Math Help - Cal 2

[neil7818]

---

Members don't see this ad.

Hi everyone,

I needed some quick help for a math problem and didnt know where to go.

Can someone prove why n!/3^n has no upper boundary? It's in regards to monotone sequences from Calculus 2.

It's probably very easy I'm just very slow on the uptake. Any and all help would be appreciated (the sooner the better though), I'll be up til like 4-5 so feel free to PM or respond, you'd be a life saver!

Here's the problem again - prove why "n factorial" divided by "Three to the N" has no upper boundary.

Anyone up for a challenge?

Thanks again,
Neil

 

[Nuel]

This is rather conceptual. First write the equation as n!((1/3)^n). The sequence { (1/3)^n} should have a limit that exists since its series exists (geometric here) and consequently an upper bound. However n! just keeps increasing and does not converge (the series), and also the sequence. Therefore one can intuit that the infinite sequence {n!(1/3)^n} has no upper bound since |x| < 1.
If you write the sequence as a continuous function f(x), where x is greater than 0, the limit of the function as x approaches infinity should approach infinity.

I tried to think up an harcore proof for this thing, but my memory is "weak" on the sequence and series thing.
Blessings upon you.

 

[doctorvenkman]

Originally posted by Nuel


I tried to think up an harcore proof for this thing, but my memory is "weak" on the sequence and series thing.
Blessings upon you.

Prove this rigorously by using induction on n.
If you need help you can email me.

 

[dan0909]

Nuel...is that MEGAMAN????


Hahahah...thats great mann

 

[Nuel]

Originally posted by dan0909
Nuel...is that MEGAMAN????


Hahahah...thats great mann

Yeah, it is megaman. I stole the avatar off some website 🙂. And to doctork, I don't know any rigorous proof to the question as at now, but I know there are proofs. Offer yours please, I will offer another intuitive one.

 

[Nuel]

On doctorvenkman's suggestion I offer my inductive proof. The limit of sequence A👎 (with discrete values) of (n+1)!(1/3)^(n+1) tends to infinity. If we reduce {A(n+1)}, we get (n+3)/(An/3) where {A👎} is the regular sequence. If the subsequent approaches infinity (which it does) the sequence actually does not have an upper bound. Just do the math and apply limit rules. My best shot now 😎