limit problem

[yossigi]

Hi, I'm having trouble computing the following limit:
0 < c < 1 is a constant

lim (1 - (1 - c^n)^(2n))/(1-c)^(2n)
n--> infinity

I -think- it's going to 0 but I'm not sure. I was able to prove that both numerator and denominator are converting to 0, so it's possible to use l'Hôpital's rule, but that doesn't seem to help much.
Additionally, I want to prove a stronger argument: for every polynomial p(n) the following holds:

lim (1 - (1 - c^n)^(2n))/(1-c)^(2n) < 1/p(n)
n--> infinity

Does anyone have any idea?

 

[yossigi]

oh, small mistake, I meant I think the limit goes to infinity and is larger than any polynomial.
Sorry guys!

 

[yossigi]

Ahh! I keep inverting the fraction by mistake. first post is the correct one :/

 

[galactus]

Is this the limit you mean:

\(\displaystyle \lim_{n\to \infty}\frac{\left(1-(1-c^{n})^{2n}\right)}{(1-c)^{2n}}, \;\ 0<c<1\)


I played around with this and found that if c is in the interval (0, .382], it converges to 0.

If it is in [.382, 1), then it diverges.

There is something ab out .382.

This is an approximation. It can be taken out more decimal places.

 

[yossigi]

Yes, that's exactly it.

Really? that sounds weird, but maybe you're right.
Got any idea how to solve it?
I'm having trouble with this problem for some time now..

 

[galactus]

Additionally, I want to prove a stronger argument: for every polynomial p(n) the following holds:

lim (1 - (1 - c^n)^(2n))/(1-c)^(2n) < 1/p(n)
n--> infinity

Does anyone have any idea?

Maybe I am missing something, but as \(\displaystyle n\to \infty\), this does not make much sense.

You can not have an 'n' on the right side of the inequality as well.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

As \(\displaystyle n\to \infty\), the numerator behaves like \(\displaystyle 2n\cdot c^{n}, \;\ (\text{Bernoulli inequality})\)**

So, the problem is like \(\displaystyle 2n\left(\frac{c^{n}}{(1-c)^{2n}}\right)=2n\left(\frac{c}{(1-c)^{2}}\right)^{n}\)

If c is between 0 and 1, the numerator heads toward 0 as long as \(\displaystyle \frac{c}{(1-c)^{2}}<1\)

Which leads to:

\(\displaystyle c<(1-c)^{2}\)

\(\displaystyle c^{2}-3c+1>0\)

\(\displaystyle c<\frac{3-\sqrt{5}}{2}\approx .382\)

That is where the .382 comes from.

** The Bernoulli inequality can be mighty handy. It says that \(\displaystyle (1+x)^{n}\geq 1+nx\)

If the exponent, n, is even, then the inequality is valid for all real x.

 

[yossigi]

Yes well, what I meant is that there exists k such that for any k' > k fraction(k') < 1/p(k).

I don't understand what you did there - where did the first inequality came from? (1/ (1 - c)^2 < 1)

Also, how did the second inequality followed?
if you multiply by (1 - c)^2 you get
(1 - c)^2 > 1

 

[galactus]

Sorry, yos, I had a typo. Good catch.

I amended and emended the post .