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! :)
 A: It would be best to proceed as Peter does starting with the series representation of $\ln(1-x)$.
However, to address your argument:
You wish to compute
$$
\sum_{n=1}^\infty {1\over 2^{2n+1} n}=
{1\over2} \sum_{n=1}^\infty { 1\over n \, 4^n }
$$
(this was your first error, the sum starts at $n=1$).
Using the Geometric series:
$$\tag{1}
\sum_{n=1}^\infty (x/4)^n= {x/4\over 1-(x/4)}= {1\over 1-(x/4)}-1 
$$
(your sum of the series was your second error).
If $|x|<4$:
$$
\sum_{n=1}^\infty\int_0^x (t/4)^n\,dt=  \int_0^x {1\over 1-(t/4)}-1\, dt 
$$
(note, you need to take definite integrals).
Whence
$$
\sum_{n=1}^\infty{x^{n+1}\over 4^n (n+1)}=  -4\ln|1-(x/4)|-x
$$
Substituting $x=1$ gives:
$$
\sum_{n=1}^\infty{1\over 4^n (n+1)}=  -4\ln(3/4)-1\approx0.15073
$$
So
$$\eqalign{
{1\over2} \sum_{n=1}^\infty { 1\over n \, 4^n }
&={1\over2} \sum_{n=0}^\infty { 1\over( n +1)\, 4^{n+1} }\cr
&={1\over8} \sum_{n=0}^\infty { 1\over( n +1)\, 4^{n } }\cr
&={1\over8} [{1}   -4\ln(3/4)-1]\cr
&={1\over8} [   -4\ln(3/4) ]\cr
&=  {  \ln(4/3)\over 2 }\cr
}
$$
A: 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
$$\frac{1}{2}\frac{{\frac{1}{4}}}{{{{\left( {1 - \frac{1}{4}} \right)}^2}}} = \frac{1}{2}\sum\limits_{n = 1}^\infty  {n{{\left( {\frac{1}{4}} \right)}^n}} $$
$$\frac{2}{9} = \frac{1}{2}\sum\limits_{n = 0}^\infty  {n{{\left( {\frac{1}{4}} \right)}^n}} $$
