The difference $$\log (2) - \sum\limits_{n=1}^{100} \frac {1} {2^n \cdot n}\ \ \ \ \ \ \text {is}$$
$(1)$ less than $0.$
$(2)$ greater than $1.$
$(3)$ less than $\frac {1} {2^{100} \cdot 101}.$
$(4)$ greater than $\frac {1} {2^{100} \cdot 101}.$
It is clear that $(2)$ is false because $\log (2) - \sum\limits_{n=1}^{100} \frac {1} {2^n \cdot n} < \log (2) < 1.$ What other options are true?
Any help will be highly appreciated. Thank you very much.
With the help of Ragib Zaman I finally able to do it. Here it is $:$
We first observe that $$\log (2) = - \log \left ( \frac 1 2 \right ) = \sum\limits_{n=1}^{\infty} \frac {1} {2^n \cdot n}.$$ So $$\log (2) - \sum\limits_{n=1}^{100} \frac {1} {2^n \cdot n} = \sum\limits_{n=101}^{\infty} \frac {1} {2^n \cdot n} < \frac {1} {101} \sum\limits_{n=101}^{\infty} \frac {1} {2^n} = \frac {1} {2^{100} \cdot 101}.$$