The Petersburg game and fair game (Feller Vol.1)

Question

(Feller Vol.1, P.252, Petersburg's paradox) Let $(X_k)$ be mutually independent random variables, each of which assume $2, 2^2, 2^3, …$ with probability $2^{-1}, 2^{-2}, 2^{-3}, …$. We define $S_n = X_1 + \cdots + X_n$. Let $e_n$ be the accumulated entrance fees, and $S_n$ be the accumulated gains. The author says that the game is fair if for every $\epsilon >0$,
$$P(|\frac{S_n}{e_n} -1|> \epsilon) \to 0.$$
Then, he claims that in the Petersburg game, $e_n = n \log_2 (n)$ makes it fair. The proof is given as follows:

First, he introduces truncated variables $U_k$ and $V_k$. If $n \log_2(n) \ge X_k$, $U_k=X_k$, $V_k =0$. On the other hand, if $n \log_2(n) < X_k$, $U_k = 0$, $V_k = X_k$. Then, it follows that
$$P(|\frac{S_n}{e_n} -1|> \epsilon) \le P(|U_1+ \cdots + U_n – e_n| > e_n \epsilon) + P(V_1 + \cdots + V_n \not=0).$$
The inequality holds since if neither of two events on the right does occur, the event on the left cannot happen. Thus, it is sufficient to prove that two probability on the right converges to $0$. The author uses the following inequality
$$P(V_1 + \cdots + V_n \not= 0) \le n \sum_{i=1}^n P(V_1 \not= 0) \le P(X_1 > n \log_2(n) ) \le \frac{2}{\log_2(n)} \to 0.$$

However, I don't understand how the last inequality holds. I was thinking about using Chebyshev's inequality, or the fact that $P(X_1 > n \log_2(n)) = P(X_1 \ge 2^{r+1})$ assuming that $r$ is the largest integer satisfying $2^{r} \le n \log_2(n)$. But, I failed both attempts.

I would appreciate if you give some help, and please let me know if you need more context.

Answer 1

Your second suggestion works. Let $r$ be the greatest natural number satisfying $2^r \leq n \log n$. Then $X_1 > n \log n$ implies $X_1 \geq 2^{r+1}$. So, $$P(X_1 > n \log n) \leq P(X_1 \geq 2^{r+1}) = \sum_{k=r+1}^\infty \frac{1}{2^k}= \frac{1}{2^r} <\frac{2}{n\log n},$$ where the last inequality is just rearranging $2^{r+1} > n \log n$.