The difference $$\log(2)-\sum_{n=1}^{100}\frac{1}{2^n n}$$ is
(1) less than $0$
(2) greater than $1$
(3) less than $\frac{1}{2^{100}101}$
(4) greater than $\frac{1}{2^{100}101}$
My Attempt:
I know $$\log(2)=1-1/2+1/3-1/4+\dots\tag{1}$$
Let us call $S=\sum_{n=1}^{100}\frac{1}{2^n n}$ so $S=\frac{1}{2}+\frac{1}{8}+\frac{1}{24}$
How should I move from here?
Best Answer
Use the fact that\begin{align}\log(2)&=-\log\left(\frac12\right)\\&=\frac12+\frac1{2\times2^2}+\frac1{3\times2^3}+\cdots\\&=\sum_{n=1}^\infty\frac1{n2^n}.\end{align}