prove that the following sequence converges:

 

[math]x_{k+2} =\frac{2}{x_{k+1} + x_{k}} ,x_{1},x_{2}>0[/math]

For [math]x_1 = x_2 = 1[/math] it does not.

For [math]x_1 = x_2 = 1[/math] it does not.

 

For [math]x_1 = x_2 = 1[/math] we have [math]x_k = 1 \ \forall k[/math] which most certainly converges to 1.

Oops. Right *blush*

For [math]x_1 = x_2 = 1[/math] we have [math]x_k = 1 \ \forall k[/math] which most certainly converges to 1.

 

Can we consider that answer as a proof??

Can we consider that answer as a proof??

 

No. It merely shows that the proposed counter example fails.

Personally - and with no proof or basis - intuitively I think all specified cases converge towards 1. The way the sequence ebbs and flows I cannot envisage any two starting points going off to infinity - but (without maths that is beyond me) there is always a possibility that two starters end up with an oscillating pattern that never calms down

Personally - and with no proof or basis - intuitively I think all specified cases converge towards 1. The way the sequence ebbs and flows I cannot envisage any two starting points going off to infinity - but (without maths that is beyond me) there is always a possibility that two starters end up with an oscillating pattern that never calms down

 

It is pretty easy to see that if the limit exists then the limit must be 1.

It is pretty easy to see that if the limit exists then the limit must be 1.

 

How do we prove it exists??

Kavlas, what is the definition of convergence?

Kavlas, what is the definition of convergence?

 

A sequence [math]x_n[/math] converges iff there exists, a, such that :[math] lim_{n\rightarrow\infty} x_{n}= a[/math]

What I always did with these kind of problems is take that x_k is equal to x_k+n or x_k-n when k goes to infinity,and n doesn't.

 

if x_k is equal to some x as x goes to k,than from your equation we get: [math]x=\frac{2}{x+x}[/math]

[math] x=\frac{1}{x}[/math]

[math]x^2=1[/math]

[math]x=1[/math]

 

you can also use the Ratio test which states that [math]x_n[/math] converes if [math]\lim_{n\to \infty}\ \frac{x_{n+1}}{x_n}<1[/math]

Edited August 28, 201114 yr by anonimnystefy