Find what the given sum is equal to I am given the following sum:
$$\sum_{n=0}^\infty{\frac{1}{2^{n+2}(n+2)}}$$
How do I find what is it equal to? It sort of looks like a logarithm, but not exactly, so how do I proceed to solve this problem?
 A: HINT:
Note that 
$$\int_0^x t^{n+1}\,dt=\frac{x^{n+2}}{n+2} \tag 1$$
Now, sum both sides over $n$ and let $x=1/2$.
SPOILER ALERT:  Scroll over the highlighted area to reveal the solution

For $|x|<1$, the series $\sum_{n=0}^\infty \frac{x^{n+2}}{n+2}$ converges.  So, for $|x|<1$, we sum both sides of $(1)$ to obtain $$\sum_{n=0}^\infty \frac{x^{n+2}}{n+2}=\sum_{n=0}^\infty \int_0^x t^{n+1}\,dt \tag 2$$For all $x_0<1$ and $|x|\le x_0$, the uniform convergence of the series $\sum_{n=0}^\infty t^n$ allows us to interchange the summation and the integration in $(2)$.  Proceeding, we find that $$\begin{align}\\\sum_{n=0}^\infty \frac{x^{n+2}}{n+2}&=\int_0^x \sum_{n=0}^\infty t^{n+1}\,dt\\\\&=\int_0^x \frac{t}{1-t}\,dt\\\\&=-x-\log(1-x)\tag 3\\\end{align}$$ Letting $x=1/2$ in $(3)$ yields the coveted result $$\sum_{n=0}^\infty \frac{1}{(n+2)2^{n+2}}=\log(2)-\frac12$$And we are done! 

A: If you know that
$$-\log(1-x)=\sum_{k=1}^\infty\frac{x^k}k$$
you can set $x=\frac12$ and deduce the first term. Then
$$\frac12+\sum_{k=2}^\infty\frac1{2^kk}=\log2.$$
