An unfair "fair game." This is problem 2.2.8 from Durrett's Probability Theory and Examples 4th edition, I am using the version of this book that can be found on his website.

Let $p_k=\frac{1}{2^k k (k+1)}, \ k=1,2,\dots$ and $p_0=1-\sum_{k\geq 1}p_k.$
  $$
\sum\limits_{k=1}^\infty 2^k p_k = \left(1-\frac{1}{2}\right)+\left(\frac{1}{2}- \frac{1}{3}\right)+\dots = 1
$$
  so if we let $X_1, X_2, \dots$ be i.i.d. with $P(X_n = -1)=p_0$ and $P(X_n=2^k-1)=p_k$ for $k\geq 1$ then $E(X_n)=0.$ Let $S_n=X_1 + \dots + X_n.$ Use the Weak Law for Triangular Arrays to conclude that 
  $$
\frac{S_n}{n/\log_2(n)} \rightarrow -1 \ \text{in probability}.
$$

The weak law of triangular arrays is as follows:

For each $n$ let $X_{n,k}, 1\leq k \leq n$, be independent. Let $b_n>0$ with $b_n \rightarrow \infty$, and let $\overline{X_{n,k}}=X_{n,k}1_{(|X_{n,k}|\leq b_n)}. $ Suppose that as $n\rightarrow \infty$
(i) $\sum_{k=1}^n P(|X_{n,k}|>b_n) \rightarrow 0$
(ii) $b_n^{-2} \sum_{k=1}^n E(\overline{X_{n,k}}^2)\rightarrow 0.$
If we let $S_n=X_{n,1}+\dots +X_{n,n}$ and put $a_n=\sum_{k=1}^n E(\overline{X_{n,k}})$ then
  $$\frac{S_n-a_n}{b_n}\rightarrow 0 \ \text{in probability}.
$$

I am letting $b_n=n/\log_2(n)$. Durrett suggests using something else, but this seems more natural to me. I have that all the $X_n$ are independent and that $b_n\rightarrow \infty$ as $n\rightarrow 0$. What I am having a hard time checking is why (i) and (ii) hold for our $X_n$. I see that $X_n=-1,1,3,7,\dots$, in case that helps.
I appreciate any help.
 A: 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,
$$\begin{align}
\dfrac{S_n-a_n}{b_n}
&\to \dfrac{S_n-\frac{n}{log_2n}}{2^{m(n)}}\\\\
&\to \dfrac{S_n-\frac{n}{log_2n}}{\frac{n}{log_2n^{3/2}}}\to 0\\\\
&\Rightarrow \dfrac{S_n}{n/log_2n}\to -1
\end{align}$$
