Math Is Fun Forum

mathkeep

Member

Registered: 2009-08-02

Posts: 18

Limit

here is a question of limit...

...where n tends to infinity
actually my limit is coming out to be 0...i thought in this way...
write the limit as :

so the further ratios will make 1 smaller and smaller....thus making the fraction almost zero...
but i want a rigorous proof..

juriguen

Member

Registered: 2009-07-05

Posts: 59

Re: Limit

Hi!

I think your proof is rigorous enough. However, here you have a similar demo, based on Stirling's formula (approximates n!)
http://planetmath.org/encyclopedia/AsymptoticBoundsForFactorial.html

Jose

---

“Make everything as simple as possible, but not simpler.” -- Albert Einstein

Riad Zaidan

Guest

Re: Limit

Hi mathkeep ,
The solution as I think  is as follows:
n!/(n^n)=(n/n)((n-1)/n)((n-2)/n)...(2/n)(1/n)
but n-1<n means that           ((n-1)/n)<1
     n-2<n  means  that         ((n-2)/n)<1
     n-3<n  means  that         ((n-3)/n)<1
                                        .
                                        .
                                        .
                                     ( 2/n)<1
therefore  ((n-1)/n)((n-2)/n)...(2/n)<1
so        (n/n)((n-1)/n)((n-2)/n)...(2/n)<1
         (n/n)((n-1)/n)((n-2)/n)...(2/n)(1/n)<1(1/n) so
            (n/n)((n-1)/n)((n-2)/n)...(2/n)(1/n)<(1/n)

            0 <n!/(n^n)<(1/n)
but lim 0 =0 and lim (1/n) = 0 both as n tends to  ∞
and by squezing theorem we have lim (n!/(n^n))=0 as n tends to ∞ .
So the required limit is 0
W.B.W
Riad Zaidan

mathkeep

Member

Registered: 2009-08-02

Posts: 18

Re: Limit

hi Riad Zaidan!
thanks for the proof..got convinced from that easily..
n juriguen thanks for the link too

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Limit

I think a much easier proof would be to just note that

Note that this is just using the fact that:

And then making this swap for each n, n-1, n-2, ... except for 1.

This shows that

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

soroban

Member

Registered: 2007-03-09

Posts: 452

Re: Limit

. .

.

Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Limit

Unfortunately soroban, the question was to prove that it converges to 0, not just to show convergence.

---

"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

mathkeep

Member

Registered: 2009-08-02

Posts: 18

Re: Limit

I think a much easier proof would be to just note that

Note that this is just using the fact that:

And then making this swap for each n, n-1, n-2, ... except for 1.

This shows that

yeah a nice proof..thanks Ricky..

mathkeep

Member

Registered: 2009-08-02

Posts: 18

Re: Limit

Unfortunately soroban, the question was to prove that it converges to 0, not just to show convergence.

hmm..well Sorobon ...Ricky is right..anyways...that may help me in some other question :D

soroban

Member

Registered: 2007-03-09

Posts: 452

Re: Limit

.
Absolutely right, Ricky.

Thank you for the heads-up.
.

riad zaidan

Guest

Re: Limit

Hi soroban
The question is the convergence of a sequence not of a series , and the convergence of a sequence does not mean that the corresponding series is convergent.
As an example the sequence 1/n  converges to 0  but the series ∑(1/n) diverges.
w.b.w
Riad Zaidan

Sarah12

Guest

Re: Limit

Wow That is A Big Problem.:|

johnkv

Member

Registered: 2009-08-07

Posts: 4

Re: Limit

fine go ahead