This is a problem from K.L. Chung's "A Course in Probability Theory". I have been asked to prove by giving an example that if a sequence of random variables $\{X_n\}$ converges to $0$ in probability then $\frac{S_n}{n}$ may not converge to $0$ in probability (where $\{S_n\}$ denotes the partial sum sequence).
I have started with the sequence (as per the hint given) $\{X_n\}$ such that $X_n$ assumes values $2^n$ and $0$ with probabilities $\frac{1}{n}$ and $1-\frac{1}{n}$ respectively. It is easy to show that this sequence converges to $0$ in probability. However I am not being able to prove that $\frac{S_n}{n}$ doesn't converge to $0$ in probability. Can someone give a hint or a direction to proceed in?
We want to show that $\frac{S_n}{n}$ doesn't converge to $0$ in probability. For this, it is enough to show that $\mathcal{P}\{\frac{S_n}{n}\geq 1\}=\mathcal{P}\{S_n\geq n\}>C$ for some positive constant $C$ independent of $n$ (or at least for every $n$ greater than some fixed natural number). For this, we note that if $S_n<n$ holds then $X_i=0$ for each $i>\log_2 n$, because, if $S_n<n$, then the sum cannot consist of terms bigger than $n$. This and the fact that the $X_i$'s are independent random variables, together imply that \begin{equation*} \mathcal{P}\big\{S_n<n\big\}\leq\mathcal{P}\big\{X_i=0:\log_2 n<i\leq n\big\}=\prod\limits_{\lceil\log_2 n\rceil}^n\Bigg(1-\frac{1}{i}\Bigg)=\frac{\lceil\log_2 n\rceil-1}{n}. \end{equation*} The right-hand term converges to $0$ as $n\to\infty$. We can see this by computing the limit $\lim\limits_{x\to\infty}\frac{\log_2 x -1}{x}$. The expression being in $\frac{\infty}{\infty}$ indeterminate form, we can apply l'Hopital's rule to write \begin{align*} \lim\limits_{x\to\infty}\frac{\log_2 x -1}{x}=\lim\limits_{x\to\infty}\frac{1}{x\ln 2}=0 \end{align*} Thus we see that $\mathcal{P}\{S_n<n\}\to 0$ as $n\to\infty$. Thus if we choose $C=0.5$, say, then there exists a natural number $N_0$ such that for all natural numbers $n\geq N_0$, $\mathcal{P}\{S_n<n\}<C$, or in other words $\mathcal{P}\{S_n\geq n\}>1-C=1-0.5=0.5=C$ which is exactly what we wanted to show. Hence $\frac{S_n}{n}$ does not converge to $0$ in probability. Hope that helps.