I tried to solve it and I couldn't. If anyone can help me how to resolve or a hint where to start.
$\mathbf{7.7.29}$
Suppose $\{X_n, n > 1\}$ is iid with common distribution described as follows. Define
$$ p_k =\displaystyle{\frac{1}{2^kk(k+1)}}, \hspace{.5cm} k \geq 1, $$
and $p_0 = 1 - \displaystyle{\sum_{i = 1}^\infty} p_k$. Suppose
$$ P[X_n = 2^k-1] = p_k, \hspace{.5cm} k \geq 1 $$
and $P[X_n = -1] = p_0$· Observe that
$$ \displaystyle{\sum_{i = 1}^\infty} 2^kp_k = \displaystyle{\sum_{i = 1}^\infty} \bigg( \frac{1}{k}-\frac{1}{k+1} \bigg) = 1, $$
and that $E(X_n) = 0$. For $S_n = \displaystyle{\sum_{i = 1}^n}, n > 1$, prove
$$ \displaystyle{\frac{S_n}{n/\log_2 n}} \overset{P}\longrightarrow -1 $$