Problem: Suppose that the random variables $X_1,X_2,\dots$ have finite variances and nonzero means. Show the following statement: $$\text{if }\lim\limits_{n\to\infty}\frac{\sigma_n^2}{E[X_n]^2}=0\text{ then }\frac{X_n}{E[X_n]}\overset{P}\longrightarrow1.$$
Attempt: Fix $\varepsilon>0$. An application of Chebyshev's inequality yields $$P\left(\left\vert \frac{X_n}{E[X_n]}-1\right\vert\geq\varepsilon\right)=P\left(\left\vert X_n-E[X_n]\right\vert\geq\varepsilon \vert E[X_n]\vert\right)\leq\frac{\sigma_n^2}{\varepsilon^2 E[X_n]^2}\overset{n\to\infty}\longrightarrow0,$$ where in the last step we used the assumption that $\lim\limits_{n\to\infty}\frac{\sigma_n^2}{E[X_n]^2}=0.$ Since $\varepsilon>0$ was arbitrary it follows that $\frac{X_n}{E[X_n]}\overset{P}\longrightarrow1$ as $n\to\infty.$
Do you agree with my attempt at a proof of the above exercise?
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