mjhjmtu

I am trying to find what this value will converge to
$$sum_n = 1^ inftyfracn^2n!$$

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
beginalign
sum_n = 1^ inftyfracn^2n!&=sum_n = 1^ inftyfracn(n-1)+nn!
\\&=sum_n = 1^ inftyfracn(n-1)n!+sum_n = 1^ inftyfracnn!
\\&=sum_n = 2^ inftyfrac1(n-2)!+sum_n = 1^ inftyfrac1(n-1)!
endalignCan you take it from here?

Just to give a slightly different approach,

$$sum_n=1^inftyn^2over n!=sum_n=1^inftynover(n-1)!=sum_m=0^inftym+1over m!=sum_m=0^inftymover m!+e=sum_m=1^inftymover m!+e=sum_m=1^infty1over(m-1)!+e=sum_k=0^infty1over k!+e=e+e$$

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.

$$e^x-1=sum_ngeq1fracx^nn!$$
Taking $d/dx$ on both sides,
$$e^x=sum_ngeq1fracnn!x^n-1$$
Multiplying both sides by $x$,
$$xe^x=sum_ngeq1fracnn!x^n$$
Then taking $d/dx$ on both sides again,
$$(x+1)e^x=sum_ngeq1fracn^2n!x^n-1$$
Then plug in $x=1$:
$$sum_ngeq1fracn^2n!=2e$$

Edit

This is a really neat trick that is widely used. Whenever you see an $n^k$ in the numerator, think applying the $xfracddx$ operator $k$ times. Example:
$$e^x-1=sum_ngeq1fracx^nn!$$
Apply $xfracddx$:
$$xe^x=sum_ngeq1fracxn!x^n$$
Apply $xfracddx$:
$$x(x+1)e^x=sum_ngeq1fracn^2n!x^n$$
The pattern continues:
$$left(xfracddxright)^k[e^x-1]=sum_ngeq1fracn^kn!x^n$$

A similar thing can be done with integration. Example:

Evaluate $$S=sum_ngeq0frac(-1)^n(2n+2)(2n+1)$$
Start by recalling that (use geometric series)
$$frac11+t^2=sum_ngeq0(-1)^nt^2n$$
Then integrate both sides from $0$ to $x$ to get
$$arctan x=sum_ngeq0frac(-1)^n2n+1x^2n+1$$
integrate both sides from $0$ to $1$ now to produce
$$S=sum_ngeq0frac(-1)^n(2n+2)(2n+1)=fracpi4-frac12log2$$