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 cdf of $Y_n$ is given by
$$
P(Y_n\le y)=\frac{\lfloor ny\rfloor}n
$$
for $0\le y\le 1$, where $\lfloor\cdot\rfloor$ is the floor function. We have that $\lfloor ny\rfloor/ny\to1$ as $n\to\infty$. Hence, $P(Y_n\le y)\to y$ as $n\to\infty$ which is the cdf of the continuous uniform distribution on $[0,1]$.
Alternatively, we can use the moment generating functions. The moment generating function of $Y_n$ is given by
$$
\frac{e^{t/n}[e^t-1]}{n(e^{t/n}-1)}.
$$
for $t\ne0$. We have that $e^{t/n}\to1$ as $n\to\infty$ and $n(e^{t/n}-1)\to t$ as $n\to\infty$. We obtain
$$
\frac{e^{t/n}[e^t-1]}{n(e^{t/n}-1)}\to\frac{e^t-1}t
$$
as $n\to\infty$ which is the moment generating function of the continuous uniform distribution on $[0,1]$.
