Convergence in distribution of a product

Question

I'm kinda lost in this kind of problems, so I apologize if this is to easy.

Let $V_n\sim\operatorname{Exp}(n)$. I've already prove that $V_n \rightarrow 1$ in probability, $e^{-V_n}\rightarrow 1$ in probability but I cant manage to prove that $nV_{n}\rightarrow V$ in distribution where $V\sim\operatorname{Exp}(1)$.

Hope you guys could help, thanks so much in advance. :)

Answer 1

$P\{V_n >t\}=e^{-nt}$ by definition of exponential distribution. Hence $P\{nV_n >t\}=P\{V_n >\frac t n\}=e^{-t}=P\{V>t\}$. This means $nV_n$ has the same distribution as $V$ for every $n$. Surely, $nV_n \to V$ in distribution.