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I ran across this confounding limit I am wondering about. It is as follows:

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

I played around with this on Maple and found that if c is less than approximately .382 (but

greater than 0), it converges to 0. If c is greater than .382 (but less than 1), it

diverges. What is it about .382?.

.382 is an approximation. By playing around more I could have taken it out to more decimal places.

The actual problem asks to prove that the above limit is < $\frac{1}{p(n)}$, where p(n) is a polynomial.

I was mainly wondering how to solve the limit and why .382 is so significant.

Thank you all very much. You are always a big help.

  • 0
    I see geometric series and the binomial theorem.2011-03-17
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    Actually, playing around myself with Maple the turning point appears to be at c=1/2.2011-03-17
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    One cannot compare the limit of something when $n\to+\infty$ to something which depends on $n$, so there is a problem with the sentence of the post which begins by *The actual problem*.2011-03-17
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    Yes, I reaalize that. That 1/p(n) did not make much sense to me either, but that is all I was given. Besides that, I thought it was curious because of the .382. Thank you.2011-03-17

3 Answers 3

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Edited to use Christian Blatter's simpler formula

$0.38196...$ is $(3-\sqrt{5})/2$, which is a root of the equation $x^2-3x+1=0$.

When $n$ is large, your numerator is about $2nc^n$, so your expression is about

$2n(c/(1-c)^2)^n$

For $c \in (0,1)$, this tends to zero if and only if $c/(1-c)^2 < 1$, giving:

$c < (1-c)^2$
$c^2-3c+1>0$
$c < (3-\sqrt{5})/2$.

  • 0
    Correct me if I'm wrong on this, but $(.38196...)^2 + .38196... - 1 < .4^2 + .4 - 1 = -.44 < 0$.2011-03-17
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    @Alex: I edited my answer while you posted, so it's not there any more. But I said .38196... was the *square* of the solution to $x^2+x-1=0$.2011-03-17
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    Thank you, Tony and Alex. But, may I ask one more dumb question?. How do you know the numerator behaves like $2n\cdot c^{n}$?. Now that you mention it, I certainly see that this is the case, but how did you know that it is specifically $2n\cdot c^{n}$?.2011-03-17
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    Nevermind, it is Bernoulli. I knew that:) Thanks again.2011-03-17
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Hint: Use Bernoulli's inequality to estimate $(1-c^n)^{2n}$. You will then find out that the critical value is $c_0:=(3-\sqrt{5})/2\doteq 0.382$. For $c_0\leq c<1$ an estimate in the other direction is required. Maybe a suitable Taylor approximation of $(1+x)^{2n}$ for small $|x|$ helps.

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    Thank you Christian, I will look into it.2011-03-17
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My previous comment was wrong of course. One can solve the problem using L'Hopital's rule. $0.38196$ appears as the solution to the equation $x^2-3x+1=0$ and equals $\frac{3-\sqrt{5}}{2}$ to be precise.

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    Differentiating with respect to $n$, while doable, seems messy.2011-03-17
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    You are right, it is.2011-03-17
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    Yes, I tried L'Hopital, but the derivatives turned into a monster so I did not proceed much further.2011-03-17