Thursday, 10 August 2017

real analysis - Show forallvarepsilon>0,lim_{ntoinfty}frac{#{text{positive divisors of
n}}}{n^varepsilon}=0



Show that ε>0,



lim



I'm trying to solve this problem for a long time, but I'm really stuck I have totally no idea where to start.
I tried replacing \varepsilon by \frac{1}{k} where k is a natural number, and show the statement is true for all k by induction, but I haven't succeeded and it doesn't seem promising.




If you give me any advice or comment, I would greatly appreciate.



Thank you.


Answer



The solution for this problem falls out from the following fact.



\begin{equation} e^{O(\frac{log(n)}{loglog(n)})}=d(n) \end{equation}




So,



\begin{equation} \lim_{n->\infty}\frac{d(n)}{n^\epsilon}=\frac{e^{\frac{Clog(n)}{loglog(n)}}}{n^\epsilon} \end{equation}
For some constant C.



Therefore,
\begin{equation} \lim_{n->\infty}\frac{e^{\frac{Clog(n)}{loglog(n)}}}{n^\epsilon}=\lim_{n->\infty}\frac{n^\frac{C}{loglog(n)}}{n^\epsilon}=\lim_{n->\infty}n^{\frac{C}{loglog(n)}-\epsilon} \end{equation}



You can add a nice proof regarding epsilon and deltas to show from here that



\begin{equation} \lim_{n->\infty}\frac{C}{loglog(n)}-\epsilon=-\epsilon\\and\\ \lim_{n->\infty}n^{-\epsilon}=0\\ implies\\ \lim_{n->\infty}n^{\frac{C}{loglog(n)}-\epsilon}=0 \end{equation}



This completes our proof. Many websites have the first point written and I believe you may be able to find out more about it from other posts.


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