3
$\begingroup$

Let $(a_1) = 1$ and $$(a_{n+1})= \frac{4+3a_n}{3+2a_n}$$ It is required to show that the following recursively defined sequence converges. I know one way to show this converges. Define $$f(x)= \frac{4+3x}{3+2x}$$ on the set of non-negative reals(which is closed hence complete). It is easy to show that $f'$ exists and $|f'(x)|<1$ hence it is a contraction and by the fixed point theorem has a unique fixed point. Is there any easy way to show that this converges using only the basic theorems on limits. All my attempts have failed so far. Any solutions will be highly appreciated.

$\endgroup$
  • 1
    $\begingroup$ Replace $a_n$ and $a_{n+1} $with $l$...this is the limit. The sequence is monotone and bounded... $\endgroup$ – dmtri Aug 20 '18 at 10:34
  • 1
    $\begingroup$ Can you provide a detailed solution? $\endgroup$ – John Mitchell Aug 20 '18 at 10:35
  • $\begingroup$ I am afraid not. I am at beach right now and it is really painful to write Mathjax with smartphone...If there is a limit it should be sqrt(2). $\endgroup$ – dmtri Aug 20 '18 at 10:39
2
$\begingroup$

$$(a_{n+1})= \frac{4+3a_n}{3+2a_n}= 1+\frac{1+a_n}{3+2a_n}$$ the sequence is bounded increasing so the limit exist and

$$l= \frac{4+3l}{3+2l} \Rightarrow 3l+2l^2=4+3l \Rightarrow l^2=2 \Rightarrow l=\pm \sqrt{2}$$ as the limit is positive so $l=\sqrt{2}$

$\endgroup$
  • $\begingroup$ That the function is increasing doesn't necessarily mean the sequence is increasing. $\endgroup$ – Bernard Aug 20 '18 at 11:12
  • $\begingroup$ @Bernard Thank you, i have edited it $\endgroup$ – The_lost Aug 20 '18 at 11:16
0
$\begingroup$

This is a homographic sequence. The strategy is to express $a_{n+1}-a_n$ in function of $a_n-a_{n-1}$.

First rewrite $f(x)$ in canonical form (obtained by Euclidean division of $3x+4$ by $2x+3$: $$\frac{4+3x}{3+2x}=\frac32-\frac1{2(3+2x)}.$$ We obtain \begin{align} a_{n+1}-a_n&=\biggl(\frac32-\frac1{2(3+2a_n)}\biggr)-\biggl(\frac32-\frac1{2(3+2a_{n-1})}\biggr)\\&=\frac1{2(3+2a_{n-1})}-\frac1{2(3+2a_n)} %\\[1ex] =\frac{a_n-a_{n-1}}{(3+2a_{n-1})(3+2a_n)}. \end{align} As $a_1>0$, it is easy to see all $a_n$ are positive. from this relation, we draw two conclusions:

  • $a_{n+1}-a_n$ and $a_n-a_{n-1}$ have the same sign for all $n$, hence the sequence is monotonic, and ac tually increasing since $a_2>a_1$.
  • $0<a_{n+1}-a_n <\frac19(a_n-a_{n-1})$, so that $a_{n+1}-a_n <\frac1{9^{n-1}}(a_2-a_{1})$ and $$a_n-a_1=\sum_{k=1}^{n-1}(a_{k+1}-a_k)<\sum_{k=1}^{n-1}\frac1{9^{k-1}}(a_2-a_1)<\frac98(a_2-a_1)=\frac9{20}$$ and finally $\; a_n=(a_n-a_1)+a_1<\frac{29}{20}$ is a bounded increasing sequence. Therefore it converges.

By continuity, the limit is a non-negative root of the equation $$x=\frac{3x+4}{2x+3}\iff x^2=2.$$

$\endgroup$

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.