Computing $\sum_{1}^{\infty}\frac{1}{2^{2n+1}n} $ with a power series- What did I do wrong? The requested sum: $\sum_{1}^{\infty}\frac{1}{2^{2n+1}n}=
\frac{1}{2}\sum_{0}^{\infty}\frac{1}{4^nn}$
My helper- this power series: $
\sum_{0}^{\infty}\frac{x^{n}}{4^n}=\frac{1}{1-\frac{4}{x}}
$
Integration due to uniform convergence: $ \int \sum_{0}^{\infty}\frac{x^{n}}{4^n}=\int \frac{1}{1-\frac{4}{x}}$
This is what I get:$\sum_{0}^{\infty}\frac{x^{n+1}}{4^{n+1}(n+1)}=-4\ln(4-x)$
Or: $
\sum_{1}^{\infty}\frac{x^{n}}{4^{n}n}=-4\ln(4-x)
$
Finally what we need is: $\frac{1}{2}\sum_{1}^{\infty}\frac{x^{n}}{4^{n}n}=-2\ln(4-x)$
Now I want to plug-in $x=1$ and get the requested result, but what bothers me is that this is a positive series and I get a negative number in the RHS, obivously something's wrong.
Please tell me what's wrong with the steps described above.
Thanks! :)

The problem with your method is that you're using primitives and not definite integrals:
$$\sum\limits_{n = 0}^\infty  {\frac{1}{{n + 1}}{{\left( {\frac{x}{4}} \right)}^{n + 1}}}  =  - \log \left( {1 - \frac{x}{4}} \right) =  - \log \left( {4 - x} \right) + \log 4 = \int\limits_0^x {\frac{1}{{4 - t}}dt} $$

Since you know
$$-\log \left( {1 - x} \right) = \sum\limits_{n = 1}^\infty  {\frac{{{x^n}}}{n}} $$
You simply plug in $1/4$. You get
$$\eqalign{
  &  - \frac{1}{2}\log \left( {1 - \frac{1}{4}} \right) = \frac{1}{2} \sum\limits_{n = 1}^\infty  {\frac{1}{{n{4^n}}}}   \cr 
  & \frac{1}{2}\log \left( {\frac{4}{3}} \right) =  \frac{1}{2}\sum\limits_{n = 1}^\infty  {\frac{1}{{n{4^n}}}}  \cr} $$

You should be thinking about differentiating, not integrating. You have
$$\frac{1}{2}\sum\limits_{n = 1}^\infty  {n{{\left( {\frac{1}{4}} \right)}^n}} $$
So you might want to find 
$$\frac{1}{2}\sum\limits_{n = 1}^\infty  {n{x^n}}  = f\left( x \right)$$
Adn the plug in $1/4$. Use
$$\eqalign{
  & F\left( x \right) = \frac{1}{{1 - x}} = \sum\limits_{n = 0}^\infty  {{x^n}}   \cr 
  & x\frac{d}{{dx}}F\left( x \right) = x\frac{d}{{dx}}\frac{1}{{1 - x}} = x\frac{d}{{dx}}\sum\limits_{n = 0}^\infty  {{x^n}}   \cr 
  & xF'\left( x \right) = \frac{x}{{{{\left( {1 - x} \right)}^2}}} = \sum\limits_{n = 1}^\infty  {n{x^n}}   \cr 
  & \frac{1}{2}xF'\left( x \right) = \frac{1}{2}\frac{x}{{{{\left( {1 - x} \right)}^2}}} = \frac{1}{2}\sum\limits_{n = 1}^\infty  {n{x^n}}  \cr} $$
Now plug in $1/4$ to get
$$\frac{1}{2}\frac{{\frac{1}{4}}}{{{{\left( {1 - \frac{1}{4}} \right)}^2}}} = \frac{1}{2}\sum\limits_{n = 1}^\infty  {n{{\left( {\frac{1}{4}} \right)}^n}} $$
$$\frac{2}{9} = \frac{1}{2}\sum\limits_{n = 0}^\infty  {n{{\left( {\frac{1}{4}} \right)}^n}} $$
