# 2 monotone bounded questions

• Nov 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
• Nov 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.
• Nov 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.

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.
• Nov 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

$\chi$ $\sigma$
• Nov 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????