# Sum of infinite series $\sum_{n=1}^{\infty}\frac{1}{(n+1)!n}2^n$

I have a problem finding a infinite sum of this series:

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

It should be done using derivatives and integrals, like for example:

$$\sum_{n=1}^{\infty}\frac{t^n}{n}=\sum_{n=0}^{\infty}\frac{t^{n+1}}{n+1}=\sum_{n=0}^{\infty}\int_{0}^{t}s^nds=\int_{0}^{t}\sum_{n=0}^{\infty}s^nds=\int_{0}^{t}\frac{1}{1-s}ds=-\ln(1-t)$$

But of course if there's any other way of doing it, it's fine.

All I could think about is writing it as:

$$\sum_{n=1}^{\infty}\frac{2^n}{(n+1)\space n!\space n}$$ but then I'm stuck.

Thanks very much for any help!

$\displaystyle\sum_{n=1}^\infty \frac{t^n}{n(n+1)!} = \sum_{n=0}^\infty \frac{t^{n+1}}{(n+1)(n+2)!} = \sum_{n=0}^\infty \int_0^t \frac{s^n}{(n+2)!} ds= \int_0^t\sum_{n=0}^\infty \frac{s^n}{(n+2)!}ds = \int_0^t\frac{1}{s^2}\sum_{n=0}^\infty \frac{s^{n+2}}{(n+2)!} ds$
• Yes, thank you very much. It was actually simple, but what got me confused is that we have $2^n$ and not e.g. $x^n$, but now I realize that this $t$ is actually any number. Again, thanks so much! – mAlex Jun 24 '14 at 12:02