Show-that-n-1-n-n-2-3-e-

Question Number 7771 by FilupSmith last updated on 14/Sep/16

Show that :

\underset{n = 1}{\sum^{\infty}} \frac{n}{(n + 2)!} = 3 - e

Commented by FilupSmith last updated on 14/Sep/16

Amazing!

Commented by sou1618 last updated on 14/Sep/16

S = \underset{n = 1}{\sum^{\infty}} \frac{n}{(n + 2)!}

\frac{1}{(n + 1)!} - \frac{1}{(n + 2)!} = \frac{n + 1}{(n + 2)!}

S = \underset{n = 1}{\sum^{\infty}} (\frac{1}{(n + 1)!} - \frac{2}{(n + 2)!})

g (x) = e^{x}

g (x) = \underset{i = 0}{\sum^{\infty}} \frac{g^{(i)} (0)}{i!} x^{i}

g (1) = e = \underset{i = 0}{\sum^{\infty}} \frac{1}{i!}

S = \underset{n = 0}{\sum^{\infty}} \frac{1}{(n + 2)!} - \underset{n = 0}{\sum^{\infty}} \frac{2}{(n + 3)!}

S = {(\underset{n = 0}{\sum^{\infty}} \frac{1}{n!}) - (\frac{1}{0!} + \frac{1}{1!})} - 2 {(\underset{n = 0}{\sum^{\infty}} \frac{1}{n!}) - (\frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!})}

0! = 1

S = (e - 2) - 2 (e - \frac{5}{2})

S = 3 - e

Facebook Tweet Pin