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−>∞n−ϵ=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.
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