triclino Posted August 11, 2011 Share Posted August 11, 2011 prove that the following sequence converges: [math]x_{k+2} =\frac{2}{x_{k+1} + x_{k}} ,x_{1},x_{2}>0[/math] Link to comment Share on other sites More sharing options...

Shadow Posted August 14, 2011 Share Posted August 14, 2011 For [math]x_1 = x_2 = 1[/math] it does not. Link to comment Share on other sites More sharing options...

DrRocket Posted August 14, 2011 Share Posted August 14, 2011 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. Link to comment Share on other sites More sharing options...

Shadow Posted August 16, 2011 Share Posted August 16, 2011 Oops. Right *blush* Link to comment Share on other sites More sharing options...

triclino Posted August 19, 2011 Author Share Posted August 19, 2011 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?? Link to comment Share on other sites More sharing options...

DrRocket Posted August 19, 2011 Share Posted August 19, 2011 Can we consider that answer as a proof?? No. It merely shows that the proposed counter example fails. Link to comment Share on other sites More sharing options...

imatfaal Posted August 19, 2011 Share Posted August 19, 2011 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 Link to comment Share on other sites More sharing options...

DrRocket Posted August 19, 2011 Share Posted August 19, 2011 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. Link to comment Share on other sites More sharing options...

kavlas Posted August 27, 2011 Share Posted August 27, 2011 It is pretty easy to see that if the limit exists then the limit must be 1. How do we prove it exists?? Link to comment Share on other sites More sharing options...

DJBruce Posted August 27, 2011 Share Posted August 27, 2011 Kavlas, what is the definition of convergence? Link to comment Share on other sites More sharing options...

kavlas Posted August 27, 2011 Share Posted August 27, 2011 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] Link to comment Share on other sites More sharing options...

anonimnystefy Posted August 28, 2011 Share Posted August 28, 2011 (edited) 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, 2011 by anonimnystefy Link to comment Share on other sites More sharing options...

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