diverges. 

The result is easy if $a_n\geq 0$ . But I don't know what to  do with an arbitrary $a_n$ (negatives and positives at time). I appreciate a suggestion.
Thanks!
 A: *

*$\sum n\cdot a_{n}$ converges $\implies \sum a_{n}$ converges.  This follows from what is called as the Abel's Test. Just take your $a_{n} =na_{n}$ and your $b_{n}=\left\{\frac{1}{n}\right\}$

A: Let $\displaystyle\sum_{n=1}^{\infty} c_n$ be a convergent series.  
Since $\displaystyle\sum_{n=1}^{\infty} c_n$ has bounded partial sums, $\displaystyle\sum_{n=1}^{\infty}\frac{c_n}{n}$ converges by Dirichlet's test.
A: Assume that $\sum\limits_nna_n$ converges and consider $$A_n=\sum\limits_{k\geqslant n}ka_k.$$ Then the series defining $A_n$ converges for every $n$, and $A_n\to0$ when $n\to\infty$, in particular $(A_n)$ is bounded. For every $n$, $$\sum\limits_{k=1}^na_k=\sum\limits_{k=1}^n\frac1k(A_k-A_{k+1})=\sum\limits_{k=1}^n\frac1kA_k-\sum\limits_{k=2}^{n+1}\frac1{k-1}A_k=A_1-\sum\limits_{k=2}^n\frac1{k(k-1)}A_k-\frac1nA_{n+1}.$$
The sum on the RHS converges absolutely when $n\to\infty$ since $(A_k)$ is bounded and the last term on the RHS converges to zero for the same reason, hence the series $\sum\limits_na_n$ converges and its sum is $$\sum\limits_{n\geqslant1}a_n=A_1-\sum\limits_{n\geqslant2}\frac1{n(n-1)}A_n.$$ This hands-on approach can be used to (re)discover Abel's test.
