Limit of integration of harmonic mean Let $I_n=\displaystyle\int_{[0,1]^n}\displaystyle\frac{n}{\frac{1}{x_1}+\cdots+\frac{1}{x_n}}dx_1\cdots dx_n$. Show that $\lim_{n\rightarrow\infty}I_n$ exists.
What I have tried: can use mean inequality to derive upper bound for $I_n$, yielding $0 \leq\limsup_{n\rightarrow\infty}I_n \leq\displaystyle \frac{1}{e}$, but this is not enough.
 A: Let $H_n(x)$ denote the harmonic mean. Set $g(t)=\exp(-2/t)$ $(t > 0)$ and $g(0)=0$. Then $g$ is convex on $[0,1]$. Since $[0,1]^n$ has measure $1$ we have
$$
g(I_n) \le \int_{[0,1]^n} g(H_n(x)) dx = \Big(\int_0^1 g(nt) dt\Big)^n =(\ast).
$$
As $\exp(s) \ge s+1$ we have
$$
g(nt) \le 1/(1+2/(nt))= t/(t+2/n).
$$
Now
$$
(\ast) \le \Big(\int_0^1\frac{t}{t+2/n}dt\Big)^n = \Big(1-\frac{2}{n}\log(1+\frac{n}{2}) \big)^n \to 0 ~~ (n \to \infty).
$$
Thus $g(I_n) \to 0$, which implies $I_n \to 0$ as $g(t) > 0$ in $(0,1]$. Hope everything is correct, please check.
A: A simple proof using Strong Law of Large Numbers: Let $(X_i)$ be i.i.d. with uniform distribution on $(0,1)$. Then $(\frac  1 {X_i})$ is i.i.d, non-negative, with infinite expectation. This implies, by Strong Law of Large Numbers, that $\frac {\frac 1 {X_1}+\frac 1 {X_2}+\cdots+\frac 1 {X_n}} n  \to \infty$ with probability $1$. It now follows that the integrand in the given integral converse almost surely to $0$ and it takes values in $[0,1]$. By DCT it follows that $I_n \to 0$.
