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