Does $n/\Sigma_{i=0}^n(1/X_i)$ converge to $0$ in probability for $X_i$ iid standard uniformly random variables? Suppose $X_i \sim\operatorname{uniform}[0,1]$ and that they are iid. Does $n/\Sigma_{i=0}^n(1/X_i)$ converge to $0$ in probability? A simulation seems to indicate that it does. But as the expected value of $1/X_i$ does not exist, I cannot use LLN. Any hints?
 A: Let $$Y_n=\frac{1}{\sum_{i=0}^n(1/X_i)}$$
Then, for any $0<a<1$, $$P(Y_n>a)=P(\sum_{i=0}^n X_i^{-1}<a^{-1})\le P( X_i^{-1}<a^{-1}; \forall i)=\prod_{i=1}^n P(X_i>a)=(1-a)^n$$
This tends to $0$ as $n\to \infty$, so...
A: Let
$$
Z =\frac{1}{ \sum_{i=0}^n \frac{1}{X_i}}
$$
Observe that
$$
P\left(\sum_{i=0}^n \frac{1}{X_i}\geq n\right) =1
$$
Then,
$$
P\left(Z\leq \frac{1}{n}\right) = 1
$$
Moreover, $P(Z<0)=1$.
Now let $k> 0$, $\varepsilon>0$ and Notice that $\lim_{n\rightarrow\infty} P(|Z-k|\geq \varepsilon) = 0$ because as $n$ gets larger, $Z-k$ gets arbitrarily close to $0$.
As a result, $Z$ converges in probability to $0$.
A: (does not work and should be discarded)
For the modified problem: Let $Z_{n}=\sum^{n}_{i=0}\frac{1}{X_{i}}, X_{i}\sim U(0,1)$. Then we have the following inequalities using pigeon hole principle:
$$
P(X_{(1)}\ge \frac{1}{x})=P(\frac{1}{X_{(1)}}\le x)\le P(Z_{n}/n\le x)\le P(\frac{1}{X_{(n)}}\le x)=P(X_{(n)}\ge \frac{1}{x})
$$
which gives us
$$
(1-\frac{1}{x})^{n}\le P(\frac{Z_{n}}{n}\le x)\le 1-x^{-n}
$$
This does not give us any useful info when $x>1$, since then we get a useless $(0,1)$ bound. A direct $R$-simulation showed the mean did not diminish when $n$ goes from $10^{3}$ to $10^{5}$, and the histogram is tilted towards the right. But it could also be because the $n$ is still relatively small.
