# Convergence of $\sum (n - 1)a_n$ given convergence of $\sum a_n$

I haven't studied basic calculus/analysis for a while so I apologise if this is obvious. In the course of solving another problem I have found that a sum of the form $$\sum_{n=1}^{\infty}(n - 1)a_n$$ converges. I also know that the sum $\sum_{n=1}^{\infty}a_n$ converges, and also that each term $a_n$ is non-negative. Can I then conclude that $\sum_{n=1}^{\infty}na_n$ converges? I have a feeling this is true but I'm not really sure why.

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If $a_n \rightarrow a$ and $b_n \rightarrow b$, then $(a+b)_n \rightarrow a + b.$ Your partial sums can certainly be written in this form. –  mjqxxxx Oct 18 '12 at 20:46
This is however not exactly matching the title of the question. Knowing just that $\sum a_n$ converges does not allow to conclude that $\sum(n-1)a_n$ converges (e.g. if $a_n=\frac1{(n-1)^2}$ for $n>1$). –  Hagen von Eitzen Oct 18 '12 at 20:49

For $n>1$ you have $0\le a_n\le(n-1)a_n$, and $\sum_{n\ge 1}(n-1)a_n$ converges, so by the comparison test $\sum_{n\ge 1}a_n$ converges. The convergence is absolute, since these are non-negative series, so
$$\sum_{n\ge 1}na_n=\sum_{n\ge 1}(n-1)a_n+\sum_{n\ge 1}a_n$$