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I have a function $f(k)$ defined on the set of natural numbers and I managed to show that $f(k)>n-\binom n k(1-n^{-2/3})^{k(k-1)/2}$ for all integers $n\ge k$. I am hoping to get a further estimation that $f(k)>(\frac{k}{3\log k})^{3/2}$ for large $k$.

I am not sure how to do this, but with a graph plotter, I set $n=k^{3/2}$ and from the asymptotic behavior of $(k^{3/2}-\binom {k^{3/2}} k(1-k^{-1})^{k(k-1)/2})/(\frac{k}{3\log k})^{3/2}$ in the graph it seemed to work.

But I couldn't work out how to prove it.

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What is $t$ here? –  Hagen von Eitzen Jan 21 '13 at 15:36
    
Sorry it was me being completely stupid. That was a typo. –  Montez Jan 21 '13 at 15:37
    
I think a better way to go is to find the minimum of $\binom n k(1-n^{-2/3})^{k(k-1)/2}$ to get an upper bound for the RHS and therefore $f(k)$. This could give a scale factor for $n$ in terms of $k$. I tried playing around with this, but I ended up taking derivatives using Sterling's approximation, and I am not sure how valid that is. Also, the resulting equation, even if it is valid, looks painful to solve. It could be used to derive a simple scale factor, though. –  Ron Gordon Jan 21 '13 at 16:04

1 Answer 1

No, it's not true. With $n \approx k^{3/2}$, $n - {n \choose k} (1 - n^{-2/3})^{k(k-1)/2} < 0$ for large $k$ (in fact for any $k \ge 4$).

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