# summation problem

• Nov 1st 2010, 12:56 PM
casey
summation problem
The problem is the summation of (n/(n+1)!) from n=1 to infinity.
any help that can be given is greatly appreciated!
• Nov 1st 2010, 01:10 PM
chisigma
Is...

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

... so that...

$\displaystyle \sum_{n=1}^{\infty} \frac{n}{(n+1)!} = \sum_{n=1}^{\infty} \frac{1}{n!} - \sum_{n=1}^{\infty} \frac{1}{(n+1)!} = \sum_{n=1}^{\infty} \frac{1}{n!} - \sum_{n=2}^{\infty} \frac{1}{n!} = 1$ (2)

Kind regards

$\chi$ $\sigma$
• Nov 1st 2010, 01:24 PM
casey
sorry I have not done a summation in a long time but how do you know that the summation of 1/n! from one to infinity minus the summation of 1/n! from 2 to infinity is 1?
thanks so much for the help
• Nov 1st 2010, 01:26 PM
mr fantastic
Quote:

Originally Posted by casey
sorry I have not done a summation in a long time but how do you know that the summation of 1/n! from one to infinity minus the summation of 1/n! from 2 to infinity is 1?
thanks so much for the help

Write out the first few terms and simplify.