Are the two sequences decreasing, inferiorly bounded, and converging to 0?

[Francolino]

Let be $ (a_n) $ and $ (b_n) $ two sequences that verify: $$ a_{n+1} = \sqrt{a_{n}\cdot b_{n}}, \quad b_{n+1} = \frac {b_{0}}{a_{0}}\sqrt{a_{n}\cdot b_{n}}, \quad a_{0} > b_{0} > 0 $$ Indicate if the following statements are true:

(1) Both sequences are decreasing.
(2) Both sequences are inferiorly bounded.
(3) Both sequences converge to 0.

Well, I tried to do something similar to what I was suggested to do in http://mathhelpboards.com/calculus-10/convergence-sequence-15868.html. So I took polar coordinates:

Using that: $$ \left\{\begin{matrix}
a_{n} = r_{n}\cos(\theta_{n}) \\
b_{n} = r_{n}\sin(\theta_{n})
\end{matrix}\right. $$ Then: $$ \left\{\begin{matrix}
r_{n+1}\cos(\theta_{n+1}) = \sqrt{r_{n}\cos(\theta_{n})} \\
r_{n+1}\sin(\theta_{n+1}) = \left (\frac{b_{0}}{a_{0}} \right )\sqrt{r_{n}\sin(\theta_{n})}
\end{matrix}\right. $$ Squaring both equations: $$ \left\{\begin{matrix}
r_{n+1}^2\cos^2(\theta_{n+1}) = r_{n}\cos(\theta_{n}) \\
r_{n+1}^2\sin^2(\theta_{n+1}) = \left (\frac{b_{0}}{a_{0}} \right )^2 r_{n}\sin(\theta_{n})
\end{matrix}\right. $$ Summing both equations I get: $$ r_{n+1}^2 = r_{n}\left [ \cos(\theta_{n}) + \left (\frac{b_{0}}{a_{0}} \right )^2\sin(\theta_{n}) \right ] $$ But I got stuck there. :/

 

[Opalg]

Polar coordinates are not going to be much help here. I think that the key idea will be to use proof by induction.

Can you see for example whether it is true that $a_{n+1} < a_n$, or that $b_n <a_n$?

 

[johng1]

Hi,
I think you want to prove these statements directly. Here's an outline of a proof:

View attachment 4570

 

[johng1]

Hi again,
Now that you have digested my previous post, I hope, here's an alternate way to prove the results:

View attachment 4571