TITLE: A summation question ,Is $S \to \ln 2$?
QUESTION [9 upvotes]: 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)

REPLY [4 votes]: 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
$$
f(1/2)=-\ln\Bigl(1-\frac{1}{2}\Bigr)=\ln2
$$