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Can someone teach me how to solve the following series. I have no idea how to deal with factorial.

$$\sum_2^\infty \frac{n-1}{n!}$$


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Hint: $\sum\limits_{n=0}^\infty \dfrac 1{n!} = \mathrm{e}$ – Graham Kemp Apr 22 '14 at 4:20
up vote 8 down vote accepted

You can compute this using a few tricks: First, since the series converges by the Ratio Test, you can split it up as the difference of two sums (each of these sums also will converge by the Ratio Test). Then, you can re-index one of the sums to make all of the terms in the second sum line up nicely with the terms in the first, so that they all cancel (this is the idea behind telescoping sums, but is able to be written a bit more explicitly since all series involved converge), and using all of these yields a very nice result.

\begin{align*} \sum_{n=2}^\infty \frac{n-1}{n!} &= \sum_{n=2}^\infty \frac{n}{n!} - \sum_{n=2}^\infty \frac{1}{n!}\\ &= \sum_{n=2}^\infty \frac{1}{(n-1)!} - \sum_{n=2}^\infty \frac{1}{n!}\\ &= \sum_{n=1}^\infty \frac{1}{n!} - \sum_{n=2}^\infty \frac{1}{n!}\\ &= \frac{1}{1!} = 1 \end{align*}

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I feel this is an overly complicated way to say it is telescopic...(and consider that the nth term goes to $0$) – chubakueno Apr 22 '14 at 5:13
This may be a complicated way to say it is telescoping, but the reason I did it this way was to demonstrate a few of the other techniques one does not often see used with telescoping series. – Nicholas Stull Apr 22 '14 at 12:23

$$\sum_2^\infty \frac{n-1}{n!}$$

$$=\sum_2^\infty \frac{1}{(n-1)!}-\frac{1}{n!}$$

As you plug in values of $n$, terms start cancelling out and the required summation is hence $=1$

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@NotNotLogical, by putting a '$=$' in the next line. Click the edit button and see how it works... I also discovered it accidentally. – Apurv Apr 22 '14 at 4:30
@NotNotLogical: It works because putting one or more dashes or equals signs on the next line is one of the ways to format a heading in Markdown. Prefixing the line with a # would work just as well. – Ilmari Karonen Apr 22 '14 at 5:54
$\verb={\Huge\sum}=$ yields ${\Huge\sum}$ $\verb={\huge\sum}=$ yields ${\huge\sum}$ $\verb={\LARGE\sum}=$ yields ${\LARGE\sum}$ $\verb={\Large\sum}=$ yields ${\Large\sum}$ $\verb={\large\sum}=$ yields ${\large\sum}$ $\verb={\small\sum}=$ yields ${\small\sum}$ $\verb={\tiny\sum}=$ yields ${\tiny\sum}$ – Felix Marin Apr 22 '14 at 6:18
Why did you try to get the LaTeX so large? – Did May 18 '14 at 10:21
@Did why do you think he tried? If you read the comments, you'll understand that he didn't. – Ruslan May 18 '14 at 13:02

Hint: $$\frac{n-1}{n!} = \frac{1}{(n-1)!} - \frac{1}{n!}$$

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