Using Chebyshev to derive an upper bound for Coupon Collecter's Problem I'm TA'ing a course and have trouble solving an exercise.

Let $X$ be a RV defined to be the number of trials required to collect at least one of each type of coupon (of which there are $n$). Then $E[X] = nH_n = n \ln(n) + O(n)$ and $\sigma^2_X = n^2 \sum_{i=1}^n \frac{1}{i^2} - nH_n$ with $\lim_{n \to \infty} \frac{\sigma^2_X}{n^2} = \frac{\pi^2}{6}$. The exercise asks how Chebyshev's inequality can be used to upper bound the probability that $X &gt; \beta n \ln(n)$ for some $\beta &gt; 1$.

I tried this:
\begin{align*}
Pr[X&gt;\beta n \ln(n)] &amp;\leq Pr[|X-\mu_X| \geq (\beta - 1)n \ln n] \\
&amp;\leq Pr\left[|X-\mu_X| \geq (\beta - 1) \left( \frac{\sqrt 6}{\pi} \sigma_X \right) \ln n\right] (1) \\
&amp;\leq \frac{\pi}{6(\beta -1)^2 \ln^2 n} = O\left(\frac{1}{\ln^2n}\right)
\end{align*}

where (1) follows because $n &gt; \frac{\sigma_X \sqrt 6}{\pi}$. However I believe the first inequality is incorrect because I'm discarding the $-  O(n)$ from $\mu_X$ in deriving an upper bound.

I also tried something else:
\begin{align*}
Pr[X&gt;\beta n \ln(n)] &amp;\leq Pr[|X-\mu_X| \geq \beta n \ln n - nH_n] \\
&amp;\leq Pr[|X-\mu_X| \geq (\beta -1)nH_n] \\
&amp;= Pr\left[|X-\mu_X| \geq \frac{6(\beta - 1)H_n}{\pi^2n} \left(n^2 \sum_{i=1}^\infty \frac{1}{i^2} - n H_n \right)\right]
\end{align*}

but I'm not sure where to go from here. Thanks for any help!
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 $$\frac{\sigma_X^2}{(\beta n \ln n - \mu_X)^2} = O\left(\frac{1}{\ln^2n}\right).$$