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 $\lim_{n \to \infty}p_n=0$
 A: 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 $p_n \rightarrow 0$.
A: As in my comment, we have $\binom{n}{j} \leq n^j$, so
$$p_n = 2^{-n} \sum_{j=0}^{\frac{a \sqrt{n} - 1}{2}} \binom{n}{j} \leq 2^{-n} \sum_{j=0}^{a\sqrt{n}} n^j \leq 2^{-n} \cdot n \cdot n^{a\sqrt{n}} = 2^{-n + (a\sqrt{n} + 1)\log_2 n}$$
but $(a\sqrt{n} + 1) \log_2 n = o(n)$, so $-n + (a\sqrt{n} + 1)\log_2 n \to -\infty$ and hence $p_n \to 0$.
A: Hint: $$(1+x)^n=\sum_{j=0}^{n}x^j\binom{n}{j}$$
Put x= 1
Now your  limit becomes
 $$L<\lim_{n \to \infty} 2^{{a\sqrt n -1 \over 2}-n}$$ or 
$L</=0$. and Observe L can't be negative so L=0.
A: This is true for
$n^a-1$ for
$0 < a < 1$.
More generally, let
$p_n
=\frac{1}{2^n} \sum_{j=0}^{f(n)}\binom{n}{j}
$
where
$\dfrac{f(n)\ln(n)}{n}
\to 0
$
(you will see where the
$\ln(n)$ comes from later).
$\binom{n}{j}
=\dfrac{\prod_{i=0}^{j-1} (n-i)}{j!}
\lt\dfrac{\prod_{i=0}^{j-1} n}{j!}
=\dfrac{n^j}{j!}
$
so
$\begin{array}\\
p_n
&=\frac{1}{2^n} \sum_{j=0}^{f(n)}\binom{n}{j}\\
&<\frac{1}{2^n} \sum_{j=0}^{f(n)}\dfrac{n^j}{j!}\\
&<\frac{n^{f(n)}}{2^n} \sum_{j=0}^{f(n)}\dfrac{1}{j!}\\
&<ee^{f(n)\ln(n)-n\ln(2)}\\
&\to 0\\
\end{array}
$
since
$f(n)\ln(n) < cn
$ for any $c > 0$
so
$f(n)\ln(n)-n\ln(2)
\lt n(c-\ln(2))
\to -\infty
$
for $c < \ln(2)
$.
