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($$Y_{n}$ $\le$ $u$) = $u$ which is equal to 
$\int_0^u$ 1 dU.
Any suggestions?
 A: The cdf of $Y_n$ is given by
$$
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
$$
as $n\to\infty$ which is the moment generating function of the continuous uniform distribution on $[0,1]$.
A: First, it's necessary to be a bit more careful with the notation. When you say "Is the CDF of $Y_n$ for every integer $n \ge 1$... this is confusing since $n$ indexes a sequence of random variables $Y_n$, and thus should be avoided being used to also denote other arbitrary integers.
To reframe this problem in a clearer context, here we have a sequence of random variables $Y_n$. For each $n$, the probability density changes, and we have $P(Y_n = \frac{i}{n}) = \frac{1}{n}$. So, for each $n$, $Y_n$ is a discrete random variable with uniform probability mass distributed over the finite sample space $\{1, \dots, n\}.$
You are on the right track with the CDF of $Y_n$, for each $n$. The CDF, call it $G_n$ for each $n$, is $$G_n(x) := P(Y_n \le u) = \sum_{k=1}^{\lfloor u \cdot n\rfloor} \frac{1}{n} $$
where $\lfloor u n \rfloor$ is the floor of $u n $, i.e. we round $u n$ down to the nearest integer.
Now we want to consider if this sequence of CDFs, $G_n$, converges to the CDF of a uniform $[0,1]$ random variable. We will denote the CDF of a standard uniform random variable as $F$. As you mentioned, $F(x) := \int^x_0 1 dt.$ By the limit definition of a definite integral, we have
\begin{align}
F(x) &= \int^x_0 1 dt \\\ &= \lim_{n\rightarrow\infty} \sum^n_{i=1} 1 \cdot \Delta x_i \\\
&= \lim_{n\rightarrow\infty} \sum^n_{i=1} 1 \cdot \frac {1}{n} \\\
&= \lim_{n\rightarrow\infty} \sum_{k=1}^{\lfloor u \cdot n \rfloor} \frac{1}{n}
\end{align}
where $\Delta x_i$ is the difference between points in any partition $x_0 = 0 < x_1 < \dots < x_{n-1} < x_n = 1$ of $[0,1]$. Clearly $x_0 = 0 < x_1 = \frac {1}{n} < \dots < x_{n-1} = \frac {n-1}{n} < x_n = \frac{n}{n}$ is such a partition, with difference $\Delta x_i = \frac {1}{n}$. Therefore we have desired CDF convergence.
A: 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}
