# calculate the sum $\sum 2^{-n} (\frac 1 n -\frac 1 {n+1})$ [duplicate]

calculate the sum $$\sum 2^{-n} (\frac 1 n -\frac 1 {n+1})$$

well, I need this because it show up in an integral, here my attemp:

$$\sum 2^{-n} (\frac 1 n -\frac 1 {n+1}) = \sum 2^{-n} \frac{1}{(n)(n+1)}$$

and I know that $$\sum_{n=1}^{\infty}2^{-n} = 1$$

The power series for $$\log(1-x)$$ is $$-\sum_{n\geq 1}\frac{x^n}{n}$$. In particular $$\sum_{n\geq 1}\frac{1}{2^nn}=\log 2$$. It follows that

$$\sum_{n\geq 1}\frac{1}{2^n(n+1)} = 2\sum_{m\geq 2}\frac{1}{2^mm}=2(\log 2-1/2).$$

Combining these gives

$$\sum_{n\geq 1}2^{-n}\left(\frac{1}{n}-\frac{1}{n+1}\right) = 1-\log 2 = 0.3068\cdots$$

$$\ln(1-x)=-\sum_{n=1}^{\infty} \frac{x^n}{n}.~~~(1)$$ Integrate (1) w.,r.x both sides from $$x=0$$ to $$x=1/2$$ to get get $$-\int_{0}^{1/2} \ln(1-x) dx =\left .\sum_{n=1}^{\infty} \frac{x^{n+1}}{n(n+1)}\right|_{0}^{1/2}.$$ $$\Rightarrow \sum_{n=1}^{\infty} 2^{-n} \left (\frac{1}{n}-\frac{1}{n+1} \right)=2\int_{1}^{1/2} \ln t~ dt=1-\ln2.$$