Solution
= Solution
Independence and the geometric variance formula yield
$$
\begin{aligned}
\operatorname{var}(T_n)
&=\sum_{j=1}^n
\frac{1-j/n}{(j/n)^2}\\
&=\boxed{\sum_{j=1}^n\frac{n(n-j)}{j^2}}.
\end{aligned}
$$
Since $n(n-j)\leq n^2$,
$$
\operatorname{var}(T_n)
\leq n^2\sum_{j=1}^n\frac1{j^2}
\leq\boxed{Cn^2}.
$$