<s>[INST] Find the limit of a sequence $p_n=\frac{1}{2^n} \sum_{j=0}^{\frac{a\sqrt{n}-1}{2}}\binom{n}{j}$ Given sequence $p_n=\frac{1}{2^n} \sum_{j=0}^{\frac{a\sqrt{n}-1}{2}}\binom{n}{j}$ where a is some natural number.
Show $\lim_{n \to \infty}p_n=0$
 [/INST] This is a direct consequence of the central limit theorem!
EDITED: With probabilistic intuition, $p_n\rightarrow 0$ is obvious. It is just about tail probability.
Detailed proofs
(Proof using the Chebyshev inequality)
$p_n$ is the probability
\begin{align}
\mathbb{P}\left(X \leq \frac{a\sqrt{n}-1}{2}\right)
\end{align}
where $X$ follows the binomial random variable $B(n,1/2)$. Then, the Chebyshev inequality gives
\begin{align}
\mathbb{P}\left(X \leq \frac{a\sqrt{n}-1}{2}\right) &= \mathbb{P}\left(X-\frac{n}{2} \leq \frac{a\sqrt{n}-1}{2}-\frac{n}{2}\right)\\
&\leq \mathbb{P}\left(\left| X-\frac{n}{2}\right| \geq \frac{n}{2}-\frac{a\sqrt{n}-1}{2}\right)\\
&\leq \frac{n/4}{\left(\frac{n}{2}-\frac{a\sqrt{n}-1}{2}\right)^2} \rightarrow 0.
\end{align}
(Proof using the central limit theorem)
Consider a sequence of i.i.d. Bernoulli random variables $X_1, X_2, \ldots, X_n$. Note that
\begin{align}
p_n &= \mathbb{P}\left(X_1 + X_2 + \ldots +X_n \leq \frac{a\sqrt{n}-1}{2}\right)\\
&= \mathbb{P}\left(\frac{X_1 + X_2 + \ldots +X_n-\frac{n}{2}}{\sqrt{n}} \leq \frac{\frac{a\sqrt{n}-1}{2}-\frac{n}{2}}{\sqrt{n}}\right)
\end{align}
On the other hand, by the central limit theorem, the distribution of
\begin{align}
\frac{X_1 + X_2 + \ldots +X_n-\frac{n}{2}}{\sqrt{n}}
\end{align}
converges to the normal distribution $\mathcal{N}(0,1/2)$. Hence $p_n \rightarrow 0$.
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