Is it possible to choose $M$ (depend on $n$) such that $\frac{\log(1+M^2)}{\log n} \to 2$ and $n(1-\frac{2}{\pi}\tan^{-1}(M)) \to 0$ as $n\to\infty$
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Yes it is possible. Could you explain what ou tried or, at least, how you approached the problem ? – Claude Leibovici Nov 17 '15 at 08:00
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What i tried did not get me anywhere. Realize that when $n$ is big, the term $1$ has no contribution at all, so I get rid of it to get $2\frac{\log M}{\log n}$. Now, I am trying to figure out a function $M(n)$ that is increasing at most as fast as $n^{1-\epsilon}$ for some $\epsilon>0$ that make the problem works. I got stuck right here. – Yan Wong Nov 17 '15 at 08:44
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note that $\tan^{-1}$ is bounded between $-\pi/2$ and $\pi/2$ – MGP Nov 17 '15 at 09:28
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thank you guys. I solved the problem. In fact, i made a mistake thinking that I should choose $M$ that increase slightly slower than $n$. In fact, I need it increase faster. – Yan Wong Nov 17 '15 at 20:42