A friend left this teaser for me. He asked me to first compute:
$$ \lim_{n \to \infty} \frac{\binom{2n}{n}}{2^{2n}}$$
Using Stirling's approximation (and another method), I got the answer as $0$. Then he gave the following "proof" for me to check:
Let $X_i$ be iid Bernoulli$(1/2)$. Then we have
$$ P\{\text{n out of 2n $X_i$ are 1} \} = P\{ \sum_{i=1}^{2n}X_i = n\} = \frac{\binom{2n}{n}}{2^{2n}}$$ $$ = P\left\{\frac{\sum_{i=1}^{2n}X_i}{2n} =\frac{1}{2}\right \}$$ $$ = P\left\{\frac{\sum_{i=1}^{2n}(X_i - \frac{1}{2})}{2n} = 0\right \}$$
Denote $E_n := \frac{\sum_{i=1}^{2n}(X_i - \frac{1}{2})}{2n}$
He argues according to Weak law of large numbers, $\forall \epsilon > 0$ $$ \lim_{n \to \infty} P(|E_n| < \epsilon) = 1 \Rightarrow \lim_{n \to \infty} P(|E_n| =0) = 1 \Rightarrow \lim_{n \to \infty} P(E_n =0) = 1$$
Hence the limit must be $1 \neq 0$.
I, for some strange reason, was unable to refute his proof although I suspect that the bug is in the last step somewhere. I would appreciate any help/hints on this.
$\lim_{n\to \infty} P(|E_n| < \epsilon) = 1$ does not imply $\lim_{n\to \infty}P(|E_n| = 0)= 1$
Take deterministic $E_n = \frac{1}{n}$ for example.
To answer your question in the comment:
Let $E_n$ be the sum of i.i.d $X_i$. If $X_i$ are continuous variables, then $P(E_n = 0) = 0, \forall n$, thus $\lim_n P(E_n = 0) = 0$
If $EX_i = 0$ and $E(X_i^2) = \sigma^2 < \infty$, then $P(E_n = 0) = P(\frac{E_n}{\sqrt{n}} = 0) \to P(\mathbb{N}(0, \sigma^2) = 0) = 0$ by central limit theorem(Remark that the cdf of standard normal distribution is continuous everywhere, thus we have the pointwise convergence of cdf)
We don't always have a zero limit, since if $X_i = 0$ deterministically, then the limit is $1$.
Btw, since Bernoulli is of finite variance, now we have a true probabilistic proof