sequences and series - Show that $sum_{n=0}^{infty} dfrac{n}{2^{n+1}} = 1$

My Work

I felt the best way to go about this problem was to compare it to a well known MacLaurin series. I noticed it resembled the reciprocal of the absolute value of the MacLaurin series of $\ln(1+x)$ where $x = \dfrac12$ but I had the problem of the $n$ in the numerator still. None of the well known series have an $n$ in the numerator.

The Well Known MacLaurin Series

My Question

Can someone give me a hint as to which well known MacLaurin series this resembles? Once I have that I know the solution!

Answer

If you want to directly proceed via MacLaurin series, try the MacLaurin series of $\dfrac{x^2}{(1-x)^2}$, which unfortunately is not there in your list. Else, proceed as follows:
Let $S_m = \displaystyle \sum_{n=1}^m \dfrac{n}{2^{n+1}}$.
\begin{align}

S_m & = \dfrac14 + \dfrac28 + \dfrac3{16} + \dfrac4{32} + \cdots + \dfrac{m-1}{2^m} + \dfrac{m}{2^{m+1}} &\spadesuit\\
\dfrac{S_m}2 & = \,\,\,\,\,\,\,\,\,\,\,\dfrac18 + \dfrac2{16} + \dfrac3{32} + \cdots + \dfrac{m-2}{2^m} + \dfrac{m-1}{2^{m+1}} + \dfrac{m}{2^{m+2}} & \diamondsuit
\end{align}
$\spadesuit-\diamondsuit$ now gives us
$$\dfrac{S_m}2 = \dfrac14 + \dfrac18 + \cdots + \dfrac1{2^{m+1}} - \dfrac{m}{2^{m+2}}$$
I trust you can conclude from this making use of the first MacLaurin series.