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I've been trying to figure out the following sum for a while now:

$$\sum_{k=1}^{\infty} \frac{1}{(k+1)(k-1)!}\left(1 - \frac{2}{k}\right)$$

I'm pretty sure that this doesn't evaluate to $0$.

As $k$ increases the term tends to $0$, but the first few terms add up to give a non-zero number.

I'm just having trouble figuring out how to find that number. Any help would be appreciated.

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  • $\begingroup$ You can write the interior of the sum as $\frac{k-2}{(k+1)!}$. Im not sure if this can help. $\endgroup$ – Masacroso Jun 3 '16 at 1:35
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Hint: Note that $1-\frac{2}{k}=\frac{(k+1)-3}{k}$. So our sum is $$\sum_1^\infty \frac{1}{k!} -3\sum_1^\infty \frac{1}{(k+1)!}.$$ Each sum is a fairly close relative of $e$.

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  • $\begingroup$ According to math.stackexchange.com/questions/1054175/…. Wouldn't it be e - 1? Since we are starting at 1 not 0. $\endgroup$ – The_Questioner Jun 3 '16 at 1:49
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    $\begingroup$ @The_Questioner: The first sum is $e-1$. The second is $e-2$. So our original series has sum $(e-1)-3(e-2)$. $\endgroup$ – André Nicolas Jun 3 '16 at 1:51
  • $\begingroup$ @The_Questioner I can confirm using a much less elegant method that 5-2e is the correct answer $\endgroup$ – Chill2Macht Jun 3 '16 at 1:57
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Let's try splitting up the series (since if it converges to a finite number, it will be absolutely convergent, since all terms are positive).

$$\sum_{k=1}^{\infty} \frac{1}{(k+1)(k-1)!}(1 - \frac{2}{k}) = \sum_{k=1}^{\infty} \frac{1}{(k+1)(k-1)!} - 2 \sum_{k=1}^{\infty} \frac{1}{(k+1)(k)(k-1)!} \\ = \sum_{k=1}^{\infty} \frac{k}{(k+1)(k)(k-1)!} - 2 \sum_{k=1}^{\infty} \frac{1}{(k+1)!} \\ = \sum_{k=1}^{\infty} \frac{k}{(k+1)!} - 2 \left[\sum_{k=0}^{\infty} \frac{1}{(k+2)!} + \frac{1}{0!} + \frac{1}{1!} -2 \right] = \\ \sum_{k=0}^{\infty} \frac{k+1}{(k+2)!} -2\left[ \sum_{k=0}^{\infty}\frac{1^k}{k!} -2 \right] = \sum_{k=0}^{\infty}\frac{k}{(k+2)!} + \sum_{k=0}^{\infty} \frac{1}{(k+2)!} - 2[e^1 -2]$$

Using the same technique as before:

$$ = \sum_{k=0}^{\infty}\frac{k}{(k+2)!} + [e^1 -2] - 2[e^1 -2] \\= \sum_{k=1}^{\infty} \frac{ k }{(k+2)!} -[e^1 -2] \\ = \sum_{k=1}^{\infty} \frac{ k }{(k+2)!} + \sum_{k=0}^{\infty} \frac{2}{(k+2)!} - \sum_{k=0}^{\infty} \frac{2}{(k+2)!} -[e^1 -2] \\ = \sum_{k=0}^{\infty} \frac{k+2}{(k+2)!} - 2 \sum_{k=0}^{\infty} \frac{1}{(k+2)!} - [e^1 -2]$$

Using a second time our computation that $\sum_{k=0}^{\infty} \frac{1}{(k+2)!} = e^1 -2$ we get $$= \sum_{k=0}^{\infty} \frac{k+2}{(k+2)!} - 3[e^1 -2] \\ = \sum_{k=0}^{\infty} \frac{1}{(k+1)!} - 3[e^1 -2] \\ = \sum_{k=0}^{\infty}\frac{1}{(k+1)!} + 1 -1 -3[e^1-2] \\ = \sum_{k=0}^{\infty} \frac{1}{k!} -1 - 3[e -2] \\ = e-1 -3e +6 \\ \\= -2e+5 $$

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    $\begingroup$ Who. Thanks for the work you put in. I understand your method $\endgroup$ – The_Questioner Jun 3 '16 at 2:10
  • $\begingroup$ sorry about the mess - I started before Andre Nicolas posted his answer, so I didn't see the more elegant solution before I was already 80% of the way through, at which point I figured I might as well finish. $\endgroup$ – Chill2Macht Jun 3 '16 at 2:13

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