Using Maclaurin series, finding the value of an infinite sum...

I want to find $$\sum\limits_{n = 1}^\infty {{{{{({1 \over 2})}^n}} \over {n(n + 1)}}}$$

The book that has this problem in says to use the Maclaurin series for $(1 - x)\log (1 - x)$. I don't understand how you are supposed to do this at all. Any help you can offer would be greatly appreciated.

• If you start following the book's hint, what do you get as the MacLaurin series for $(1-x)\log (1-x)$? Dec 17 '15 at 14:17

You have for $x\in ]-1,1[$:
$$-\log(1-x)=\sum_{k\geq 1}\frac{x^k}{k}$$
Multiplying by $(1-x)$ gives $$-(1-x)\log(1-x)=\sum_{k\geq 1}\frac{x^k}{k}-\sum_{k\geq 1}\frac{x^{k+1}}{k}=x-\sum_{k\geq 2}\frac{x^k}{k(k-1)}$$
Hence $$\sum_{n\geq 1}\frac{x^{n+1}}{n(n+1)}=x+(1-x)\log(1-x)$$ It is easy to finish.
$$\sum_{n\geq 1}\frac{1}{2^n n(n+1)} = \sum_{n\geq 1}\frac{1}{n 2^n}-\sum_{n\geq 1}\frac{1}{2^n(n+1)} = \log(2)-\left(2\log 2-1\right) = \color{red}{1-\log 2}$$ since: $$\sum_{n\geq 1}\frac{1}{(n+1) 2^n} = \int_{0}^{1}\sum_{n\geq 1}\frac{x^n}{2^n}\,dx = \int_{0}^{1}\frac{x\,dx}{2-x}$$ and a similar argument shows that $\sum_{n\geq 1}\frac{1}{n 2^n}=\log 2$.