2 monotone bounded questions

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November 14th 2009, 09:33 AM
osudude

2 monotone bounded questions

1) let x_1 > 1 and x_(n+1) := (2-1) / x_n for n c X. Show that x_n is decreasing and bounded below by 2. find the limit.

2) let x_n := 1/(1^2) + 1/(2^2) + ... + 1/(n^2) for each n c N. Prove that x_n is increasing and bounded and hence converges. [Hint: note that if k>= 2, then 1/(k^2) =< 1/(k(k-1)) = 1/(k-1) - 1/k]

THANK YOU! :D
November 14th 2009, 04:44 PM
Focus

Quote:

Originally Posted by osudude

1) let x_1 > 1 and x_(n+1) := (2-1) / x_n for n c X.

Do you really mean (2-1)? Why not just say 1/x_n?

Also please post as to where the problem is, and an attempt at a solution.
November 14th 2009, 05:24 PM
Plato

Quote:

Originally Posted by Focus

Do you really mean (2-1)? Why not just say 1/x_n?
Also please post as to where the problem is, and an attempt at a solution.

Focus, I applaud your effort in this thread.
However, please review this other thread.
It has been my considerable experience that when someone is so wanting in being able to post in correct notation, that person will not understand any reply given.
November 14th 2009, 08:58 PM
chisigma

I apologize for my intervention but the first proposed 'quiz' is of remarkable interest... also because it has do be 'decrypted' (Wink) ...

Under the hypothesis that is ...

$x_{n+1}=\frac{1}{x_{n}}$ , $x_{1}=a$ (1)

... the solution is 'trivial' ...

$x_{1}=a , x_{2}=\frac{1}{a}, x_{3}=a, x_{4}=\frac{1}{a}, \dots$ (2)

... and the $\lim_{n \rightarrow \infty} x_{n}$ clearly doesn't exist...

However under the hypothesis that is...

$x_{n+1}=2-\frac{1}{x_{n}}$ , $x_{1}=a$ (2)

... the proposed problem is very nice! (Rofl)... The first terms of the sequence in this case are ...

$x_{1}=a , x_{2}=2-\frac{1}{a}, x_{3}=2-\frac{1}{2-\frac{1}{a}} , x_{4}= 2-\frac{1}{2-\frac{1}{2-\frac{1}{a}}}, \dots$ (4)

The expressions that are in (4) are called 'continued fractions' and they interpret one of the most 'sggestive' chapter of Math. If we denote with $x$ the $\lim_{n \rightarrow \infty} x_{n}$ , at first finding $x$ seems to be an 'hard' eneterprise. By observing with care (4) however You 'discover' that is...

$x=2- \frac{1}{x}$ (5)

... the only 'valid' solution of which is $x=1$ . Therefore is...

$\lim_{n \rightarrow \infty} x_{n}=1$ (5)

... no matter which is $x_{1}$ (Wink) ...

Kind regards

$\chi$ $\sigma$
November 15th 2009, 12:05 PM
osubiomathchick

Osumath dude. I am taking 547 this quarter as well I have completed 1...and am working on 2! Let me know what you find!

Here is my 1:
Lemma: If a>1 then 2-(1/a)>1.
Proof 0<1/a<1...then -1/a>-1 2+(-1/a)>2-1=1. by induction xsubn>1 for all n in N.

We wish to show if a>1 then 2-1/a<a. Which will be a polynomial, and therefore 1<a and 0<(a-1)^2 is always greater than 0.
Now we know that xsubn is bounded and monotone.

To find the limit solve for L in L+1/L=2.

Anyone how does this abbreviated proof sound????