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I need help with proving the partial sum formula for the following series with induction: $$ \sum^\infty_{n=1}\frac{n}{(n+1)!}$$ I found a partial sum formula for the series: $$S_n=1-\dfrac{1}{(n+1)!}$$ I am stuck on proving the partial sums with induction.

My attempt is:

For the base case:

$S_1=\dfrac{1}{2}$ Next assume true for k: $$S_k=\dfrac{k}{(k+1)!}$$ However, at this point I am not sure how to proceed with proving $k+1$, if someone could give me some pointers that would be awesome.

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Let $a_k = \frac{k}{(k+1)!}$ be the $k^{\mathrm{th}}$ term. Your hypothesis is that $S_n = 1-\frac{1}{(n+1)!}$ for all $n$. Given your inductive hypothesis, it is enough to show that $$ S_{k+1} = S_k + a_{k+1} = 1-\frac{1}{(k+1)!} + \frac{k+1}{(k+2)!} = 1-\frac{1}{(k+2)!}.$$ This is a calculation that I suspect you can perform.

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  • $\begingroup$ Makes sense, thanks for the fast reply! $\endgroup$ – modimagni Apr 15 '14 at 0:43
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    $\begingroup$ The hypothesis is that $$S_n = 1 - \frac{1}{(n+1)!}$$ for all $n \leq k$. If the OP had it for all $n$, they'd be done (i.e., that would mean the induction argument was already finished). $\endgroup$ – Nicholas Stull Apr 15 '14 at 0:44

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