Using Chebyshev to derive an upper bound for Coupon Collecter's Problem
As you should have suspected, there is only a thin line between what you have tried and what is rigorous.

Here it is the rigorous version of your attempt. Since $\beta n\ln n\gt \mu_X$ when $n$ is large enough, the first inequality holds for $n$ large enough.
\begin{align*}
Pr[X&gt;\beta n \ln(n)] 
&amp;\leq Pr[|X-\mu_X| \geq \beta n \ln n - \mu_X] \\
&amp;\le \frac{\sigma_X^2}{(\beta n \ln n - \mu_X)^2}.\\
\end{align*}
Now

$$\begin{aligned}
\lim_{n\to\infty}\frac{\frac{\sigma_X^2}{(\beta n \ln n - \mu_X)^2}}{\frac1{\ln^2 n}}
&amp;=\lim_{n\to\infty}{\frac{\sigma_X^2}{n^2}}\lim_{n\to\infty}\frac{(n\ln n)^2}{(\beta n \ln n - \mu_X)^2}\\
&amp;=\frac{\pi^2}{6}\lim_{n\to\infty}\frac{1}{(\beta - \frac{\mu_X}{n\ln n})^2}\\
&amp;=\frac{\pi^2}{6(\beta - 1)^2},
\end{aligned}$$
which implies $$\frac{\sigma_X^2}{(\beta n \ln n - \mu_X)^2} = O\left(\frac{1}{\ln^2n}\right).$$