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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.

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  • $\begingroup$ think about the Eulerian number $\endgroup$ – Dr. Sonnhard Graubner May 17 '16 at 20:06
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    $\begingroup$ Everybody knows $$ e=\frac 1 {0!} +\frac 1 {1!} +\frac 1 {2!} +\frac 1 {3!} +\frac 1 {4!} +\frac 1 {5!} +\cdots. $$ So you're series is \begin{align} & 0\cdot e \\ {} + {} & 2 \left( e - \frac 1 {0!} \right) \\ {} + {} & 4 \left( e - \frac 1 {0!} - \frac 1 {1!} \right) \\ {} + {} & 6 \left( e - \frac 1 {0!} - \frac 1 {1!} - \frac 1 {2!} \right) \\ {} + {} & 8 \left( e - \frac 1 {0!} - \frac 1 {1!} - \frac 1 {2!} - \frac 1 {3!} \right) \\ {} + {} & \cdots \cdots \\ \vdots\ \ \end{align} It would be amusing if one could use the "fact" that $2+4+6+8+\cdots = \dfrac{-1}6\qquad$. $\endgroup$ – Michael Hardy May 17 '16 at 20:07
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    $\begingroup$ I see I wrote "you're" where "your" should appear. Doubtless if I run for office this will be brought up. $\qquad$ $\endgroup$ – Michael Hardy May 18 '16 at 15:55
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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, $f''(1) = 3e$.

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    $\begingroup$ Clear and concise. +1 $\endgroup$ – Mark Viola May 17 '16 at 20:11
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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 $3e$.

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  • $\begingroup$ short and clear , nice answer ... $\endgroup$ – user354674 Aug 15 '16 at 12:49
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$$xe^x = \sum\frac{x^{n+1}}{n!}\\ \frac d{dx} x e^x = e^x + x e^x = \sum \frac{(n+1)x^n}{n!}\\ \frac {d^2}{dx^2} x e^x =2 e^x + x e^x =\sum \frac{n(n+1)x^{n-1}}{n!}$$

and as $x$ approaches $1$

$$3 e=\sum \frac{n(n+1)}{n!}$$

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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 $$\sum_{n=1}^{\infty} \frac{n(n+1)}{n!} = 3e$$

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    $\begingroup$ Formatting tips here. $\endgroup$ – user228113 May 17 '16 at 23:51
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    $\begingroup$ Specifically, aligned equations $\endgroup$ – Teepeemm May 18 '16 at 0:23
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$$\begin{eqnarray*} \sum_{n\geq 1}\frac{n(n+1)}{n!} &=& \sum_{n\geq 1}\frac{1}{(n-1)!}+\sum_{n\geq 1}\frac{n-1}{(n-1)!}+\sum_{n\geq 1}\frac{1}{(n-1)!}\\&=&2\sum_{m\geq 0}\frac{1}{m!}+\sum_{n\geq 2}\frac{n-1}{(n-1)!}=\color{red}{3e}.\tag{1}\end{eqnarray*}$$


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 $1$.

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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 $A(1)$.

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