1
$\begingroup$

$\text{We have two sequences}$ $(a_n), (b_n)$ where $0<b_1<a_1$ and:

$$a_{n+1}=\frac{a_n+b_n}{2} \ , \ b_{n+1}=\frac {2a_nb_n}{a_n+b_n} $$

Prove both sequences converge to the same limit and try to find the limit.

What I did: Suppose $\displaystyle\lim_{n\to\infty}a_n=a, \displaystyle\lim_{n\to\infty}b_n=b$ So $\displaystyle\lim_{n\to\infty} \frac {a_n+b_n} 2= \frac{a+b} 2 =K$

Take $a_{n+2}= \frac {a_{n+1}+b_{n+1}}{2}=\frac {\frac{a_n+b_n}{2}+\frac {2a_nb_n}{a_n+b_n}}{2}=...=X$

We know that as $n$ tends to infinity $\lim x_n= \lim x_{n+1}$ so: $X=K$ and after some algebra I get $a=b$

As for the limit, it depends on only one of the sequences, since both tend to the same limit. The limit can be any constant or $\pm\infty$.

Is this approach correct ?

I excluded the algebra because I type this manually and to make the solution easier to read.

$\endgroup$
  • $\begingroup$ It seems that $a_{n+1}$ is a geometric mean of $a_n$ and $b_n$, and $b_{n+1}$ is a harmonic mean of same two previous terms. $\endgroup$ – JiminP Mar 25 '14 at 14:33
  • $\begingroup$ @JulienGodawatta Why we easily have $b_1<\cdots< b_n<\cdots<a_n<\cdots< a_1$ ? $\endgroup$ – GinKin Mar 25 '14 at 14:38
  • $\begingroup$ @GinKin See mookid's answer. $\endgroup$ – Julien Godawatta Mar 25 '14 at 14:41
  • $\begingroup$ @JiminP how did you get this ? there's no root nor $n$ in either... $\endgroup$ – GinKin Mar 25 '14 at 14:54
  • $\begingroup$ @GinKin Oops. I meant arithmetic mean. Sorry. $\endgroup$ – JiminP Mar 26 '14 at 9:21
2
$\begingroup$
  1. Just show via an induction that $b_n\le b_{n+1} \le a_{n+1} \le a_n$: this proves that both sequences are convergent.
  2. Then take the limit in the definition and the previous inequality: you get $$A = \frac 12 (A+B) \\A\ge B$$so $A=B.$

details for 1.:

a) The inequality $$ u<v\implies \frac {u+v}2<v $$is trivial.

b)$$ u<v\implies \frac 1u > \frac 1v \\ \implies \frac 1u > \frac 12 \left(\frac 1u +\frac 1v\right) =\frac{u+v}{2uv}\implies u< \frac{2uv}{u+v} $$

c) As $0\le(\sqrt{u}-\sqrt{v})^2$, $$ \sqrt{uv}\le \frac{u+v}2\\ 4uv\le (u+v)^2\\ \frac{2uv}{u+v} \le \frac {u+v}2 $$

$\endgroup$
  • $\begingroup$ How do you show that ? $\endgroup$ – GinKin Mar 25 '14 at 14:52
  • $\begingroup$ the most simple way is via convexity. $\endgroup$ – mookid Mar 25 '14 at 14:53
  • $\begingroup$ Well this question is supposed to be answered without functions or convexity (it's early in the course). In the comments above Julien said we have $b_1<\cdots< b_n<\cdots<a_n<\cdots< a_1$, how did he get this and how it can be used ? $\endgroup$ – GinKin Mar 25 '14 at 14:56
  • $\begingroup$ explanations improved. $\endgroup$ – mookid Mar 25 '14 at 15:06
  • 1
    $\begingroup$ this is just by induction. $\endgroup$ – mookid Mar 25 '14 at 15:28
1
$\begingroup$

Firstly, in your initial solution K = a by definition, so the result a=b follows immediately.

For an estimate of the limit for large n observe that the next member of each sequence is between an and bn, and so the final result must also be between a1 and b1, and indeed between (a1 + b1)/2 and 2*(a1*b1)/(a1+b1), and so certainly not infinity.

$\endgroup$
  • $\begingroup$ So you're saying my solution is wrong or not ? $\endgroup$ – GinKin Mar 25 '14 at 15:04
  • 1
    $\begingroup$ Your assertion that the limit can be any constant is wrong. It is bounded by a1 and b1. $\endgroup$ – Dave Mear Mar 25 '14 at 15:13
  • $\begingroup$ You solution for equality assumes that the limits exist and this must be proved first. Once this is proved, however, by replacing K by a, (because by definition K = a) gives (a+b)/2 = a => a+b=2a => b=a $\endgroup$ – Dave Mear Mar 25 '14 at 15:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.