The difference $\log(2)-\sum_{n=1}^{100}\frac{1}{2^n n}$ is

Question

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?

Answer 1

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}

Answer 2

You could also determine the value of the summa by using the polylogarithm functions:

$\sum\limits_{n=1}^{\infty}\frac{1}{2^n\,n}=\sum\limits_{n=1}^{\infty}\frac{{\big(\frac{1}{2}}\big)^n}{n}=Li_1(\frac{1}{2})=\ln2$