$ \sum_{n=1}^{\infty} \frac{n^2}{ n!} $ equals
I'm not able to convert in any standard series? Any hints?
Asked
Active
Viewed 152 times
5
-
1Related : http://math.stackexchange.com/questions/576976/evaluate-the-series-lim-limits-n-to-infty-sum-limits-i-1n-fracn22 – lab bhattacharjee Mar 17 '15 at 06:53
-
@kingW3 Thanks for sharing that. But before you flag it -- How am I supposed to know if the question is asked? When I ask my question there it didnt show any similar of such this type? So unless the search of math.stackexchange is improved no way this can be stopped. – N S Mar 17 '15 at 07:52
-
2@NS: asking duplicate questions is not a crime, especially since the search function on Math.SE leaves much to be desired. On the other hand, it helps a lot for the site's overall organization to point out such duplicates when one is aware of them (which is what kingW3 did). – Willie Wong Mar 17 '15 at 08:25
-
@WillieWong I apologise if was being rude. But speaking from past experiences when I was very new to this amazing platform a few users were very hostile to me. So I apologise. Its really great to learn here. – N S Mar 17 '15 at 08:29
-
@NS As Willie said it's not something that is bad,once I voted the question I asked as a duplicate,anyway I up-voted your question since it's a nice question. – kingW3 Mar 17 '15 at 11:20
-
But thanks for marking it as duplicate. It'll be deleted within a month. slow claps I dont think this question is at all redundant. The discussions done here is totally new and enlightening. But hey who cares!! This has to be deleted as ITS a DUPLICATE!!! – N S Mar 17 '15 at 14:42
4 Answers
3
You may write, for $n =2,3,4,...$: $$ \frac{n^2}{ n!}=\frac{n^2-n+n}{ n!}=\frac{n(n-1)}{ n!}+\frac{n}{ n!}=\frac{1}{ (n-2)!}+\frac{1}{ (n-1)!} $$ and use a change of indices in the new infinite sums.
Olivier Oloa
- 120,989
2
HINT: First do the obvious cancellation:
$$\sum_{n\ge 1}\frac{n^2}{n!}=\sum_{n\ge 1}\frac{n}{(n-1)!}\;.$$
Now consider performing some familiar operations on the well-known power series expansion of $e^x$.
Added: Multiply
$$e^x=\sum_{n\ge 0}\frac{x^n}{n!}$$
by $x$ and differentiate:
$$\frac{d}{dx}\left(xe^x\right)=\frac{d}{dx}\left(\sum_{n\ge 0}\frac{x^{n+1}}{n!}\right)=\sum_{n\ge 0}\frac{(n+1)x^n}{n!}=\sum_{n\ge 1}\frac{nx^{n-1}}{(n-1)!}\;.$$
Brian M. Scott
- 616,228
-
And which familiar power series expansion is that? I'm looking but not getting any of this type. – N S Mar 17 '15 at 06:52
-
-
By writing the numerator as n-1+1 it'll ultimately get converted into the above answer by Olivier form. Is this what you meant? – N S Mar 17 '15 at 06:54
-
@NS: No. What happens if you multiply the power series that you know by $x$? – Brian M. Scott Mar 17 '15 at 06:55
-
-
-
@NS: Sorry, that comment got sent prematurely: it was supposed to say ‘multiply [it] by $x$ and differentiate’. – Brian M. Scott Mar 17 '15 at 06:58
-
Sorry for asking if its obvious but how you can get 2e from the series after differentiation? – N S Mar 17 '15 at 07:23
-
@NS: The derivative of $xe^x$ is $(x+1)x^x$; now substitute $x=1$. – Brian M. Scott Mar 17 '15 at 07:26
0
You can also use the fact that $n^2=n(n-1)+n$ and consider $$S=\sum_{n=1}^{\infty} \frac{n^2}{ n!}x^n=\sum_{n=0}^{\infty} \frac{n^2}{ n!}x^n=\sum_{n=0}^{\infty} \frac{n(n-1)}{ n!}x^n+\sum_{n=0}^{\infty} \frac{n}{ n!}x^n$$ $$S=x^2\sum_{n=0}^{\infty} \frac{n(n-1)}{ n!}x^{n-2}+x\sum_{n=0}^{\infty} \frac{n}{ n!}x^{n-1}$$ and recognize that the first summation is the second derivative of $e^x$ and that the second summation is the first derivative of $e^x$ and you know them. Then, at the end, replace $x$ by $1$.
Claude Leibovici
- 260,315
0
In general, $~\displaystyle\sum_{n=1}^\infty\frac{n^k}{n!}~=~B_k\cdot e,~$ where $B_k$ is the $k^{th}$ Bell number, and the identity in question
is called Dobinski's formula. They are all integers for integer values of k, and the sequence they
form can be found on OEIS.
Lucian
- 48,334
- 2
- 83
- 154