If the sequence$\{na_n\}$ converges and $\sum_{n=1}^\infty n(a_n-a_{n-1})$ converges, prove $\sum_{n=1}^\infty a_n$ converges.

I've been having issues with general proofs of convergence such as this one, which I'm currently trying to work on. I find them really hard to begin.

For example, for the one in the title I have $\displaystyle\sum_{n=1}^\infty n(a_n-a_{n-1}) = \sum_{n=1}^\infty na_n - \sum_{n=1}^\infty na_{n-1}$. I think this may equal zero but I'm not even sure on that. Another idea I had is that because $n$ is increasing to infinity, this means $a_n$ must be decreasing otherwise $\{na_n\}$ would be divergent. Is that correct?

I'm having a lot of issues with these, so any push in the right direction would be greatly appreciated. Thanks guys.

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Are $a_n$ decreasing or change sign? – Alex Mar 4 '13 at 3:05
The sequence $na_n$ converges, if I read correctly. This does not mean the series $\sum na_n$ converges. So you can't separate the series $\sum n(a_n-a_{n-1})$. – 1015 Mar 4 '13 at 3:05
That's correct Julien, I fixed it to make that more clear. – MangoPirate Mar 4 '13 at 3:18
No $a_n$ needs not be decreasing. And the result holds without this assumption. – 1015 Mar 4 '13 at 3:21

Work on the partial sums: $$\sum_{1}^na_k=\sum_1^n(k+1-k)a_k=\sum_1^n(k+1)a_k-\sum_1^nka_k$$ $$=\sum_2^{n+1}ka_{k-1}-\sum_1^nka_k$$ $$=\sum_1^nk(a_{k-1}-a_k)+(n+1)a_n-a_0.$$ Now the partial sum on the left converges by assumption and $$(n+1)a_n=\frac{n+1}{n}na_n\longrightarrow \lim_{n\rightarrow+\infty} na_n$$ So the series converges to $$\sum_{k=1}^{+\infty}a_k=-\sum_{k=1}^{+\infty}k(a_k-a_{k-1})+\lim_{n\rightarrow +\infty}na_n-a_0.$$
The analogy with $\int f(t)dt$ which becomes, after integration by parts, $xf(x)-\int tf'(t)dt$. Here $x=n$, $f(x)=a_n$ and $f'(x)=a_n-a_{n-1}$. – 1015 Mar 4 '13 at 3:27