Find the sum of the series $\displaystyle \sum$$\displaystyle \frac{3k}{(k+1)!}$ from k=1 to infinity.

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- Nov 22nd 2009, 09:13 AMfriday616finding the sum of a series
Find the sum of the series $\displaystyle \sum$$\displaystyle \frac{3k}{(k+1)!}$ from k=1 to infinity.

- Nov 22nd 2009, 09:40 AMgalactus
Compare to the series for e at x=1 and maybe you can see something.

$\displaystyle \frac{k}{(k+1)!}=\frac{1}{k!}-\frac{1}{k!(k+1)}$ - Nov 22nd 2009, 09:41 AMchisigma
The constant 3 can be 'pulled out' from the summation so that is...

$\displaystyle 3\cdot \sum_{k=1}^{\infty} \frac {k}{(k+1)!} = 3\cdot \sum_{n=2}^{\infty} \frac{n-1}{n!}= 3\cdot \{\sum_{n=1}^{\infty} \frac{1}{n!} - \sum_{n=2}^{\infty} \frac{1}{n!}\} = 3$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$