Convergence in probability: The inverse of the simple mean

I have a question on convergence: I have to prove that $$\frac{n}{U_{n}} \longrightarrow 1$$ in probability, where $$U_{n}=\sum X_{i}$$, $$X_{i}\sim \mathrm{Exp}(1)$$ and because of this, $$U_{n}\sim \mathrm{Gamma}(n,1)$$.

This problem had two parts, the other part was to prove that $$\frac{U_{n}}{n}\longrightarrow 1$$ in probability which I proved invoking the weak law for big numbers. But this, I have no clue how to prove it.

Thanks so much for your help! :)

• So I just have to use $g(x)=x^{-1}$, and then, since $\frac{U_{n}}{n}$ converges in probability to 1, then $g( \frac{U_{n}}{n})= \frac{n}{U_{n}}$ converges in probability to $g(1)=1$ ? That simple? Thanks! :) – user1trill Nov 28 '18 at 22:11
• Yes. Ideally, you define $g(0)=0$ and note that the set of discontinuity points is indeed a $P_X$-zero-set :) (as $U_n$ can be $0$) – Stockfish Nov 29 '18 at 10:09