Determining convergency of random variable So, let $X_n \in U(0,n)$ and let $Y_n = \frac{X_n}{1-X_n}$, where for all n, $X_n$ and $Y_n$ are supposed to be independent. We need to examine to which random variable this converges.
We examine convergency in distribution, meaning $F_{X_n} \to F_{X}$, n tends to $\infty$, in probability, meaning $\lim_{n \to +\infty} P \{ |X_n-X|>\epsilon \} = 0$, almost sure, meaning $P\{\lim_{n\to \infty} X_n=X\}=1$ and in mid of order two if $\lim_{n \to \infty} E|X_n|^2 = 0$. P is probability, E is mathematical expectation, and F is function of distribution, which means it is linked with probability with the law $F_{X}(t)=P\{X \leq t\}$.
I have tried first to see what I am dealing with, or to what this converges in distribution. What I got is:
$F_{Y_n}(t)=\{1-\frac{1}{n},\ t \in (-1,0)\\
              (\frac{2}{n}-1)\frac{t}{n(1+t)}+1-\frac{1}{n}, t\in (-\infty, -\frac{n}{n-1}) \bigcup [0,+\infty)\\
              \frac{1}{n}, t \in (-\frac{n}{n-1},-1]\}$
Obviously, this converges to a function equal 1 everywhere!(except point 1) so it can not be a function of distribution of any variable, so this does not converge in distribution thus not in any sense. But, am I right is the question for you. 
I defined everything because I never read any probability text in english, so terminology I don't know, and because there are maybe some other types of convergence.

No, I was not right. Watch this:
$\frac{X_n}{1-X-n} = \frac{-X_n}{X_n-1} = \frac{-X_n-1+1}{X_n-1}=\frac{-X_n+1}{X_n-1}-\frac{1}{X_n-1} = -1+\frac{1}{1-X_n}$\
and define:
$Z_n = \frac{1}{1-X_n}$
It is not hard to proove that $Z_n \to 0$ in distribution, so, $X_n \to -1$ in distribution, so in probability(as -1 is constant), and so on.
