Infinite sum

[man111]

Finding sum of [tex]\displaystyle \sum^{\infty}_{n=1}\frac{(n^2+1)}{(n+2)n!}[/tex]

[Davids]

Let's denote the sum $S$. We have:
$S = \sum_{n=1}^\infty \frac{1}{n!}\biggl[\frac{n^2+1}{n+2}\biggr] = \sum_{n=1}^\infty \frac{1}{n!}\biggl[n - 2 + \frac{5}{n+2}\biggr] =$
$= \underbrace{\sum_{n=1}^\infty \frac{n}{n!}} _{S_1} - 2\underbrace{\sum_{n=1}^\infty \frac{1}{n!}}_{S_2} + 5\underbrace{\sum_{n=1}^\infty \frac{1}{(n+2)n!}}_{S_3}$

We consider:

a) $S_1 = \sum_{n=1}^\infty \frac{n}{n!} = \sum_{n=1}^\infty \frac{1}{(n-1)!} = \sum_{n=0}^\infty \frac{1}{n!} = e$

b) $S_2 = \sum_{n=1}^\infty \frac{1}{n!} = -1 + \sum_{n=0}^\infty \frac{1}{n!} = e - 1$

c) $S_3 = \sum_{n=1}^\infty \frac{1}{(n+2)n!} = \sum_{n=1}^\infty \frac{n + 1}{(n+2)!} =$
$= \sum_{n=3}^\infty \frac{n - 1}{n!} = \sum_{n=3}^\infty \frac{1}{(n-1)!} - \sum_{n=3}^\infty \frac{1}{n!} =$
$= \sum_{n=2}^\infty \frac{1}{n!} - \sum_{n=3}^\infty \frac{1}{n!} = \frac{1}{2}$ by telescoping.

So finally $S = S_1 - 2S_2 + 5S_3 = e - 2(e-1) + \frac{5}{2} = \frac{9}{2} - e$

[vezni]

$S=\sum_{n=1}^{\infty}\frac{n^2+1}{(n+2)n!}=\sum_{n=3}^{\infty}\frac{(n-2)^2+1}{n(n-2)!}=\sum_{n=3}^{\infty}\frac{((n-2)^2+1)(n-1)}{n!}=$
[tex]=\sum_{n=3}^{\infty}\frac{n^3-5n^2+9n-5}{n!}=-5\sum_{n=3}^{\infty}\frac{1}{n!}+9\sum_{n=3}^{\infty}\frac{n}{n!}-5\sum_{n=3}^{\infty}\frac{n^2}{n!}+\sum_{n=3}^{\infty}\frac{n^3}{n!}=
-5S_0+9S_1-5S_2+S_3[/tex].
$\sum_{n=0}^{\infty}\frac{n^k}{n!}=f_k(1)$, където $f_k(x)=\left(x\frac{d}{dx}\right)^{k}e^x$ (тоест прилагаме $k$-пъти оператора $x\frac{d}{dx}$ към $e^x$). Това следва от [tex]e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}[/tex].
$f_0=e^x,f_1=xe^x,f_2=(x^2+x),f_3=(x^3+3x^2+x)e^x$
Оттук се намира $S_0=e-\frac 52,S_1=e-2,S_2=2e-3,S_3=5e-5$ и следователно $S=\frac 92-e$.