Showing that a sequence of random variables with $\mathcal{L}(X_{n})$ (I.e., law) uniform on $[-n,n]$ does not converge at all Let $(X_{n})_{n\geq 1}$ be a sequence of real valued random variables with $\mathcal{L}(X_{n})$ (that is, law or distribution) uniform on $[-n,n]$. In what sense(s) do $X_{n}$ converge to a random variable $X$?
We are told that the answer is none. To prove this, are we supposed to show that $X_{n}$ does not converge under the weakest form of convergence possible (convergence in distribution/law)? And if we do, is that sufficient to show that there is no convergence? Also, how does the fact that the r.v.'s are uniformly distributed give us that?

Convergence in distribution means
$$\lim_{n\rightarrow \infty}F_{X_n}(x) = F_X(x)$$
where the RHS is a distribution function, and the equality to hold for every $x$ for which  $F_X(x)$ is continuous.
Our distribution function is  
$$F_{X_n}(x)= \begin{cases}0&\text{if $x<-n$}\\\frac{x+n}{2n}&\text{if $x\in[-n,n]$} \\1&\text{if $x> n$.}\end{cases}$$
The first and the last branch are not defined as $n\rightarrow \infty$, since there is no $x$ lower than "minus infinity", or higher than "plus infinity". Then
$$\lim_{n\rightarrow \infty}F_{X_n}(x) = \lim_{n\rightarrow \infty}\Big (\frac{x}{2n} + \frac 12\Big) = 0+\frac 12$$
The constant function $1/2$ does not satisfy the properties of a distribution function, specifically
$$\lim_{x\rightarrow -\infty}\frac 12 =\frac 12 \neq 0,\;\;\lim_{x\rightarrow \infty}\frac 12 =\frac 12 \neq 1$$
Carefully note that $n$ and $x$ do not "go together" at plus/minus infinity. To study convergence we first "send $n$ to infinity", and then we examine the behavior of the limiting function we have obtained, as its argument passes over.
