CDF of a random variable Consider a random variable $Y_{n}$ such that,
$P$($Y_{n}$ = $\frac{i}{n}$) = $\frac{1}{n}$ , $i$ = 1, 2,..., $n$.
Is the cdf of $Y_n$ for every integer $n \ge 1$ simply $\sum_{k=1}^i \frac{1}{n}$ = $\frac{i}{n}$?  
Also, how do you show that for any $u$ $\in$ $R$  the $\lim_{n\to\infty} P($$Y_{n}$ $\le$ $u$) = P(U $\le$ $u$) 
Where U is uniform $[0,1]$.
My thought was that if $P($$Y_{n}$ $\le$ $\frac{i}{n}$) = $\frac{i}{n}$, then $P($$Y_{n}$ $\le$ $u$) = $u$ which is equal to 
$\int_0^u$ 1 dU.
Any suggestions?

The cumulative distribution function (CDF) is obtained by integration of the density:
\begin{aligned}
F_{Y_n}(t) & = \int_{-\infty}^t \sum_{i=1}^n P\!\left(Y_n=\frac{i}{n}\right) \delta_{i/n}(x) \, dx \\
& = \underset{i\leq n t}{\sum_{i=1}^n} \frac{1}{n} \\
& = \frac{1}{n} \min(n,\max (1,\lfloor n t \rfloor)) \, ,
\end{aligned}
where $\delta$ is the Dirac delta, and $\lfloor \cdot \rfloor$ is the floor function. When $n$ goes to infinity, this CDF tends towards the CDF of the uniform distribution over $[0,1]$:
\begin{aligned}
\lim_{n\rightarrow +\infty} F_{Y_n} (t) & = \min(1,\max (0, t )) \\
& = F_U (t) \, .
\end{aligned}
