Thread: sum of the infinite series!

1. sum of the infinite series!

We have to find the value of the sum: -
[ (n)/(n+1)! ] when n goes from 1 <= n <= infinity...

I know that converges, and that approaches to some approximate value, but not able to figure out the exact value of it...I think that converges to .99999 or 1.000...please help!

Thanks!

2. Originally Posted by vikramtiwari
We have to find the value of the sum: -
[ (n)/(n+1)! ] when n goes from 1 <= n <= infinity...

I know that converges, and that approaches to some approximate value, but not able to figure out the exact value of it...I think that converges to .99999 or 1.000...please help!

Thanks!

Method 1: (Telescoping)

$
\sum_{1}^{\infty} \frac{n}{(n+1)!} = \sum_{1}^{\infty} \left(\frac{n+1 - 1}{(n+1)!}\right) = \sum_{1}^{\infty} \left(\frac{1}{n!} - \frac{1}{(n+1)!}\right)$

You should be able to continue from here

Method 2: (Power series)
If you know e^x power series,

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

Differentiate on both sides and then set x=1.

$\left(\sum_{0}^{\infty} \frac{x^{n}}{(n+1)!}\right)' \bigg{|}_{x=1}= \sum_{1}^{\infty} \frac{nx^{n-1}}{(n+1)!}\bigg{|}_{x=1} = \sum_{1}^{\infty} \frac{n}{(n+1)!}$
$\left(\frac{e^x - 1}{x}\right)'\bigg{|}_{x=1} = \frac{xe^x - (e^x - 1)}{x^2}\bigg{|}_{x=1} = \frac{e - e + 1}1 = 1$
So $\sum_{1}^{\infty} \frac{n}{(n+1)!} = 1$ Q.E.D