Series that sums to ln2...but how?

Question

I have this series:

$$\sum_{n=1}^\infty \frac{1}{2^n \cdot n}$$

According to Wikipedia this series converges to $\ln2$, but I can't seem to prove this result. Any idea?

Answer 1

$\sum_{n>0} \dfrac{x^n}n=-\log(1-x)$ for $|x|<1$, which you can prove by integrating the series $\sum_{n>0}x^{n-1}=\sum_{n\ge0}x^{n}=\dfrac{1}{1-x}$, also for $|x|<1$.

Then for $x=\frac12$, you get

$$\sum_{n>0} \dfrac{1}{2^nn}=-\log(1-\frac12)=\log2$$

Answer 2

If you go here you can check that $$\mathrm{ln}\left(\frac{n+1}{n}\right)=\sum_{k=1}^{\infty}\frac{1}{k(n+1)^k} $$ Now take n=1.