Jtdylktuy

I am trying to find what this value will converge to
$$sum_{n = 1}^{ infty}frac{n^2}{n!}$$

I tried using the Taylor series for $e^x$ but couldn’t figure out how to manipulate it to get the above expression, can someone help me out.

Edit: I have seen the solution, the manipulation required didn’t come to me, is there any resource that you guys can tell me about where I can find/practice more questions like this?

One may write
begin{align}
sum_{n = 1}^{ infty}frac{n^2}{n!}&=sum_{n = 1}^{ infty}frac{n(n-1)+n}{n!}
\\&=sum_{n = 1}^{ infty}frac{n(n-1)}{n!}+sum_{n = 1}^{ infty}frac{n}{n!}
\\&=sum_{n = 2}^{ infty}frac{1}{(n-2)!}+sum_{n = 1}^{ infty}frac{1}{(n-1)!}
end{align}Can you take it from here?

Just to give a slightly different approach,

$$sum_{n=1}^infty{n^2over n!}=sum_{n=1}^infty{nover(n-1)!}=sum_{m=0}^infty{m+1over m!}=sum_{m=0}^infty{mover m!}+e=sum_{m=1}^infty{mover m!}+e=sum_{m=1}^infty{1over(m-1)!}+e=sum_{k=0}^infty{1over k!}+e=e+e$$

The trick in reading this is to note what changes across each equal sign as you proceed from left to right and understand what justifies the equality for each change.

$$e^x-1=sum_{ngeq1}frac{x^n}{n!}$$
Taking $d/dx$ on both sides,
$$e^x=sum_{ngeq1}frac{n}{n!}x^{n-1}$$
Multiplying both sides by $x$,
$$xe^x=sum_{ngeq1}frac{n}{n!}x^n$$
Then taking $d/dx$ on both sides again,
$$(x+1)e^x=sum_{ngeq1}frac{n^2}{n!}x^{n-1}$$
Then plug in $x=1$:
$$sum_{ngeq1}frac{n^2}{n!}=2e$$

Edit

This is a really neat trick that is widely used. Whenever you see an $n^k$ in the numerator, think applying the $xfrac{d}{dx}$ operator $k$ times. Example:
$$e^x-1=sum_{ngeq1}frac{x^n}{n!}$$
Apply $xfrac{d}{dx}$:
$$xe^x=sum_{ngeq1}frac{x}{n!}x^n$$
Apply $xfrac{d}{dx}$:
$$x(x+1)e^x=sum_{ngeq1}frac{n^2}{n!}x^n$$
The pattern continues:
$$left(xfrac{d}{dx}right)^k[e^x-1]=sum_{ngeq1}frac{n^k}{n!}x^n$$

A similar thing can be done with integration. Example:

Evaluate $$S=sum_{ngeq0}frac{(-1)^n}{(2n+2)(2n+1)}$$
Start by recalling that (use geometric series)
$$frac1{1+t^2}=sum_{ngeq0}(-1)^nt^{2n}$$
Then integrate both sides from $0$ to $x$ to get
$$arctan x=sum_{ngeq0}frac{(-1)^n}{2n+1}x^{2n+1}$$
integrate both sides from $0$ to $1$ now to produce
$$S=sum_{ngeq0}frac{(-1)^n}{(2n+2)(2n+1)}=fracpi4-frac12log2$$