With 100?

The green graph represents the curve $y=\frac{1}{x}$ and the brown boxes represent the sum $\sum^{m}_{k=1}\frac{1}{k}$ for $m\geq 1$ . Then, we have $\sum^{m}_{k=1}\frac{1}{k}>\int^m_1\frac{1}{x}\ dx=\ln(m).$ Thus, by substituting $m=2^n-1,$ we have $\sum^{2^n-1}_{k=1}\frac{1}{k}>\ln(2^n-1)>n\ln2.$ Hence, when $n=100,$ we see that the sum is only slightly larger than 69, which means that it must be smaller than 100. The choice $P(100)<100$ is correct.