# Limits of n!/a^n and n!/n^n

1. Sep 18, 2006

### 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 im 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 doesnt work.

2. Sep 18, 2006

### StatusX

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

3. Sep 18, 2006

### MathematicalPhysicist

how should i compare it to 1/n?
i mean n!/n^n<=n!/n but this doesnt 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 dont see how it will get me to have a constant.

4. Sep 18, 2006

### 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.

5. Sep 18, 2006

### 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, im quite sure i cant 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 doesnt depend on b_n-1, how do i do it?

btw, for 0<a<1 it converges doesnt it?

6. Sep 18, 2006

### StatusX

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.

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.

No, it diverges faster.

Last edited: Sep 18, 2006
7. Sep 18, 2006

### dextercioby

Use Stirling's approximation formula.

Daniel.

8. Sep 18, 2006

### MathematicalPhysicist

but b_n+1/b_n=n+1/a>n/a
but n/a isnt a constant, shouldnt r be a constant?

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?

9. Sep 18, 2006

### 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.

10. Sep 18, 2006

### MathematicalPhysicist

so if r=2 then b_n+1/b_n>=2.
but i dont 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?

11. Sep 18, 2006

### 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).

12. Sep 18, 2006

### MathematicalPhysicist

but then, shouldnt 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?

13. Sep 18, 2006

### 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.