Thursday, 10 August 2017

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



Show that ε>0,



limn#{positive divisors of n}nε=0



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 ε by 1k 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.



eO(log(n)loglog(n))=d(n)




So,



limn>d(n)nϵ=eClog(n)loglog(n)nϵ


For some constant C.



Therefore,
limn>eClog(n)loglog(n)nϵ=limn>nCloglog(n)nϵ=limn>nCloglog(n)ϵ



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



limn>Cloglog(n)ϵ=ϵandlimn>nϵ=0implieslimn>nCloglog(n)ϵ=0



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.


No comments:

Post a Comment

real analysis - How to find limhrightarrow0fracsin(ha)h

How to find limh0sin(ha)h without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...