Asymptotic analysis : Theory

Question

how do you prove that when the limit of n approaches towards positive infinity while n^2/(log n)! We tried to used Stirling theorem but this may not work due to the fact that it may or may not exist on all intervals.

Answer 1

If $n$ is a power of $2$, write $n=2^k$ and apply Stirling's approximation to $k!$ to show that $$\begin{align} \frac{n^2}{\log^2 n} &= \frac{2^{2k}}{k!} \sim_{n\to\infty} \frac{2^{2k}}{\sqrt{2\pi k}\left(\frac{k}{e}\right)^k} = \frac{1}{\sqrt{2\pi k}}\cdot \frac{2^{(2+\log e)k}}{2^{k\log k}} = 2^{- k\log k + (2+\log e)k - \frac{1}{2}\log k + O(1) } \\ &= 2^{-k\log k + o(k\log k)}\xrightarrow[n\to\infty]{} 0. \end{align}$$

If $n$ is not a power of two, then $(\log n)!$ should be interpreted as $\Gamma(\log n+1)$. Then use $\Gamma(\log n+1) \geq \Gamma(\lceil \log n \rceil)$ (the $\Gamma$ function is increasing) and do the above for $k = \lceil \log n \rceil$.