Calculating the sum of $\sum\frac{n^2-2}{n!}$

Question

I want to calculate the sum of $\sum_{n=0}^{+\infty}\frac{n^2-2}{n!}$.

This is what I have done so far:

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

And here I don't know how to deal with the $\frac{n}{(n-1)!} $. Any tips?

EDIT: One of the answers recommends to write down the sum as follows:

$$\sum_{n=0}^{+\infty}\frac{n^2-2}{n!}=\sum_{n=0}^{+\infty}\frac{n(n-1)}{n!} + \sum_{n=0}^{+\infty}\frac{n}{n!}-2\sum_{n=0}^{+\infty}\frac{1}{n!}$$

Which equals to:

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

But here I have negative factorials. What should I do next? Or can I just say that $\sum_{n=0}^{+\infty}\frac{1}{(n-1)!}=e$?

To avoid problems like $(-1)!$, it seems better to begin the sum at $n=1$:$$\sum_{n=0}^\infty\frac {n^2}{n!}=\sum_{n=1}^\infty\frac{n^2}{n!}$$ — ajotatxe, Dec 27 '16 at 13:49
Once you know what is $\sum\frac{1}{n!}$, this is essentially a duplicate of this question. — , Dec 27 '16 at 14:55

robjohn · Answer 1 · 2016-12-27T16:57:13.303

5

Hint: Write $$ \frac{n^2-2}{n!}=\frac{n(n-1)}{n!}+\frac{n}{n!}-\frac2{n!} $$

Note that by throwing out terms which are zero, $$ \sum_{n=0}^\infty\frac{n(n-1)}{n!}=\sum_{n=2}^\infty\frac{n(n-1)}{n!} $$ and $$ \sum_{n=0}^\infty\frac{n}{n!}=\sum_{n=1}^\infty\frac{n}{n!} $$

edited Dec 27 '16 at 16:57

answered Dec 27 '16 at 13:47

robjohn

353,833

You guys are something else, how am I supposed to compete with this!? XD Good answer, +1 – Simply Beautiful Art Dec 27 '16 at 13:48
@robjohn I edited the answer, could you take a look? – user401855 Dec 27 '16 at 14:05
1

@user401855: I have added a couple of "notes". – robjohn Dec 27 '16 at 14:16
So basically the sum is $0$? – user401855 Dec 27 '16 at 14:57
1

Yes, that's correct. – robjohn Dec 27 '16 at 16:48

score 2 · Accepted Answer · answered Dec 27 '16 at 17:11

I thought it might be instructive to present a way forward that can be applied to a wide class of problems.

The Taylor series representation of he exponential function is given by

$$\bbox[5px,border:2px solid #C0A000]{e^x=\sum_{n=0}^\infty \frac{x^n}{n!}} \tag 1$$

Differentiating $(1)$ term-by-term, we see that

$$\frac{d\,e^x}{dx}=e^x=\sum_{n=0}^\infty \frac{n x^{n-1}}{n!} \tag2$$

whereby multiplying $(2)$ by $x$ reveals

$$xe^x =\sum_{n=0}^\infty \frac{nx^n}{n!} \tag 3$$

Differentiating $(3)$, multiplying by $x$, and subtracting $2e^x$, we obtain

$$(x^2+x-2)e^x=\sum_{n=0}^\infty \frac{(n^2-2)\,x^{n}}{n!} \tag 4$$

Finally, setting $x=1$ in $(4)$ yields

$$\bbox[5px,border:2px solid #C0A000]{\sum_{n=0}\frac{n^2-2}{n!}=0}$$

and we are done!

score 1 · Answer 3 · edited Apr 13 '17 at 12:20

1

As per this post, we have

$$\sum_{n=0}^\infty\frac{n^p}{n!}=B_pe$$

where $B_n$ is the $n$th Bell number.

edited Apr 13 '17 at 12:20

Community

1

answered Dec 27 '16 at 13:51

Simply Beautiful Art

76,603

1

A link to Stirling Numbers of the Second Kind might be helpful, too. – robjohn Dec 27 '16 at 13:59
I just realized that I had posted this answer, which is closely related to your answer. – robjohn Dec 28 '16 at 07:50
@robjohn Nice find man – Simply Beautiful Art Dec 28 '16 at 12:48

score 0 · Answer 4 · answered Dec 27 '16 at 13:50

0

Adjust indices so that you have $\frac {n+1}{n!}$ and separate that out into $\frac n{n!}+\frac1{n!}$.

answered Dec 27 '16 at 13:50

AlgorithmsX

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Calculating the sum of $\sum\frac{n^2-2}{n!}$

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