Approximate solution: factorial and exponentials

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.

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.

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*}