The definition of Big-O is $$u_n = O(v_n) \iff (\exists c \in \mathbb{R}^{*})\,\, (\exists N \in \mathbb{N}) \,\, n > N \implies u_n < c \, v_n$$


Based on that I am trying to find the upper bound for this function $\log(\binom{n}{\log n})$, given that $\binom{n}{k} \le \frac{n^k}{k!} \le (\frac{n\cdot e}{k})^k$

What I've done so far

What I've done so far is $\log(\binom{n}{\log n}) \le \log\left(\left(\frac{n\cdot e}{\log n}\right)^{\log n}\right) $ but it doesn't seem to lead anywhere.

Any ideas?

  • 1
    $\begingroup$ Use Stirling's fomrula $\endgroup$ – Mike Hawk Apr 2 at 14:42

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