According to GR9768, problem 37: $$\sum_{k=1}^{+\infty} \frac{k^2}{k!} = 2e$$ Can someone please explain how to get started in showing that?
Asked
Active
Viewed 169 times
5
-
oops.. apparently that site hosting the sample also has solutions hidden in it. – Nov 18 '16 at 19:14
-
See also: What's the value of $\sum\limits_{k=1}^{\infty}\frac{k^2}{k!}$?. Other questions linked there might be of interest, too. – Martin Sleziak Jul 21 '17 at 09:41
3 Answers
9
$$\sum \limits_{k = 1}^\infty \frac{k^2}{k!} = \sum \limits_{k = 1}^\infty \frac{k}{(k - 1)!} = \sum \limits_{k = 0}^\infty \frac{k + 1}{k!} = \sum \limits_{k = 1}^\infty \frac{1}{(k - 1)!} + \sum \limits_{k = 0}^\infty \frac{1}{k!} = 2 \sum \limits_{k = 0}^\infty \frac{1}{k!} = 2e$$
Dominik
- 20,241
-
I wonder...how efficiently can use this to solve for $$\sum_{n=0}^\infty\frac{n^k}{n!}$$ – Simply Beautiful Art Mar 01 '17 at 17:18
-
@SimplyBeautifulArt it cannot. Look at the format of your series, $n$ is the summatory variable, but here the only thing you have control over in your desired series is $k$, unlike in the series for $e^x$ where you control the base rather than the exponent. – Adam Hughes Apr 11 '17 at 14:49
-
@AdamHughes I suppose you can get a recursive relationship for each value of $k$ though. – Simply Beautiful Art Apr 11 '17 at 15:29
9
Write the series
$$e^{x}=\sum_{k=0}^\infty {x^k\over k!}$$
Now apply the derivative and multiply by $x$ to get
$$x{d\over dx}(e^x) = xe^x = \sum_{k=0}^\infty {kx^k\over k!}$$
Now do it again
$$x{d\over dx} (xe^x) = e^x + xe^x = \sum_{k=0}^\infty {k^2x^k\over k!}$$
Now let $x=1$
$$e^1+1\cdot e^1 =2e = \sum_{k=0}^\infty {k^2\over k!}$$
Adam Hughes
- 37,795
- 10
- 61
- 85
-
Perhaps you could see my more general answer, which takes the essence of your answer to arbitrary naturals. – Simply Beautiful Art Mar 01 '17 at 17:17
-
1@SimplyBeautifulArt I'm well aware of the method you use, but it was wildly inappropriate for the op's context (tagged as GRE question) so I opted for an approach that he would be most easily able to understand. Generalities are nice for oneself, but don't always help explain something to a student without the right background. – Adam Hughes Mar 01 '17 at 17:22
-
1
Consider the following:
$$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}$$
Let $x\mapsto e^x$:
$$e^{e^x}=\sum_{n=0}^\infty\frac{e^{nx}}{n!}$$
It thus follows that
$$\frac{d^k}{dx^k}e^{e^x}\bigg|_{x=0}=\sum_{n=0}^\infty\frac{n^k}{n!}$$
And by applying Faà di Bruno's formula, we find that
$$eB_k=\sum_{n=0}^\infty\frac{n^k}{n!}$$
Where $B_k$ is the $k$th Bell number.
This is famously known as Dobiński's formula.
Simply Beautiful Art
- 76,603