How do I think of the sign of the difference $\log(2) - \sum_{n=1}^{100}\frac{1}{n 2^n}$ ? Is the difference less than or greater than $\frac{1}{2^{100}.101}$?
I thought of writing the series expansion for $\log(2)$ as $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}$, so now I am looking at the difference $\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n} - \sum_{n=1}^{100} \frac{1}{n2^n}$, but how to proceed next?
Will it be helpful to write the series $\sum_{n=1}^{100}\frac{1}{n 2^n}$ in terms of logarithms, can we write it?
Well the above comments helped me in knowing that $\log 2 = \sum_{n=1}^{\infty} \frac{1}{n2^n}$, and this link - Simplify log expression with infinite series $\log x - \sum_{i=1}^{\infty} \frac{x^i}{i}$
is similar to the question above, from which I think that the difference is greater than $0$ and is equal to $\sum_{n=101}^{\infty} \frac{1}{n2^n} > \frac{1}{101 . 2^{101}}$