Find the sum of $2^{-x}/x$

Question

Task is the following: find the $\sum_{x=1}^{+∞} \frac{2^{-x}}x$

I don't even know how to proceed. I know that $\sum_{x=1}^{+∞} 2^{-x} = 1$. However, is it useful here?

Answer 1

Here's a neat trick. Note that

$$ \frac{\mathrm{d}}{\mathrm{d}\alpha} \sum_{x=1}^\infty \frac{2^{-\alpha x}}{x} = -\ln(2)\sum_{x=1}^\infty 2^{-\alpha x} = -\ln(2)\frac{2^{-\alpha}}{1-2^{-\alpha}}$$

Now integrate this from $\alpha=1$ to $\alpha=\infty$:

$$ - \sum_{x=1}^\infty \frac{2^{-x}}{x}= \left. -\ln\left(1 - 2^{-\alpha}\right) \right|_{\alpha =1}^\infty = -\ln(2)$$

Answer 2

Hint 1: It suffices to recognize this sum as a Taylor series

Hint 2: It is in fact sufficient to know that for all $x \in (-1,1)$, we have $$ \sum_{x=1}^\infty z^{x-1} = \frac{1}{1-z} $$ Note that for $z = 1/2$, this is your statement. It follows that $$ \int_{0}^t \frac{1}{1-z}\,dz = \int_{0}^t \sum_{x=1}^\infty z^{x-1}dz = \sum_{x=1}^\infty \int_{0}^t z^{x-1}dz = \sum_{x=1}^\infty \frac{t^x}{x} $$

Answer 3

According to Taylor's series,

$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots, ~~~\forall x \in (-1,1].$$

Let $x=-\dfrac{1}{2}.$ Then $$\ln \frac{1}{2}=\ln\left(1-\frac{1}{2}\right)=-\dfrac{1}{2}-\frac{1}{2 \cdot 2^2}-\frac{1}{3 \cdot 2^3}-\frac{1}{4 \cdot 2^4}+\cdots=-\sum_{n=1}^{\infty}\frac{1}{n\cdot 2^n} $$ Therefore $$\sum_{n=1}^{\infty}\frac{1}{n\cdot 2^n}=-\ln\frac{1}{2}=\ln2.$$