Let $(X_n)$ be an indepdendent sequence of real random variables distributed thus:
$$P(X_n=n)=P(X_n=-n)=\frac{1}{2} \frac{1}{n \ln(n+1)},$$
$$P(X_n=0)=1- \frac{1}{n \ln(n+1)} \hspace{1cm} n\in\mathbb{N}$$
With the aid of Theorem 10.2 show that this sequence obeys the weak law of large numbers. Hint: The function $x\mapsto \frac{x}{\ln(x+1)}$ is strictly increasing on the interval $[1,\infty)$.
First I notice that $P(X_1=0)<0$ so I suppose we mean $n\geq 2$. Second we see that $E(X_n)=0$ and that $E(X^2_n)=\frac{n}{\ln(n+1)}$ for all $n$. Hence the condition
$$ \lim_{n\to \infty} \frac{1}{n^2} \sum_{i=1}^{n} V(X_i) =0$$
of Theorem 10.2 is statisfied and the WLLN holds. But why do we need the hint for?
The fact that $\frac n {\ln (n+1)}$ is increasing shows that $\frac 1 {n^{2}} \sum\limits_{k=1}^{n}\frac k {\ln (k+1)} \leq \frac 1 {n^{2}} n\frac n {\ln (n+1)} =\frac 1 {ln (n+1)} \to 0$.