Calculus – Proving Two Sequences Converge to the Same Limit

calculus

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

Best Answer

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

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