Stuck finding the sum of two series This is the first time I ever make a post on Stack Exchange (and the last time I stop lurking it!), so apologies in advance if I caused any problems.
I am completely stuck at finding the sum of the two following series:
$$\sum_{n=1}^∞(2n+1)x^n$$
$$\sum_{n=0}^∞\frac{x^n}{(n+1)2^n}$$
I understand that I should be using this series to find out their value: $$\sum_{n=0}^∞x^n=\frac{1}{1-x}$$
I already used it for a couple of simpler series, but I cannot get my head around the two ones above
 A: For example:
$$\frac1{1-x}=\sum_{n=0}^\infty x^n\implies \frac1{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1}\implies\frac x{(1-x)^2}=\sum_{n=1}^\infty nx^n\;,\;\;|x|<1$$
and now do some calculations with $\;(2n+1)x^n=2nx^n+x^n\;$, with the help of arithmetic of series (limits)
For the other one integrate:
$$\frac1{1-x}=\sum_{n=0}^\infty x^n\implies -\log(1-x)=\sum_{n=0}^\infty\frac{x^{n+1}}{n+1}\;,\;\;|x|<1$$
You may also want to consider $\;\frac x2\;$ in the last part...
A: *

*Set $f(x) = \sum_{n=1}^\infty (2n+1)x^n$. Then 
$$
f(t^2) 
= \sum_{n=1}^\infty (2n+1)t^{2n} 
= \sum_{n=1}^\infty \frac{d}{dt}t^{2n+1}
= \frac{d}{dt} \sum_{n=1}^\infty t^{2n+1}
= \frac{d}{dt} \left( t \sum_{n=1}^\infty (t^2)^n \right)
= \frac{d}{dt} \left( \frac{t^3}{1-t^2} \right) \\
= \frac{3t^2(1-t^2)-t^3(-2t)}{(1-t^2)^2}
= \frac{3t^2-t^4}{(1-t^2)^2}.
$$
Thus,
$$
f(x) = \frac{3x-x^2}{(1-x)^2}.
$$

*Set $g(x) = \sum_{n=0}^\infty \frac{x^n}{(n+1)2^n}.$ Then,
$$
\frac{d}{dx} \left( x \, g(x) \right) 
= \frac{d}{dx} \left( 2 \sum_{n=0}^\infty \frac{(x/2)^{n+1}}{(n+1)} \right)
= 2 \sum_{n=0}^\infty \frac{1}{2} (x/2)^n
= \frac{1}{1-x/2}
= \frac{2}{2-x}.
$$
Therefore, for some constant $C$,
$$x \, g(x) = C - 2 \ln(2-x).$$
The constant can be determined by $0 = 0 \, g(0) = C - 2 \ln 2,$ i.e. $C = 2 \ln 2.$
Thus,
$$g(x) = \frac{2 \ln 2}{x} - 2 \frac{\ln(2-x)}{x} = 2 \frac{\ln 2 - \ln(2-x)}{x} = 2 \frac{\ln(2/(2-x))}{x}.$$
