Find $a_1$ so that $a_{n+1}=\frac{1}{4-3a_n}\ ,n\ge1$ is convergent

Let $$\left\{ a_n \right\}$$ be a recursive sequence such that $$a_{n+1}=\frac{1}{4-3a_n}\quad,n\ge1$$ Determine for which $$a_1$$ the sequence converges and in case of convergence find its limit.

The problem is from the book 'Problems in Mathematical Analysis I by W.J.Kaczor'.

• $${\sqrt{x^2+y}} = {x+{\cfrac {y}{2x+{\cfrac {y}{2x+{\cfrac {y}{2x+{\ddots }}}}}}}}$$ Sep 30, 2018 at 20:09

The Moebius transformation $$T:\quad\bar{\mathbb C}\to\bar{\mathbb C},\qquad x\mapsto T(x):={1\over 4-3x}$$ has the two fixed points $$1$$ and $${1\over3}$$. We therefore introduce a new complex projective coordinate $$z$$ via $$z:={x-{1\over3}\over x-1},\qquad{\rm resp.},\qquad x={z-{1\over3}\over z-1}\ .$$ In terms of this coordinate $$T$$ appears as $${\displaystyle \hat T(z)={z\over3}}$$ (with fixed points $$0$$ and $$\infty$$), so that $$\bigl(\hat T\bigr)^{\circ n}(z)={z\over 3^n}\ .$$ It follows that for all initial points $$z\ne\infty$$ we have $$\lim_{n\to\infty}\bigl(\hat T\bigr)^{\circ n}(z)=0\ .$$ In terms of the original variable $$x$$ this means that for all initial points $$x\ne1$$ we have $$\lim_{n\to\infty}T^{\circ n}(x)={1\over3}\ .$$ There is, however, the following caveat: The above argument refers to the domain $$\bar{\mathbb C}$$; but maybe you want to exclude $$x=\infty$$ as a generic point. In terms of the coordinate $$z$$ this is the point $$z_*=1$$. For all initial values $$z_k=3^k$$ $$(k\geq1)$$ we have $$\bigl(\hat T\bigr)^{\circ k}z_k=z_*$$. This implies that in the original formulation of the problem you have $$T^{\circ k}(x_k)=\infty$$ (i.e., you "accidentally" hit $$\infty$$ after finitely many steps) for all initial points $$x_k=\bigl(3^k-{1\over3}\bigr)/(3^k-1)$$ $$(k\geq1)$$.

• Nice solution :-) (+1) upvote
– user798113
Dec 24, 2020 at 20:05

Suppose that this sequence does converge to A. Then we must have $$A= \frac{1}{4- 3A}$$. Then $$A(4- 3A)= 4A- 3A^2= 1$$. $$3A^2- 4A+ 1= (3A- 1)(A- 1)= 0$$. A is either 1 or 1/3.

If $$a_1> 1$$ the sequence clearly converges to 1. If $$a_1\le 1/3$$ if clearly converges to $$\frac{1}{3}$$. It's a little harder to show, but still true, that if $$1/3< a_1< 1$$ then the sequence converges to 1/3: if $$\frac{1}{3}< a< 1$$ then $$1< 3a< 3$$ so that 0< 3a-1< 2. But for $$\frac{1}{3}< a< 1$$, $$a- 1< 0$$. That is, 3a-1 is positive while a- 1 is negative so that $$(3a- 1)(a- 1)= 3a^2- 4a+ 1< 0$$. Then $$3a^2- 4a= a(3a- 4)> 1$$ and $$a> \frac{1}{3a- 4}$$. That is, for $$\frac{1}{3}< a_1< 1$$ the sequence is decreasing to $$\frac{1}{3}$$.

• what if $a_1 = 4/3$? Sep 30, 2018 at 16:31
• Moreover, what if $a_1 =2$? I'm finding the sequence converges to $1/3$. Sep 30, 2018 at 16:52
• Let $\,a_1\in\mathbb{R}\,$. 1) How to show it converge for all $\,a_1\,$. 2) "If $\,a_1$... the sequence clearly converges to ..."; is it that clear?! Thanks. Sep 30, 2018 at 16:53