$Let\ X_1,\ X_2,\ ...\ be\ independent\ r.v.s\ with\ X_n\ \sim Uniform(-n,\ 3n)\ where\ n\ =\ 1,\ 2,\ ... $
$$Let\ S_N\ =\ \dfrac{1}{\sqrt{N}}\ \begin{align}\sum_{n=1}^{N}\ \dfrac{X_n}{n}\ \end{align}\ for\ N = 1, 2, ..\infty\\ $$ $Let\ F_N\ be\ the\ distribution\ function\ of\ S_N\\ Also\ let\ \phi\ denote\ the\ distribution\ function\ of\ a\ standard\ normal\ random\ variable.$ $$\lim_{N \to \infty} F_N(0)\ \leq\ \phi\ (0)?\ \\ \lim_{N\to\infty} F_N(1)\ \leq \phi\ (1)?$$ $How\ to\ prove\ that\ the\ above\ conditions\ are\ true?$
Let \begin{equation}\tag{1} Y_n = \dfrac{X_n - n}{3n} \end{equation} And by a simple computation we can see, \begin{equation}\tag{2} Y_n\ is\ i.i.d.\ \sim U(-\dfrac{2}{3},\ \dfrac{2}{3}) \end{equation} Also from (1), we can write \begin{equation}\tag{3} X_n = 3nY_n + n \end{equation} Now, \begin{equation}\tag{4} S_N = \dfrac{1}{\sqrt{N}}\ \sum_{n=1}^{N}\ \dfrac{X_n}{n}\\ From\ (3)\ we\ can\ write,\\ = \dfrac{1}{\sqrt{N}}\ \sum_{n=1}^{N}\ \dfrac{n + 3nY_n}{n}\\ = \dfrac{1}{\sqrt{N}}\ \sum_{n=1}^{N}\ 1 + 3Y_n\\ = 3\dfrac{1}{\sqrt{N}}\ \sum_{n=1}^{N}\ Y_n\ +\ \sqrt{N}\\ \end{equation} By Central Limit Theorem, the first term tends to a normal distribution and the whole R.H.S. tends to, $As\ (N\ \rightarrow\ \infty)\\$ \begin{equation}\tag{5} S_N = 3\dfrac{1}{\sqrt{N}}\ \sum_{n=1}^{N}\ Y_n\ +\ \sqrt{N}\ \longrightarrow\ \infty\\ \end{equation} $\because\ S_N\ \rightarrow\ \infty$, it implies
\begin{equation}\tag{6} \lim_{N \to \infty}\ F_N(0) = 0\\ \lim_{N \to \infty}\ F_N(1) = 0 \end{equation}
By definition we know, a standard normal random variable is a normally distributed random variable with mean $\mu$ = 0 and standard deviation $\sigma$ = 1 . It is denoted by the letter Z. From Z-distribution table we know, \begin{equation}\tag{7} P(Z\leqslant 0) = \phi(0) = 0.5\\ P(Z\leqslant 1) = \phi(1) = 0.84134 \end{equation} Hence by comparing (6) and (7) we see both conditions in the question are true.