Proof that $\sum\limits_{n=1}^\infty\frac1{n2^n}=\ln2 $?

Question

Using approximation with a spreadsheet, I see that:

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

Is there a proof of this?

Answer 1

Analytic functions have Taylor series. Around $x=0$ these are called Maclaurin series. For $\ln(1+x)$ specifically it is called the Mercator series and is given by:

$$ \ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots $$

This follows from integrating the geometric series termwise:

$$ \frac{1}{1+x}=1-x+x^2-x^3+\cdots $$

Both series are valid for (at least) $|x|<1$. Plugging $x=-\frac{1}{2}$ into the Mercator series yields

$$ \ln(1/2)=-\frac{1}{2}-\frac{1}{2\cdot2^2}-\frac{1}{3\cdot2^3}-\frac{1}{4\cdot2^4}-\cdots $$

Multiplying by $-1$ yields $\displaystyle\ln2=\sum_{n=1}^{\infty}\frac{1}{n2^n}$ as desired.