Im reading through a paper and I came across a statement I don't quite understand. We have a positive real number $W$ and a positive real number $k$. The statement says:
If $x = {\pi(W^{1/k}) \choose k}$ then $x = W^{1 + o(1)}$ by the prime number theorem.
Ive tried toying around with the definition of ${n \choose k}$ while using $\pi(x) \sim x/\log x$ but I cant quite get the relation stated. Any hints or help would be appreciated!
It suffices to show $\frac{\log x}{\log W}\sim 1$. The following inequality can be easily proved using the taylor series for $e^x$. $$\left(\frac{n}{k}\right)^k\leq \binom{n}{k}<\left(\frac{ne}{k}\right)^k$$ Taking $\log_W$ of everything, $$k\frac{\log\left(\frac{n}{k}\right)}{\log W}\leq \frac{\log\binom{n}{k}}{\log W}<k\frac{\log\left(\frac{n}{k}\right)+1}{\log W}$$ Clearly, $k/\log W=o(1)$ in $W$, so $$\frac{\log\binom{n}{k}}{\log W}\sim k\frac{\log\left(\frac{n}{k}\right)}{\log W}$$ Setting $n=\pi(W^{1/k})\sim \frac{k W^{1/k}}{\log W}$, $$\frac{\log x}{\log W}\sim k\frac{\log\left(\frac{n}{k}\right)}{\log W}\sim1-\frac{\log\log W}{\log W}\sim 1$$