A summation question ,Is $S \to \ln 2$? I am in doubt with this question .
let $S=\dfrac{\dfrac12}{1} +\dfrac{(\dfrac12)^2}{2}+\dfrac{(\dfrac12)^3}{3}+\dfrac{(\dfrac12)^4}{4}+\dfrac{(\dfrac12)^5}{5}+...$ Is it converge to $\ln 2$ ?  
I tried this 
$$x=\dfrac12  \to 1+x+x^2+x^3+x^4+...\sim\dfrac{1}{1-x}\to 2$$ by integration wrt x we have $$\int (1+x+x^2+x^3+x^4+...)dx=\int (\dfrac{1}{1-x})dx \to\\
x+\dfrac{x^2}{2}+\dfrac{x^3}{3}+\dfrac{x^4}{4}+...=-\ln(1-x)$$then put $x=0.5$ $$\dfrac{\dfrac12}{1} +\dfrac{(\dfrac12)^2}{2}+\dfrac{(\dfrac12)^3}{3}+\dfrac{(\dfrac12)^4}{4}+\dfrac{(\dfrac12)^5}{5}+..\sim -\ln(0.5)=\ln 2$$ now my question is : Is my work true  ?
I am thankful for you hint,guide,idea   or solutions.  (I forgot some technics of calculus)
 A: Yes. 
This series was already known to Jacob Bernoulli (Gourdon and Sebah http://plouffe.fr/simon/articles/log2.pdf, formula 14) and can be written 
$$S=\sum_{k=0}^\infty \frac{1}{k+1}\left(\frac{1}{2}\right)^{k+1}$$
To evaluate it, we can change the identity
$$\int_0^1 x^n dx = \frac{1}{n+1}$$
into
$$\int_0^\frac{1}{2} x^n dx = \frac{1}{n+1}\left(\frac{1}{2}\right)^{n+1}$$
and then
$$\begin{align}
S&=\sum_{k=0}^\infty \frac{1}{k+1}\left(\frac{1}{2}\right)^{k+1}\\
&=\sum_{k=0}^\infty \int_0^\frac{1}{2} x^k dx \\
&=\int_0^\frac{1}{2}\left(\sum_{k=0}^\infty x^k\right) dx\\
&=\int_0^\frac{1}{2} \frac{1}{1-x} dx\\
&=-\log(1-x)|_0^\frac{1}{2}\\
&=\log(2) \\
\end{align}$$
A nice way to encode formulas like
$$\log(2)=\sum_{n=1}^\infty \frac{1}{n2^n}$$
is noting that the numerator is one so the sequence of integer denominators can represent the series. When we search the OEIS for $2,8,24,64$ (http://oeis.org/A036289) we find that
$$\sum_{n=1}^\infty \frac{1}{a(n)} = \log(2)$$
is one of the formulas given.
Your series is a base-2 BBP-type formula for $\log(2)$. The base-3 version is
$$\log(2)=\frac{2}{3} \sum_{k=0}^\infty \frac{1}{(2k+1)9^k} $$
and the sequence of denominators is OEIS http://oeis.org/A155988.
A: For completeness, you shoud show that the series for $1/(1-x)$ is uniformly convergent in $[0,1/2]$.
As
$$\left|S_n-\frac1{1-x}\right|=\left|\frac{1-x^{n+1}}{1-x}-\frac1{1-x}\right|=\left|\frac{x^{n+1}}{1-x}\right|\le\left|\frac1{2^n}\right|$$ this is ensured and you can integrate term-wise.
A: Your work is good, but it can be better justified with the theory of power series.
The given series is an instance of the power series
$$
f(x)=\sum_{k=1}^{\infty}\frac{x^k}{k}
$$
for $x=1/2$. The power series has convergence radius $1$; indeed, the ratio test gives
$$
\left|\frac{x^{k+1}/(k+1)}{x^k/k}\right|=\frac{k}{k+1}|x|\to |x|
$$
Thus you know your series converges for $|x|<1$.
The function $f$, defined over $(-1,1)$, is differentiable and
$$
f'(x)=\sum_{k=1}^{\infty}x^{k-1}=\sum_{k=0}^{\infty} x^k=\frac{1}{1-x}
$$
(geometric series). Since $f(0)=0$, you can conclude that, for $x\in(-1,1)$,
$$
f(x)=\int_0^x\frac{1}{1-t}\,dt=-\ln(1-x)
$$
Therefore
$$
f(1/2)=-\ln\Bigl(1-\frac{1}{2}\Bigr)=\ln2
$$
