Infinite sum $\sum ^{\infty }_{n=1}\frac {n( n+1)^{2}}{(n-1)!}x^{n}$ I want to calculate the following.
$$\sum ^{\infty }_{n=1}\frac {n( n+1)^{2}}{(n-1)!}x^{n}=?$$
I knew.
$$\sum ^{\infty }_{n=1}\frac {x^{n}}{(n-1)!}=e^{x}x$$
By the way, answer is WolframAlpha!
Please tell me how to solve.
 A: Hint:
Write $$n(n+1)^2=(n-1)(n-2)(n-3)+a(n-1)(n-2)+b(n-1)\ \ \ \ (1)$$
so that $$\dfrac{n(n+1)^2}{(n-1)!}x^n=x^4\cdot\dfrac{x^{n-4}}{(n-4)!}+ax^3\cdot\dfrac{x^{n-3}}{(n-3)!}+bx^2\cdot\dfrac{x^{n-2}}{(n-2)!}$$
Now put $n=2$ and $n=1$ to find $a,b$
Finally $$\sum_{r=0}^\infty\dfrac{y^r}{r!}=e^y$$
See also: Evaluate the series $\lim\limits_{n \to \infty} \sum\limits_{i=1}^n \frac{n+2}{2(n-1)!}$
A: Multiply by powers of $x$ and differentiate iteratively
\begin{align*}
\sum ^{\infty }_{n=1}\frac {x^{n-1}}{(n-1)!}&=e^x\\
\frac{d}{dx}\sum ^{\infty }_{n=1}\frac {x^{n}}{(n-1)!}&=\frac{d}{dx}(xe^x)\\
\frac{d}{dx}\sum ^{\infty }_{n=1}\frac {nx^{n+1}}{(n-1)!}&=\frac{d}{dx}\left(x^2\frac{d}{dx}(xe^x)\right)\\
\frac{d}{dx}\sum ^{\infty }_{n=1}\frac {n(n+1)x^{n+1}}{(n-1)!}&=\frac{d}{dx}\left(x\frac{d}{dx}\left(x^2\frac{d}{dx}(xe^x)\right)\right)\\
\sum ^{\infty }_{n=1}\frac {n( n+1)^{2}}{(n-1)!}x^{n}&=\frac{d}{dx}\left(x\frac{d}{dx}\left(x^2\frac{d}{dx}(xe^x)\right)\right)
\end{align*}
Don't forget to check the limits of $n$ at each iteration, in that constant terms become $0$ after differentiation. No need to worry about this now for we are multiplying $x$(s) before differentiating once every time.
