Calculus help

[vtqanh]

How do you prove that lim (r^n)/n! = 0 as n goes to infinity? (with r is a finite real number)

 

[AbsolutDealage]

YADMHFMT

Yet another do my homework for me thread. When will you people learn?

 

[Analog]

Good thing you're not one of my students!

 

[EngenZerO]

follow the models that they use in the chapter...that is how I did it

 

[DrPizza]

Originally posted by: vtqanh
How do you prove that lim (r^n)/n! = 0 as n goes to infinity? (with r is a finite real number)

It depends on the type of proof you're talking about. Do you want a rigorous proof, or just simply "show that..."

 

[Heisenberg]

I know how. I'm not going to tell, but I know how.

 

[vtqanh]

Originally posted by: DrPizza

Originally posted by: vtqanh
How do you prove that lim (r^n)/n! = 0 as n goes to infinity? (with r is a finite real number)

It depends on the type of proof you're talking about. Do you want a rigorous proof, or just simply "show that..."

a rigorous one.

 

[vtqanh]

Originally posted by: AbsolutDealage
YADMHFMT

Yet another do my homework for me thread. When will you people learn?

What are you talking about? This one was one of the question in my quiz this morning.

 

[SackOfAllTrades]

Originally posted by: vtqanh

Originally posted by: AbsolutDealage
YADMHFMT

Yet another do my homework for me thread. When will you people learn?

What are you talking about? This one was one of the question in my quiz this morning.

Vtqanh, don't worry about it...at least you didn't come up with an idiotic abbreviaton for academic threads.

 

[johnjohn320]

Originally posted by: AbsolutDealage
YADMHFMT

Yet another do my homework for me thread. When will you people learn?

When kind folks help teach them.

 

[GtPrOjEcTX]

intuitively, n! gets larger quicker than r^n. so with lim -> infinity, the denominator gets larger and the value goes to 0.

but you wanted the rigorous version...

 

[vtqanh]

Originally posted by: GtPrOjEcTX
intuitively, n! gets larger quicker than r^n. so with lim -> infinity, the denominator gets larger and the value goes to 0.

but you wanted the rigorous version...

lol. it's obvious, but it needs to be proven.

 

[paulney]

I haven't done this in ages, but here's how I would do it:

Given: lim (a^n/ n!) where n -> inf
use Lhopital (did I spell it right in English?) theorem:

(a^n/ n!)' = (n * a^(n-1)) / ( can't recall derivative of factorial here))

then you will be able to cancel out a bunch of stuff and show that factorial grows much faster than exp function, thus denominator grows to infinity and the whole fraction reaches zero.

 

[RossGr]

I think you can do this by comparison.
Edit
opps pardon me I miss read the problem.!


rn/n! = r/(n-1)!<r/n

lim n->inf of 4/n =0
Since r/(n-1)! < r/n for all n It also must go to zero.

 

[her209]

The growth of r^n is slower than n!

 

[niwi7]

a + b = c

 

[alkemyst]

You can't solve this, everyone knows a calculator gives an out of range error

 

[fawhfe]

Hehe, you could be silly and use stirlings approx. Rewrite the problem as e^ln(r^n)/e^ln(n!). This is the same as the original problem since e^(lnx)=x. Then it becomes e^(n*lnr)/e^(n*lnn-n) with the denominator coming from stirlings approx (states that ln(n!) is approx n*ln(n)-n), which holds since were talking about n->infinity. Since n->infinity, n will be insignifigant compared to n*lnn, so we discard the -n part (of the exponent in the denominator). Then we have:

e^(n*lnr)/e^(n*lnn)
=e^(n*(lnr-lnn))
In the limit n->infinity, we get e^(infinity*(constant-infinity)) which is just e^(-infinity)=0. I dont know if you know Stirling's or whether the approximation illegitimizes the proof for you (though it really shouldnt in the limit n->infinity). Considering the nature of your question, I bet using Stirling's would probably make your professor ask what drugs you're smoking, but it would probably blow his mind anyways and I would do it just for that. Hopefully someone will come up w/ a more reasonable way though that will actually get you credit.

 

[vtqanh]

Originally posted by: paulney
I haven't done this in ages, but here's how I would do it:

Given: lim (a^n/ n!) where n -> inf
use Lhopital (did I spell it right in English?) theorem:

(a^n/ n!)' = (n * a^(n-1)) / ( can't recall derivative of factorial here))

then you will be able to cancel out a bunch of stuff and show that factorial grows much faster than exp function, thus denominator grows to infinity and the whole fraction reaches zero.

Nope, can't use L'Hospital. Since this is advanced calculus, you have to derive everything from scratch. L'Hospital is way beyond this one

 

[RIGorous1]

Originally posted by: vtqanh

Originally posted by: paulney
I haven't done this in ages, but here's how I would do it:

Given: lim (a^n/ n!) where n -> inf
use Lhopital (did I spell it right in English?) theorem:

(a^n/ n!)' = (n * a^(n-1)) / ( can't recall derivative of factorial here))

then you will be able to cancel out a bunch of stuff and show that factorial grows much faster than exp function, thus denominator grows to infinity and the whole fraction reaches zero.

Nope, can't use L'Hospital. Since this is advanced calculus, you have to derive everything from scratch. L'Hospital is way beyond this one

ahahh

use delta epsilon proofs!

 

[vtqanh]

Originally posted by: fawhfe
Hehe, you could be silly and use stirlings approx. Rewrite the problem as e^ln(r^n)/e^ln(n!). This is the same as the original problem since e^(lnx)=x. Then it becomes e^(n*lnr)/e^(n*lnn-n) with the denominator coming from stirlings approx (states that ln(n!) is approx n*ln(n)-n), which holds since were talking about n->infinity. Since n->infinity, n will be insignifigant compared to n*lnn, so we discard the -n part (of the exponent in the denominator). Then we have:

e^(n*lnr)/e^(n*lnn)
=e^(n*(lnr-lnn))
In the limit n->infinity, we get e^(infinity*(constant-infinity)) which is just e^(-infinity)=0. I dont know if you know Stirling's or whether the approximation illegitimizes the proof for you (though it really shouldnt in the limit n->infinity). Considering the nature of your question, I bet using Stirling's would probably make your professor ask what drugs you're smoking, but it would probably blow his mind anyways and I would do it just for that. Hopefully someone will come up w/ a more reasonable way though that will actually get you credit.

Stirling's approximation is way beyond this problem. When Stirling did his theory, this had been proven.

 

[vtqanh]

Originally posted by: RIGorous1

Originally posted by: vtqanh

Originally posted by: paulney
I haven't done this in ages, but here's how I would do it:

Given: lim (a^n/ n!) where n -> inf
use Lhopital (did I spell it right in English?) theorem:

(a^n/ n!)' = (n * a^(n-1)) / ( can't recall derivative of factorial here))

then you will be able to cancel out a bunch of stuff and show that factorial grows much faster than exp function, thus denominator grows to infinity and the whole fraction reaches zero.

Nope, can't use L'Hospital. Since this is advanced calculus, you have to derive everything from scratch. L'Hospital is way beyond this one

Well
ahahh

use delta epsilon proofs!

well, if you can use epsilon definition to prove this, it should be the best. However, we have proven all the limit algebra (addition, multiplication, multiplication with constant, division). And also, some of the common limits have also been proven (1/n -> 0, etc...)

 

[Rob9874]

Ah.. I loved Calculus until Infinite Series. That was the main reason I never went past Calc 2 in college. Good luck.

 

[fawhfe]

Hmm... Well you could just prove that given f(n)=r^n/n!, f(n)/f(n-1)<1 for all n>M where M is any non-infinite number. That should be simple enough since you just get (r^n/n!)/(r^(n-1)/(n-1)!) which is just r/n->M=r. Then for all n>M, you end up multiplying the previous term by something less than 1 (r/n to be exact) and as you do this n->infinity times, you arrive at 0. Theres probably a more eloquent way of putting it, but that should work.

Edit--yea, look at post below. If you can cite ratio test, its easy, otherwise its a few lines of explanation. I still think Stirling's is pretty cool though and I can't imagine your prof wants your proofs to be historically accurate . Oh well... You get credit this way.

 

[agnitrate]

Ratio test.

[edit] for some reason i thought you were doing a series, anyways nevermind [/edit]

-silver