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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 $$ \sum_{n=1}^{\infty}2^{-n} = 1 $$

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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

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

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

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