How Do You Prove That Limits of Factorials Over Exponentials Approach Zero?

[MathematicalPhysicist]

i forgot how to prove that lim n!/a^n=0 lim n!/n^n=0 as n appraoches infinity.(a>1)
i mean obviously i need to use the sandwich theorem:
0<=n!/a^n
0<=n!/n^n
but I am haviing difficulty to find a good upper bound to use the theorem.
in the first case i thought to use the inequality:
a=1+h (h>0) (1+h)^n>=hn
and thus n!/a^n=n!/(1+h)^n<=n!/(hn)=(n-1)!/h
but it doesn't work.

thanks in advance.

 

[StatusX]

n!/a^n -> infinity as n -> infinity, and for the other one, compare it to 1/n.

 

[MathematicalPhysicist]

how should i compare it to 1/n?
i mean n!/n^n<=n!/n but this doesn't help.
as for n!/a^n i need to show that for every constant, there exists n0 such that for every n>=n0 n!/a^n>=M, but how do i choose the constant M?
i think that i need to use the substitution a=1+h where h>0 but i don't see how it will get me to have a constant.

 

[StatusX]

Try expanding out n!/n^n and see if you get any ideas. For the other one, look at the ratio of successive terms.

 

[MathematicalPhysicist]

then if i write:
n!/n^n=(n/n)(n-1)/n)...(1/n)
lim n/n=1
lim (n-1)/n=1
.
.
.
lim 1/n=0
and thus lim n!/n^n is 0, right?

about the other one, I am quite sure i can't use here delambert criterion for convergence because it follows only for sums.
anyway, (n+1)!/a^(n+1)*(a^n)/n!=(n+1)/a
so if we let: b_n=n!/a^n then b_n+1/b_n=(n+1)/a, then this fraction diverges, and it means that b_n+1/b_n>=M for every constant M>0
b_n+1>=Mb_n or b_n>=Mb_n-1
but now i need to show that b_n is bigger than some constant which doesn't depend on b_n-1, how do i do it?

btw, for 0<a<1 it converges doesn't it?

 

[StatusX]

then if i write:
n!/n^n=(n/n)(n-1)/n)...(1/n)
lim n/n=1
lim (n-1)/n=1
.
.
.
lim 1/n=0
and thus lim n!/n^n is 0, right?

Does that look right? What do you have in the in between region marked by ...? (don't try to answer this). And you should know better than to break up a limit into limits of factors like that. Like I said at the beginning, just try to bound the sequence above.

about the other one, I am quite sure i can't use here delambert criterion for convergence because it follows only for sums.
anyway, (n+1)!/a^(n+1)*(a^n)/n!=(n+1)/a
so if we let: b_n=n!/a^n then b_n+1/b_n=(n+1)/a, then this fraction diverges, and it means that b_n+1/b_n>=M for every constant M>0
b_n+1>=Mb_n or b_n>=Mb_n-1
but now i need to show that b_n is bigger than some constant which doesn't depend on b_n-1, how do i do it?

If the ratio between terms gets bigger than some r>1, you can bound the sequence below by the divergent sequence a*r^n for the right choice of a.

btw, for 0<a<1 it converges doesn't it?

No, it diverges faster.

 

[dextercioby]

Use Stirling's approximation formula.

Daniel.

 

[MathematicalPhysicist]

but b_n+1/b_n=n+1/a>n/a
but n/a isn't a constant, shouldn't r be a constant?

about the other limit:
n!/n^n=(n/n)(n-1/n)...(1/n)=1(1-1/n)(1-2/n)...(1/n)<=1*1*1...*1*1/n
is this correct?

 

[StatusX]

Don't set r=n/a, set r to any number greater than 1. Then eventually the original series will start growing faster than the geometric series a*r^n, and so after some n it will overtake it. This is true for any a, but if you pick a and r right you can make finding the n easier. Your answer to the second one looks right now.

 

[MathematicalPhysicist]

so if r=2 then b_n+1/b_n>=2.
but i don't understand how from b_n+1/b_n>=2=r i can conclude that
"... Then eventually the original series will start growing faster than the geometric series a*r^n...", why is that?

 

[StatusX]

The ratio between successive terms in the sequence a*r^n is just r. Intuitively, if the ratio between successive terms in one sequence is greater than in another, it is growing faster. Of course, you need to prove this, and induction is one way (start at the point the faster growing series first overtakes the other, and then show it remains greater than it after this point).

 

[MathematicalPhysicist]

but then, shouldn't i be setting a to a specific number, and i have here for any a constant, wouldn't the choice of a will impact the generality of the proof?

 

[StatusX]

If a is larger it will just take longer for the faster growing series to overtake the slower one, but it will happen eventually.

Find the point where the ratio between consecutive terms first becomes greater than 1. If the ratio is r here, then the ratio is greater than r for all terms after this point. Now if you have a geometric sequence a*r^n, try to pick a so that you can easily compare these two series after this point.