For n a natural number let $X_{n}$ have discrete uniform distribution on interval {1,2...,n} and $Y_{n} =\frac{1}{n} X_{n}$. I need to show that for all t(real number) the $\lim_{n \to \infty} M_{Y_{n}}(t) = M_{U}(t)$. U is in this case the continuous uniform distribution on [0,1]. The hint says that I don't need tp compute the mgf of $Y_{n}$ or U to prove this.
My attempt: $f_{X_{n}}=\frac{1}{n}$, $f_{Y_{n}}=\frac{1}{n^{2}}$ and $f_{U}= \frac{1}{1-0}=1$. I know that $M_{Y_{n}}(n)= \sum\limits_{n=1}^n e^{nx} \frac{1}{n^{2}}$. This limit goes to $0$ as n goes to infinity right? And since U is continuous $M_{U}(t)= \int_0^\infty e^{tX}dx$. But now I'm confused since the hint says to don't compute the mgf. So how do I proceed now?
You have the right idea, except for a little issue with notation (which is likely to have been the cause of your confusion): you should be looking at a sequence of functions of $t$, i.e. $$M_{Y_n}(t)=\sum_{k=1}^n e^{tk/n} \cdot \frac{1}{n}$$ which does not go to $0$. This is a geometric series which would converge to the mgf of $U$, but to do this without computing the mgf, you can show instead that this sum converges to $M_U(t)=\int_0^1 e^{tx} dx$ (the limits should be the boundaries of your domain, so the upper limit is not $\infty$).
To see this, notice that the $\frac{1}{n}$ plays the role of dx; there are $n$ of these. Then make the substitution $k/n=x$ (in the exponential only). Note that $x$ takes values in $\{\frac{1}{n},...,1\}$, so you should converge to an integral with limits 0 and 1 as required.