How to flip all fractions in the power series for $\ln(1 + x)$? I am trying to evaluate this using power series:
$$1 + \frac{2}{2} + \frac{3}{2^2} + \frac{4}{2^3} + \dots$$
By using the power series for $\ln(1 + x)$, I have recognized that dividing through by $x$ and setting $x = -2$ will get you this:
$$1 + \frac{2}{2} + \frac{2^2}{3} + \frac{2^3}{4} + ..$$
This seems so close, but I can't seem to figure out how to flip each fraction so that it matches.  How can I do this?
If I am on the completely wrong path and this is a coincidence, please point me in the right direction.
 A: You have $\sum_{n=1}^\infty \frac{n}{2^{n-1}}$ which is equal to $F'(\frac{1}{2})$, where $F(x) = \sum_{n=1}^\infty x^n = \frac{x}{1-x}$,
So we have $F'(x) = \frac{1}{(1-x)^2}$ and $F'(\frac{1}{2}) = 4$
Alternatively without derivatives (Axion004 idea) :
Let $S = \sum_{n=1}^\infty \frac{n}{2^{n-1}} = \sum_{n=0}^\infty \frac{n+1}{2^n} = \sum_{n=0}^\infty \frac{n}{2^n} + 2 = \frac{1}{2}\sum_{n=1}^\infty \frac{n} {2^{n-1}} + 2 = \frac{S}{2} + 2 $
So $ \frac{S}{2} = 2 $ and $S = 4$
A: We have that
$$\sum_{n=1}^\infty \frac{n}{2^{n-1}}=1 + \frac{2}{2} + \frac{3}{2^2} + \frac{4}{2^3} + \dots$$
where
\begin{align*}
\sum_{n=1}^\infty \frac{n}{2^{n-1}}
= \sum_{n=0}^\infty \frac{n+1}{2^n}
= \left( \sum_{n=0}^\infty \frac{n}{2^n}
   + \sum_{n=0}^\infty \frac{1}{2^n} \right)
&= \sum_{n=1}^\infty \frac{n}{2^n} + 2\tag{1}
\end{align*}
then by the geometric series we have that
$$
f(x)=\sum_{n=0}^\infty x^n \;\; =\;\; \frac{1}{1-x}.
$$
provided $|x|<1$ . Thus
$$xf'(x)=\sum_{n=1}^{\infty} nx^n=\frac{x}{(1-x)^2}$$
where we let $x=\frac{1}{2}$ to find
$$\sum_{n=1}^\infty \frac{n}{2^n}=2\tag{2}$$
therefore by $(1)$ and $(2)$ we find that
$$\sum_{n=1}^\infty \frac{n}{2^{n-1}}=4$$
