Solve following sum $\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n+1}2^{-n}}{n} $
I am very unsure how to evaluate this, as the sign changes I would separate them in two series, one positive and one negative   $$\frac{1}{(2n-1)(2)^{2n-1}}$$ and $$-\frac{1}{(2n)(2)^{2n}}$$ 
However I can't evaluate them right.
 A: We re-write this sum to $$\sum_{n=1}^\infty \frac{(-1)(-2)^{-n}}{n} = \sum_{n=1}^\infty \frac{-1}{(-2)^n \cdot n}$$
For $-1 \le x < 1$, define $$f(x)= \sum_{n=1}^\infty \frac{x^n}{n}$$
We have $$f'(x)= \sum_{n=1}^\infty x^{n-1} = \frac{1}{1-x}$$
This gives $$f(x)=- \ln (1-x)+C$$
It is clear that $C=0$. 
Therefore, we have $$f(x)=-\ln (1-x)$$
Plugging $x=-\frac{1}{2}$, we have $$\sum_{n=1}^\infty \frac{1}{(-2)^n \cdot n} = f(-\frac{1}{2}) = -\ln (\frac{3}{2})$$
Our desired answer is $$-f(-\frac{1}{2})=\ln (\frac{3}{2})$$
A: Begin with: 
$1 - x + x^2 - x^3 + \cdots = \dfrac{1}{1+x} = \sum_{n=0}(-1)^nx^n$ (Why is this true?)
Notice you have a term $n$ in the denominator. (So, what to do? Integrate the power series?)
So you have $\ln(1+x) = \sum_{n=0} (-1)^n \dfrac{x^{n+1}}{n+1}$, with the correct interval of convergence. (Why?)
I want to make the summation 'appear' similar to the one you gave, so I would change the index so that 
$\sum_{n=1}(-1)^{n-1}\dfrac{x^n}{n} $. The only thing left is to substitute $x = \dfrac{1}{2}$. (Confirm if 1/2 is within the interval of convergence of this power series, if yes then this series will converge.) 
A: Hint:
$$\sum_{k=0}^\infty x^k=\frac1{1-x}.$$
Then integrating,
$$\sum_{k=0}^\infty\frac{x^{k+1}}{k+1}=\ln(1-x).$$
$x$ may be negative.
A: Indirect and contrived:
As is not so well-known, the logarithm can be defined as
$$\log(t)=\lim_{n\to\infty}n(t^{1/n}-1).$$
Then with $t=1+2^{-1}$, we develop $n((1+2^{-1})^{1/n}-1)$ using the generalized binomial formula and get
$$n\left(
1+\frac1n2^{-1}+\frac1n\left(\frac1n-1\right)\frac{2^{-2}}2+
\frac1n\left(\frac1n-1\right)\left(\frac1n-2\right)\frac{2^{-3}}{3!}+
\frac1n\left(\frac1n-1\right)\left(\frac1n-2\right)\left(\frac1n-3\right)\frac{2^{-4}}{4!}+\cdots-1
\right)=$$
$$2^{-1}+\left(\frac1n-1\right)\frac{2^{-2}}2+\left(\frac1n-1\right)\left(\frac1n-2\right)\frac{2^{-3}}{3!}+\left(\frac1n-1\right)\left(\frac1n-2\right)\left(\frac1n-3\right)\frac{2^{-4}}{4!}+\cdots\to$$
$$2^{-1}-\frac{2^{-2}}2+\frac{2^{-3}}{3}-\frac{2^{-4}}{4}+\cdots=\log(1.5)$$ when $n$ goes to $\infty$.
A: Consider the expansion of the series $\ln(1+x)$
$\ln(1+x)=x-x^2/2+x^3/3-x^4/4+\cdots$$ 
Put $x=1/2=0.5$ 
Belongs in the Radius of Convergence as it is less than 1
$\ln(1.5)=0.5-0.5^2/2+0.5^3/3-\cdots$$
Which is the above-mentioned series
