Convergence in probability of this sum

Question

Consider $\{0,1\}$-valued random variables $X_1,X_2,\ldots$ with $P(X_k=1)=1/k^{1/2}$ and the sequence $$ Z_n=\frac{1}{n}\sum_{k=1}^n X_k. $$

Question: Does it hold that $Z_n\to0$ in probability, as $n\to\infty$?

My intuition says yes because $Z_n$ appears to be "of the same order" as $$ \frac{1}{n}\sum_{k=1}^n \frac{1}{k^{1/2}} = \frac{1}{n^{1/3}}\sum_{k=1}^n \frac{1}{n^{2/3}k^{1/2}} \le \frac{1}{n^{1/3}}\sum_{k=1}^n \frac{1}{k^{7/6}}\to0,\text{ as $n\to\infty$,} $$ where the latter convergence follows since $\sum_{k=1}^\infty 1/k^{7/6}<\infty$ (Riemann-Zeta function).

So far a proof eludes me, is my intuition perhaps wrong?

Edit: Note that the $X_i$ are not necessarily uncorrelated.

Answer 1

Using MArkov's inequality, \begin{align} P(Z_n > \epsilon) \le \frac{1}{\epsilon} E[Z_n] = \frac{1}{n \epsilon}\sum_{k=1}^n \frac{1}{k^{1/2}} \end{align} So the RHS is the sum you wrote in the question and therefore it converges to zero, for any $\epsilon>0$.

Answer 2

I think it even converges strongly in $L^2$.

We have (using Cauchy-Schwarz) $$ \mathbb{E} |Z_n|^2 = \frac{1}{n^2} \sum_{i,j = 1}^n \mathbb{E} (X_i X_j) \leq \frac{1}{n^2} \sum_{i,j=1}^n (\mathbb{E} X_i^2)^{1/2} (\mathbb{E} X_j^2)^{1/2} = \frac{1}{n^2} \left(\sum_{i=1}^n \frac{1}{i^{1/4}} \right)^2 \leq C \frac{1}{n^2} n^{2(1-1/4)} = C n^{-1/2}. $$