= Solution
Part (iii) and the given harmonic-sum asymptotic show that
$$
\frac{\mathbb ET_n}{n\log n}
=\frac{H_n}{\log n}\longrightarrow1.
$$
For any fixed $\varepsilon>0$, this deterministic ratio lies within $\varepsilon/2$ of $1$ for all sufficiently large $n$. The <Chebyshev inequality> and part (iv) then give
$$
\begin{aligned}
\mathbb P\left(
\left|\frac{T_n}{n\log n}-1\right|>\varepsilon
\right)
&\leq
\mathbb P\left(
|T_n-\mathbb ET_n|>\frac{\varepsilon}{2}n\log n
\right)\\
&\leq
\frac{4\operatorname{var}(T_n)}
{\varepsilon^2n^2(\log n)^2}\\
&\leq\frac{4C}{\varepsilon^2(\log n)^2}
\longrightarrow0.
\end{aligned}
$$
Thus
$$
\boxed{\frac{T_n}{n\log n}\longrightarrow1
\quad\hbox{in probability}.}
$$
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