$\text{We have two sequences}$ $(a_n), (b_n)$ 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.

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

  • $\begingroup$ 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

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.

  • $\begingroup$ So you're saying my solution is wrong or not ? $\endgroup$ – GinKin Mar 25 '14 at 15:04
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    $\begingroup$ 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

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