Approximate solution: factorial and exponentials If z= $\dbinom{200}{100}/(4^{100})$, what is the value of z?
The options are:
a. $z<1/3$
b. $1/3<z<1/2$
c. $1/2<z<2/3$
d. $2/3<z<1$
How should I go about solving these type of problems?
 A: Stirling's approximation gives $$\binom{200}{100} \approx \frac{2^{200}}{10\sqrt{\pi}},$$
so $\frac{1}{4^{100}} \binom{200}{100} \approx \frac{1}{10 \sqrt{\pi}}\approx 0.0564$ which is quite small, so a) seems appropriate.
Indeed, $\frac{1}{4^{100}} \binom{200}{100} \approx 0.0563$, so Stirling's approximation is not bad here.
A: Here is an elementary approach without estimating the value of the expression:
\begin{eqnarray*} \frac{1}{4^m}\binom{2m}{m}
& = & \frac{1}{4^m}\cdot \frac{\prod_{i=1}^m 2i \cdot \prod_{i=1}^m (2i-1) }{(m!)^2} \\
& = & \frac{1}{4^m}\cdot 4^m\frac{\prod_{i=1}^m i \cdot \prod_{i=1}^m \left(i-\frac{1}{2}\right) }{\prod_{i=1}^m i \cdot \prod_{i=1}^m i} \\
& = & \prod_{i=1}^m \left( 1 - \frac{1}{2i} \right) \\
& = & \prod^m_{i=1} \frac{2i-1}{2i} \\
& \stackrel{m=100}{<} & \frac{1}{2}\cdot\frac{3}{4}\cdot \frac{5}{6} \\
& = & \frac{5}{16}\\
& < & \frac{1}{3}\\
\end{eqnarray*}
A: If we simply want to know that $z < 1/3$, then this can be verified by very elementary means.  Note that $C(200,99) = \frac{100}{101} C(200,100) > 0.99\cdot  C(200,100),$ so then $$C(200,99) + C(200,100) + C(200,101) > 2.98 \cdot C(200,100).$$
On the other hand, $$4^{100} = 2^{200} = C(200.0) + C(200,1) + \cdots + C(200,200),$$ which is clearly greater than just the three middle terms.  So $C(200,100)/4^{100}$ is certainly less than $1/2.98$, which is awfully close to $1/3$ already.  We just need to estimate one more term like $C(200,98)$ to bring the bound below $1/3$, and the estimate need not be very precise at all.
A: Using the bounds in $(10)$ from this answer:
$$
\frac{4^n}{\sqrt{\pi\left(n+\frac13\right)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi\left(n+\frac14\right)}}
$$
This gives
$$
\frac{\binom{200}{100}}{4^{100}}\lt\frac1{\sqrt{100\pi}}\lt\frac1{17}
$$
A: Consider the normal distribution $N$ with mean $100$ and variance $50$, which approximates the binomial distribution $B$ on $200$ trials with probability $0.5$.
$$P(B=100)=\binom{200}{100}\frac1{4^{100}}=z$$
$$\approx P(N\in[99.5,100.5])=P\left(|Z|\le\frac{0.5}{\sqrt{50}}\right)\approx \frac1{\sqrt{50}}f_Z(0)\approx\frac{0.4}7=0.05714\dots$$
The actual value is $0.05634\dots$, so our approximation is good and (a) is correct.
A: From the identity $\int_{-\pi}^{\pi} \cos((2m-n)x)\cos^n x\ \mathrm{d}x = \frac{\pi}{2^{n-1}} \binom{n}{m}$, with $n=200$ and $m=100$, we have $z=\frac{1}{2\pi}\int_{-\pi}^{\pi}\cos^{200} x\ \mathrm{d}x$. Using the substitution $u=\cos(x)$ and simplification of the bounds of the integral, we find 
$$z=\frac{2}{\pi}\int_0^1\frac{u^{200}}{\sqrt{1-u^2}}\mathrm{d}u$$
We can bound the integrand, for $0<u<1$, with $\frac{u^{200}}{\sqrt{1-u^2}}<\frac{u^8}{\sqrt{1-u^2}}$. Therefore, $$\int_0^1\frac{u^{200}}{\sqrt{1-u^2}}\mathrm{d}u<\int_0^1\frac{u^8}{\sqrt{1-u^2}}\mathrm{d}u$$
Solving the integral on the right-hand-side tells us $\frac\pi2z<\frac{35\pi}{256}$. So, finally, $z<\frac{35}{128}<\frac13$, which gives us the answer.
We can also expand the integrand $\frac{u^{200}}{\sqrt{1-u^2}}$ with the binomial series and integrate termwise to find a series for $z$. However, unfortunately, the convergence is very slow and seems to be only linear.
$$\begin{aligned}z&=\frac{2}{\pi}\sum_{i=0}^\infty\frac{|\binom{-1/2}{i}|}{{2i+201}}\end{aligned}$$
