An unfair "fair game."

In the latest edition of Durrett's same book, a "hint" is included in the same problem:

Use Theorem 2.2.11 with $b_n = 2^{m(n)}$, where $m(n)=min$ { $m:2^{-m}m^{-2/3}\leq n^{-1}$ }

So we start by checking that the $b_n$ given in the hint is a valid choice for $b_n$. Note that $2^{k}-1\leq b_n=2^{m(n)} \iff k\leq m(n)$ when $n\geq 1$. So $$\begin{align} \sum_{i=1}^n P(|X_i|>b_n) &=n\sum_{k=m(n)+1}^{\infty} \dfrac{1}{2^kk(k+1)}\\\\ &\leq n\sum_{k=m(n)+1}^{\infty} \dfrac{1}{2^km(n)(m(n)+1)}\\\\ &=\dfrac{n}{2^{m(n)}m(n)(m(n)+1)}\\\\ &\leq \dfrac{1}{\sqrt{m(n)}} \end{align}$$ Where the last inequality comes from the definition of $m(n)$. Since $m(n)\to \infty$ as $n\to \infty$, this shows that the first condition of weak law for triangular array is satisfied. To check the second condition, note that $$\begin{align}E(\bar{X_i}^2) &=(-1)^2p_0+\sum_{k=1}^{m(n)}2^{2k}\dfrac{1}{2^kk(k+1)}\\\\ &\leq 1+\sum_{k=1}^{m(n)}2^{2k}\dfrac{1}{2^kk(k+1)}\\\\ &\leq 1+\sum_{k=1}^{m(n)/2}2^{k}\dfrac{1}{1(1+1)}+\sum_{k=m(n)/2}^{m(n)}2^{k}\dfrac{1}{\frac{m(n)}{2}(\frac{m(n)}{2}+1)}\\\\ &\leq 1+2\cdot \frac{m(n)}{2} + 2^{m(n)/2}\cdot \frac{4}{m(n)^2}\cdot \frac{m(n)}{2}\\\\ \end{align}$$ so $$ \dfrac{E(\bar{X_i}^2)}{b_n^2} \leq \dfrac{1+m(n)+2^{m(n)/2}\cdot2/m(n)}{2^{2m(n)}} $$ which clearly goes to $0$, hence proving the second condition. $$\begin{align} E(\bar{X_i}) &=-1\cdot (1-\sum_{k\geq 1} p_k)+\sum_{k=1}^{m(n)}(2^k-1)\cdot p_k\\\\ &=\sum_{k=1}^{\infty}p_k -1 +\sum_{k=1}^{m(n)}\dfrac{1}{k(k+1)}-\sum_{k=1}^{m(n)}p_k\\\\ &=\sum_{k=m(n)+1}^{\infty} p_k-\dfrac{1}{m(n)+1} \end{align}$$ Note that $$-\dfrac{1}{m(n)}\leq E(\bar{X_i}) = -\dfrac{1}{m(n)} + P(|X_i|\geq b_n)$$ and by previous discussions $P(|X_i|\geq b_n)$ goes to $0$. So $E(\bar{X_i})\to -\dfrac{1}{m(n)}$ as $n\to \infty$. Also, $m(n)\to \log_2n$ as $n\to \infty$. Thus by the weak law, $$\begin{align} \dfrac{S_n-a_n}{b_n} &\to \dfrac{S_n-\frac{n}{log_2n}}{2^{m(n)}}\\\\ &\to \dfrac{S_n-\frac{n}{log_2n}}{\frac{n}{log_2n^{3/2}}}\to 0\\\\ &\Rightarrow \dfrac{S_n}{n/log_2n}\to -1 \end{align}$$