Question

limit n tends to infinity and x sin 2 pi in 2 factorial n factorial is in the argument it's not and there it's an

Answer 1

Using a calculator, I found that n!n! grows substantially slower than nnnn as nn tends to infinity. I guess the limit should be 00. But I don't know how to prove it. In my textbook a hint is given that:

Set an=n!/nnan=n!/nn Set m=[n/2]m=[n/2](floor function), then an≤(1/2)m≤(1/2)n/2an≤(1/2)m≤(1/2)n/2. Then by comparing to the geometric progression, the sequence anan tends to 00.

I have trouble proving the relationship an≤(1/2)m≤(1/2)n/2an≤(1/2)m≤(1/2)n/2, (I tried to prove by considering separate cases, that is when mm is odd and when it is even) using induction gets me nowhere. Or is there other way to prove this limit? I made some search on web and used Stirling's approximation, but to no avail.