Find partial sum formula for $\frac{n^3}{n!}$ I need to find the partial sum formula for $\sum\limits_{n=0}^\infty \frac{n^3}{n!}$
I started by calculating some elements of the formula, but I could not find any possible patterns. 
Could you please help me in finding the partial sum formula? I need it in order to be able to calculate the sum of the series. If there is any easier way to do this than finding a partial sum formula, please let me know.
Thank you in advance
 A: We have that
$$\sum_{n=0}^\infty \frac{n^3}{n!}=\sum_{n=1}^\infty \frac{n^2}{(n-1)!}=
\sum_{n=1}^\infty \frac{n^2-n+n-1}{(n-1)!}+\sum_{n=1}^\infty \frac{1}{(n-1)!}=$$
$$=\sum_{n=2}^\infty \frac{n-2+2}{(n-2)!}+\sum_{n=2}^\infty \frac{1}{(n-2)!}+\sum_{n=1}^\infty \frac{1}{(n-1)!}=$$
$$=\sum_{n=3}^\infty \frac{1}{(n-3)!}+3\sum_{n=2}^\infty \frac{1}{(n-2)!}+\sum_{n=1}^\infty \frac{1}{(n-1)!}=5e$$
A: Start with $$xe^{x} = \sum_{n=0}^{\infty} \frac{x^{n+1}}{n!}$$
Differentiate, multiple by $x$ differentiate again and set $x = 1$.
I believe the answer comes out to $5e$.
A: Just another way to look at it.
Considering
$$\sum_{n=0}^\infty \frac {n^3} {n!} x^n$$ write first
$$n^3=n(n-1)(n-2)+3n(n-1)+n$$
$$\sum_{n=0}^\infty \frac {n^3} {n!} x^n=x^3\sum_{n=0}^\infty \frac {n(n-1)(n-2)} {n!} x^{n-3}+3x^2\sum_{n=0}^\infty \frac {n(n-1)} {n!} x^{n-2}+x\sum_{n=0}^\infty \frac {n} {n!} x^{n-1}$$ that is to say
$$\sum_{n=0}^\infty \frac {n^3} {n!} x^n=x^3\left(\sum_{n=0}^\infty \frac {x^n} {n!} \right)'''+3x^2\left(\sum_{n=0}^\infty \frac {x^n} {n!} \right)''+x\left(\sum_{n=0}^\infty \frac {x^n} {n!} \right)'=(x^3+3x^2+x)\,e^x$$
Now, make $x=1$ to get the result.
