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Let $S = \sum_{k=0} ^\infty \frac{1}{(k+2)k!}$. I am trying to evaluate this sum. I tried using the Taylor series of $e^x$, which is similar, but I am not sure how to deal with the $1/k+2$ factor. Any advice will be welcome.

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  • $\begingroup$ $\frac{1}{(k+2)k!} = \frac{k+1}{(k+2)!}$ $\endgroup$ – dEmigOd Feb 3 '19 at 16:32
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You have

$$xe^x = \sum_{k=0}^\infty \frac{x^{k+1}}{k!}$$

By integration $$\int_0^x te^t \ dt = \sum_{k=0}^\infty \frac{x^{k+2}}{(k+2)k!}$$

Plugging in $x=1$, you get

$$\sum_{k=0}^\infty \frac{1}{(k+2)k!} = \int_0^1 te^t \ dt=[te^t]_0^1- \int_0^1 e^t \ dt=1$$

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So we start with $\frac{1}{(k+2)k!} = \frac{k+1}{(k+2)!}$, then $\frac{k+1}{(k+2)!} = \frac{1}{(k+1)!} - \frac{1}{(k+2)!}$.

Then $$S = \sum\limits_{k=0}^{\infty} \frac{1}{(k+2)k!} = \sum\limits_{k=0}^{\infty} \left(\frac{1}{(k+1)!} - \frac{1}{(k+2)!}\right)$$

and this is telescopic.

Can you take it from here?

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  • $\begingroup$ Thanks. It helped, wish I could accept more than one answer $\endgroup$ – R. Davis Feb 3 '19 at 18:10
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Have a look at the partial sums:

$\sum_{k=0}^0 = \frac{1}{2} = \frac{2!-1}{2!}$

$\sum_{k=0}^1 = \frac{5}{6} = \frac{3!-1}{3!}$

$\sum_{k=0}^2 = \frac{23}{24} = \frac{4!-1}{4!}$

$\sum_{k=0}^3 = \frac{119}{120} = \frac{5!-1}{5!}$

Here, you can detect a pattern and conclude $\sum_{k=0}^n = \frac{(n+2)!-1}{(n+2)!} = 1 - \frac{1}{(n+2)!}$. Hence you can calculate $\lim\limits_{n \to \infty}\sum_{k=0}^n = \lim\limits_{n \to \infty} \left(1 - \frac{1}{(n+2)!}\right) = 1$.

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