The problem is the summation of (n/(n+1)!) from n=1 to infinity.

any help that can be given is greatly appreciated!

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- Nov 1st 2010, 12:56 PMcaseysummation 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 PMchisigma
Is...

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

... so that...

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Nov 1st 2010, 01:24 PMcasey
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 PMmr fantastic