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Let $\sum a_n$ be a convergent series of positive real numbers with sum $s$ and partial sums $s_n=a_1+a_2+\cdots+a_n$. Prove that $\sum na_n$ is convergent if and only if $\sum (s-s_n)$ is convergent.

I have been trying to work this out for a while. I have concluded that since all $a_n$ are positive real numbers and $s_n$ is a partial sum of $s$, that $\sum (s-s_n)$ is always convergent. However, I do not know how to link this to $\sum na_n$. All help is much appreciated!!

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up vote 1 down vote accepted

Hint: a great part of the problem is done once we justify the computation $$\sum_{n=1}^\infty \sum_{k\geqslant n}a_k=\sum_{k=1}^{\infty}\sum_{n=1}^{k}a_k=\sum_{k=1}^{+\infty}ka_k.$$

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That's more than a great part of the problem. That's it. Note your $n$ runs from $1$ to $k$, so you get $k$ and not $k-1$. Pardon me if I'm wrong. – 1015 Mar 25 '13 at 21:47
Nicely done, by the way, +1. I had embarked an Abel transformation approach... – 1015 Mar 25 '13 at 21:52
thank you so much to both of you for this. However, this is a 15mark question and I find it hard to believe the working out is so simple! Do you know how I could take it further? I think the key is in the "if and only if". – ElenaC Mar 25 '13 at 22:19
To justify the computations, first work with a finite summation. – Davide Giraudo Mar 26 '13 at 12:51
@ElenaC It is simple because there is Fubini-Tonelli for series with nonnegative coefficients underneath. If you are not allowed to use this, the proof becomes more tedious as you need to prove both directions, take partial sums, bound them above, etc...If you do all that, you deserve your 15 points. – 1015 Mar 26 '13 at 13:25

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