Find the sum of the infinite series $\sum n(n+1)/n!$ How do find the sum of the series till infinity?
$$ \frac{2}{1!}+\frac{2+4}{2!}+\frac{2+4+6}{3!}+\frac{2+4+6+8}{4!}+\cdots$$
I know that it gets reduced to $$\sum\limits_{n=1}^∞ \frac{n(n+1)}{n!}$$
But I don't know how to proceed further.
 A: As you observed $$\sum_{n=1}^{\infty} \frac{n(n+1)}{n!}$$ further reduces to,
$$\begin{align}
&\sum_{n=1}^{\infty} \frac{n(n+1)}{n(n-1)!} \\
=&\sum_{n=1}^{\infty} \frac{n+1}{(n-1)!}\\
=&\sum_{n=1}^{\infty} \frac{n}{(n-1)!}+\frac{1}{(n-1)!}\\
=&\sum_{n=1}^{\infty} \frac{(n-1)}{(n-1)!} + \frac{2}{(n-1)!}\\
=&\sum_{n=1}^{\infty} \frac{1}{(n-2)!} +\sum_{n=1}^{\infty} \frac{2}{(n-1)!},
\end{align}$$  
Both summations equal $e$(euler's number) whose expansion is given by,
$$e=\sum_{n=1}^{\infty} \frac{1}{(n-1)!}.
$$ 
So $$\sum_{n=1}^{\infty} \frac{n(n+1)}{n!} = 3e$$
A: $$\begin{eqnarray*} \sum_{n\geq 1}\frac{n(n+1)}{n!} &=& \sum_{n\geq 1}\frac{1}{(n-1)!}+\sum_{n\geq 1}\frac{n-1}{(n-1)!}+\sum_{n\geq 1}\frac{1}{(n-1)!}\\&=&2\sum_{m\geq 0}\frac{1}{m!}+\sum_{n\geq 2}\frac{n-1}{(n-1)!}=\color{red}{3e}.\tag{1}\end{eqnarray*}$$

An alternative approach.
$$ \sum_{n\geq 1}\frac{n(n+1)}{n!} = \sum_{n\geq 0}\frac{n+2}{n!} = \left.\frac{d}{dx}\left(x^2 e^x\right)\right|_{x=1}=\left.(x^2+2x)\,e^x\right|_{x=1}=\color{red}{3e}.\tag{2} $$

Yet another way.
$$\begin{eqnarray*}e^{-1}\sum_{n\geq 0}\frac{n+2}{n!}=\sum_{m\geq 0}\frac{(-1)^m}{m!}\sum_{n\geq 0}\frac{n+2}{n!} &=& \sum_{a\geq 0} \sum_{n=0}^{a}\frac{(-1)^n (a-n+2)}{(a-n)!n!}\\&=&\sum_{a\geq 0}\frac{1}{a!}\sum_{n=0}^{a}\binom{a}{n}(-1)^n (a+2-n)\tag{3}\end{eqnarray*}$$
where the innermost sum is non-zero only for $a=0$ and $a=1$ by the theory of the forward difference operator: $(a+2-n)$ is a polynomial in $n$ with degree $1$.
A: You could also consider that $$A(x)=\sum\limits_{n=1}^∞ \frac{n(n+1)}{n!}x^n=\sum\limits_{n=0}^∞ \frac{n(n+1)}{n!}x^n=\sum\limits_{n=0}^∞ \frac{n(n-1)+2n}{n!}x^n$$ $$A(x)=\sum\limits_{n=0}^∞ \frac{n(n-1)}{n!}x^n+2\sum\limits_{n=0}^∞ \frac{n}{n!}x^n=x^2\sum\limits_{n=0}^∞ \frac{n(n-1)}{n!}x^{n-2}+2x\sum\limits_{n=0}^∞ \frac{n}{n!}x^{n-1}$$ $$A(x)=x^2 \left(\sum\limits_{n=0}^∞ \frac{x^n}{n!} \right)''+2x\left(\sum\limits_{n=0}^∞ \frac{x^n}{n!} \right)'=x^2e^x+2x e^x=x(x+2)e^x$$ Now, compute $A(1)$.
A: Define $f$ by $$f(x) = \sum_{n=0}^\infty \frac{x^{n+1}}{n!}$$ for $x\in\mathbb{R}$. (It is easy to check that the radius of convergence of this function is infinite.)
In particular:


*

*For all $x\in\mathbb{R}$, $f''(x) = \sum_{n=1}^\infty \frac{(n+1)n}{n!}x^{n-1}$, so you are looking for $f''(1)$;

*For all $x\in\mathbb{R}$, $f(x) = x e^x$ using the known power series for $\exp$, so that $f''(x) = (x+2)e^x$.
Therefore, $f''(1) = 3e$.
A: Note that for $n\ge 2$ we have
$$\frac{n(n+1)}{n!}= \frac{1}{(n-2)!} + \frac{2}{(n-1)!}. $$
This lets us rewrite the sum as
$$
\frac{2}{0!} + \left(\frac{1}{0!} + \frac{2}{1!} \right)
+ \left(\frac{1}{1!} + \frac{2}{2!} \right)
+ \left(\frac{1}{2!} + \frac{2}{3!} \right)
+ \left(\frac{1}{3!} + \frac{2}{4!} \right)
+ \cdots
$$
Rearranging in the obvious way gives the sum as $3e$.
A: $$xe^x = \sum\frac{x^{n+1}}{n!}\\
\frac d{dx} x e^x = e^x + x e^x = \sum \frac{(n+1)x^n}{n!}\\
\frac {d^2}{dx^2} x e^x =2 e^x + x e^x =\sum \frac{n(n+1)x^{n-1}}{n!}$$
and as $x$ approaches $1$
$$3 e=\sum \frac{n(n+1)}{n!}$$
