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I am only looking for a hint on how to start this problem, not a full proof.

If $\sum_{k=1}^n ka_k=(n+1)/(n+2)$ for $n\in \mathbb{N}$, prove that $$\sum_{k=1}^\infty a_k=\frac{3}{4}$$ So far all that I can come up with is that $\sum_{k=1}^\infty ka_k=\lim_{n\rightarrow \infty}(n+1)/(n+2)=1$. I have no idea how to come to any conclusion about $a_k$ given this information. I have written out the terms of $\sum_{k=1}^n ka_k$ and could not draw any conclusion about $a_k$ from that.

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What you want to do is find an explicit formula for $a_k$ using the given condition. This is pretty easy to do by taking differences of consecutive partial sums.

It is clear that $na_n = S_n - S_{n-1},$ where $S_k$ denotes the $k^{th}$ partial sum, so $a_n = \frac{1}{n(n+1)(n+2)}.$ The classic way to sum is this is by noting $a_n = 1/2\left(\frac{1}{n(n+1)} - \frac{1}{(n+1)(n+2)}\right).$ You should realize though that $a_1$ does not follow this rule, so the sum will be slightly altered.

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  • $\begingroup$ So, if $s_n=\sum^n_{k=1}ka_k$, we want to look at $s_n-s_{n+1}$? $\endgroup$ – kaiserphellos Feb 14 '13 at 6:09
  • $\begingroup$ You should look at $s_n - s_{n-1}$ if you want $a_n.$ $\endgroup$ – cats Feb 14 '13 at 6:12

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