I solved it!

jac006

Golden Master
Messages
5,810
well, she is gone, (chick) but she did challenge me. She asked me this question: Let xn and yn be two sequences with the following properties:
>x_(n+1)=xn2+yn2 and y_(n+1)=2*xn*yn for any n>=1 and x1 and y1>0.
>Prove that the series zn=xn/yn converges.
and....
I found that every term zn is greater than 1 so the sum of the series
must diverge.

x(n+1)
z(n+1) = ------- > 1 (aassume this to be true)
y(n+1)

xn2 + yn2
----------- > 1
2.xn.yn

xn2 + yn2 > 2.xn.yn

xn2 - 2.xn.yn + yn2 > 0

(xn - yn)2 > 0

This is always true since a perfect square must be positive. It
follows that z(n+1) > 1.

If the zn terms are all greater than 1 the series must diverge. Yes!!!
 
Well, I'm trying not to be racist... but I am asain. I'm only in the 10th grade, but I have already studied calculus and the like. So yeah. It took me a while... but there you go. Of course, I'm not a math genius or anything. No where near Einstein or Gauss or people like them. I just work hard, I guess.
 
maths good...

spelling bad...

Asian,

Race has nothing to do with intelligence.
 
I know... but you have to admit... Korea is ranked no.2 in the world as being the best at math. I think New Zealand is no.1 so yeah. Japon and China are just behind Korea, like no.3 and 4 I think. As I said, I wasn't being racist, just going off the statistics...
 
Let xn and yn be two sequences with the following properties:
x_(n+1)=xn^2+yn^2 and y_(n+1)=2*xn*yn for any n>=1 and x1 and y1>0.
Prove that the series zn=xn/yn converges.
Let a_n be x_n/y_n. Then a_(n+1)=x_n/2y_n+y_n/2x_n=(a_n+1/a_n)/2.
But a_n>=1 for any n>=2, so 1/a_n <= a_n, so a_(n+1) <= a_n.
From the fact that a_n >0 for any n we get that (a_n) is a
decreasing sequence which also has a lower bound, so it's convergent.
 
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