Getting two answers using central limit and weak law of large numbers Let $X_i$ be i.i.d random variables each with mean 1. let $S_n=\sum_{i=1}^{n}X_i$. I have to calculate the probability $P(S_n \leq n)$ as $n$ tends to $\infty$. Now using central limit theorem I am getting answer as $\frac{1}{2}$. But by using weak law of large number the solution comes as $1$. I think I am doing some mistake in the second procedure. The steps for the second procedure are as follows : 
$$P(S_n \leq n)= P(\frac{S_n}{n} \lt 1+\epsilon)=P(\frac{S_n}{n} -1 \lt \epsilon) \geq P(|\frac{S_n}{n}-1| \lt \epsilon) \geq 1$$ when n goes to infinity. Where am I wrong ?
 A: The issue is that your first equality, that 
$P(S_n \le n) = P(\frac{S_n}{n} < 1 + \epsilon)$, is incorrect.
For any fixed epsilon, the right-hand side $P(\frac{S_n}{n} < 1 + \epsilon)$ actually is the probability $P(S_n < n + n\epsilon)$ and is not equal to $P(S_n \le n)$, because $n\epsilon \to \infty$ as $n$ goes to infinity, so you're actually looking at the probability that $S_n$ lies in a much larger range $(−\infty, n + n\epsilon)$ rather than the range $(−\infty, n]$ that $P(S_n \le n)$ corresponds to. 

In fact $P(S_n < n + n\epsilon) = 1$ just as you have proved, and this fact is no surprise in light of the central limit theorem either, because what the central limit theorem says is that $|S_n − n|$ is of the order of $\sqrt{n}$, not $n$. More precisely, we have 
$$P\left(\frac{S_n}{n} < 1 + \epsilon\right) = P(S_n < n + n \epsilon) = P\left(\frac{S_n-n}{\sqrt{n}} < \epsilon\sqrt{n}\right)$$
For any real number $c$ we have $\epsilon\sqrt{n} > c$ for sufficiently large $n$, and therefore (assuming that the $X_i$s have finite variance say $\sigma$, for the CLT to be applicable)
$$\lim_{n\to\infty}P\left(\frac{S_n-n}{\sqrt{n}} < \epsilon\sqrt{n}\right) \ge \lim_{n \to \infty} P\left(\frac{S_n-n}{\sqrt{n}} < c\right) = \Phi(c/\sigma)$$
which can be made arbitrarily close to $1$ by picking sufficiently large $c$.
A: 
The steps for the second procedure are as follows : $$P(S_n \leq n)= P(\frac{S_n}{n} \lt 1+\epsilon)=P(\frac{S_n}{n} -1 \lt \epsilon) \geq P(|\frac{S_n}{n}-1| \lt \epsilon) \geq 1$$ when n goes to infinity. Where am I wrong ?

At two places: the first = sign is wrong since the events $[S_n\leqslant n]$ and $[S_n/n\lt1+\varepsilon]$ do not coincide in general (should be $\leqslant$ instead) and the last $\geqslant1$ sign is wrong and should be replaced by something like $\lim\limits_{n\to\infty}\cdots=1$.
A correct proof is as follows. Define some random variable $Z_n$ for each $n\geqslant1$ by the identity $S_n=n+\sigma(X_1)\cdot \sqrt{n}\cdot Z_n$, then the central limit theorem asserts that $Z_n\to Z$ in distribution, where $Z$ is standard normal. In particular, $P[Z_n\leqslant0]\to P[Z\leqslant0]$. Since $P[Z\leqslant0]=\frac12$ and $[Z_n\leqslant0]=[S_n\leqslant n]$, this yields
$$
\lim_{n\to\infty}P[S_n\leqslant n]=\tfrac12.
$$
