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I've been trying to understand the result of this sum: $$\sum_{n=0}^\infty\frac1{n!(n+2)}=\frac12+\frac13+\frac18+\frac1{30}+\frac1{144}+\dots=1$$ Could you show me how to obtain 1 as result?

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    $\begingroup$ It's a telescoping series. Try computing and simplifying the partial sums into a fraction; you might be able to spot the pattern. $\endgroup$ – Theo Bendit Jul 17 '18 at 16:41
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HINT:

Note that we can write

$$\frac{1}{n!(n+2)}=\frac{(n+2)-1}{(n+2)!}=\frac{1}{(n+1)!}-\frac{1}{(n+2)!}$$

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Hint Let $f(x)=\sum_{n=0}^\infty \frac{x^n}{n!}=\exp(x)$. Note that $$ \sum_{n=0}^\infty\frac{1}{n!(n+2)}=\sum_{n=0}^\infty\frac{1}{n!}\int_{0}^1x^{n+1}\, dx=\int_{0}^1\sum _{n=0}^\infty\frac{x^{n+1}}{n!}\, dx=\int_{0}^1x\exp(x)\, dx. $$ Assumes knowledge of calculus.

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  • $\begingroup$ Can you explain how you obtained the integral form? $\endgroup$ – Szeto Jul 17 '18 at 16:53
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    $\begingroup$ @Szeto notice that $\int_0^1 x^{n+1} = \frac{1}{n+2} x^{n+2} \rvert_0^1 = \frac{1}{n+2}$ $\endgroup$ – WorldSEnder Jul 17 '18 at 20:23

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