HMMT February, 2019

Solution:

Let $S$ denote the value drawn from the hat. The probability that 2 is picked is $\frac{1}{n}$ if $n \geq 2$ and 0 if $n=1$. Thus, the total probability $X$ that 2 is picked is
$$P(S=2)=\sum_{k=2}^{\infty} \frac{2^{-k}}{k}$$
By the definition of conditional probability, $P(N=n \mid S=2)=\frac{P(N=n, S=2)}{P(S=2)}=\frac{2^{-n} / n}{X}$ if $n \geq 2$ and 0 if $n=1$. Thus the conditional expectation of $N$ is
$$\mathbb{E}[N \mid S=2]=\sum_{n=1}^{\infty} n \cdot P(N=n \mid S=2)=\sum_{n=2}^{\infty} n \cdot \frac{2^{-n} / n}{X}=\frac{1}{X} \sum_{n=2}^{\infty} 2^{-n}=\frac{1}{2 X}$$
It remains to compute $X$. Note that $\sum_{k=0}^{\infty} x^{k}=\frac{1}{1-x}$ for $|x|<1$. Integrating both sides with respect to $x$ yields
$$\sum_{k=1}^{\infty} \frac{x^{k}}{k}=-\ln (1-x)+C$$
for some constant $C$, and plugging in $x=0$ shows that $C=0$. Plugging in $x=\frac{1}{2}$ shows that $\sum_{k=1}^{\infty} \frac{2^{-k}}{k}=\ln 2$. Note that $X$ is exactly this summation but without the first term. Thus, $X=\ln 2-\frac{1}{2}$, so $\frac{1}{2 X}=\frac{1}{2 \ln 2-1}$.