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$$\sum_{k=1}^n \frac{1}{k!(k+1)}$$ I tried for this, $$\sum_{k=1}^n \frac{k+1}{(k+2)!}$$ I don't know what to do anymore.

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    $\begingroup$ Do you realize that $(k+1)k!=(k+1)!$, and $\frac{k+1}{(k+2)!}=\frac{1}{(k+2)k!}$? $\endgroup$
    – Mark Viola
    Apr 21, 2018 at 2:18
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    $\begingroup$ Do you mean $\sum_{k=1}^n \frac{1}{k!(k+2)}$? Cause that, and not the headline, simplifies into the second thing you wrote. (And has a nice closed form as Durgesh's answer shows.) $\endgroup$ Apr 21, 2018 at 2:27

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$$\sum^{n}_{k=1}\frac{(k+2)-1}{(k+2)!}=\sum^{n}_{k=1}\bigg[\frac{1}{(k+1)!}-\frac{1}{(k+2)!}\bigg]$$

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If you've taken calculus, you'd know that $\sum_{k=0}^∞ {x^k \over k!} = e^x$. So, summing $\frac1{k!}\to∞ = e$. Problem is, this includes 1+1... before the actual equation you have. So...

Final answer: $e-2$

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  • $\begingroup$ They appear to be asking for a finite sum $\sum_{k=1}^n$ $\endgroup$ Apr 21, 2018 at 3:24

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