Finding a limit of recursively given sequence

Question

Given a sequence $\{a_n\}_{n=1}^\infty$: $$a_1=2$$ $$\forall n\in \Bbb{N}:a_{n+1}=4-\frac{3}{a_n}$$ find $\lim_{n\to\infty}a_n$. Writing out first few terms i found out $a_n\to 3$ as $n\to\infty$. But i would like it to be in more precise way (the best would be the $\epsilon$-definition). So I attempted to get a direct formula for $a_n$ not depending on any other $a_i$'s but also couldn't find a suitable way to do that. Could you please give me some hints, how to go about this problem in more mathematical way?

Answer 1

One shows that the sequence is strictly increasing and bounded from above. From this conclude that the limit exists. For the limit $a$, we must have $a=4-\frac 3a$. Conclude that $a=3$ (and not $a=1$).

Answer 2

Technical point:

To show that $a_n$ is strictly increasing and bounded by $3$ note that we have

I:

$$3>x>0\implies (x-3)(x-1)<0\implies x^2-4x+3<0\implies x^2<4x-3\implies x<4-\frac 3x$$

II: $$0<x<3\implies \frac 3x>1\implies 4-\frac 3x<3$$

Answer 3

For the main question "how to find the limit", please see other answers.
As a supplement, here is a trick to obtain a direct formula for this particular $a_n$.

If $L_a$ is the limit of sequence $a_n$, it will satisfy

$$L_a = 4 - \frac{3}{L_a} \iff L_a^2 - 4L_a + 3 = (L_a-1)(L_a-3) = 0$$

In general, if you have a sequence $b_n$ whose limit $L_b$ satisfy some polynomial equation with roots $\lambda_1, \lambda_2, \ldots, \lambda_n$, you can construct auxillary sequences $c_n$ of the form $\frac{(b_n - \lambda_{i_1})(b_n-\lambda_{i_2})\cdots(b_n - \lambda_{i_p})}{(b_n - \lambda_{j_1})(b_n-\lambda_{j_2})\cdots(b_n - \lambda_{j_q})}$ and see whether any of them is easier to solve or estimate the bounds.

For the sequence at hand, let $\displaystyle\;c_n = \frac{a_n - 3}{a_n - 1} \iff a_n = \frac{c_n-3}{c_n-1}\;$, we have

$$c_{n+1} = \frac{ c_n - 3}{c_n - 1} = \frac{1 - \frac{3}{a_n}}{3 -\frac{3}{a_n}} = \frac13\frac{a_n-3}{a_n-1} = \frac13 c_n$$ Together with $c_1 = \frac{2-1}{2-3} = -1$, we obtain following explicit form for $c_n$ and hence the one for $c_n$. $$c_n = -\frac{1}{3^{n-1}} \quad\implies\quad a_n = \frac{-\frac{1}{3^{n-1}} - 3}{-\frac{1}{3^{n-1}} - 1} = \frac{3^n+1}{3^{n-1}+1} $$

Answer 4

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$\ds{a_{n + 1} = 4 - {3 \over a_{n}} = {4a_{n} - 3 \over a_{n}}}$. Lets $\ds{a_{n} = {x_{n} \over y_{n}}}$ such that $\ds{{x_{n + 1} \over y_{n + 1}} = {4x_{n} - 3y_{n} \over x_{n}}}$

Set $\ds{x_{n + 1} = 4x_{n} - 3y_{n}}$ and $\ds{y_{n + 1} = x_{n}}$ which can be written as $\ds{\pars{~\mbox{with}\ x_{1} = y_{1} = 1~}}$ \begin{align} {x_{n + 1} \choose y_{n + 1}} & = \pars{\begin{array}{rr}\ds{4} & \ds{-3} \\ \ds{1} & \ds{0}\end{array}} {x_{n} \choose y_{n}} = \pars{\begin{array}{rr}\ds{4} & \ds{-3} \\ \ds{1} & \ds{0}\end{array}}^{2} {x_{n - 1} \choose y_{n - 1}} = \pars{\begin{array}{rr}\ds{4} & \ds{-3} \\ \ds{1} & \ds{0}\end{array}}^{3} {x_{n - 2} \choose y_{n - 2}} \\[5mm] & = \cdots = \pars{\begin{array}\ds{4} & \ds{-3} \\ \ds{1} & \ds{0}\end{array}}^{n} {x_{1} \choose y_{1}} = \pars{\begin{array}\ds{4} & \ds{-3} \\ \ds{1} & \ds{0}\end{array}}^{n} {1 \choose 1} \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\, \lambda^{n}\,\mathbf{u}\,\mathbf{u}^{\mathsf{T}}{1 \choose 1} \end{align}

$\ds{\lambda}$ is the eigenvalue, with the largest magnitude, of the above mentioned matrix and $\ds{\mathbf{u}}$ is the correspondent eingenvector.

Then, $$ \lim_{n \to \infty}a_{n + 1} = {\mathbf{v}_{+}^{\mathsf{T}}\mathbf{u}\,\mathbf{u}^{\mathsf{T}}{1 \choose 1} \over \mathbf{v}_{-}^{\mathsf{T}}\mathbf{u}\,\mathbf{u}^{\mathsf{T}}{1 \choose 1}}\,,\qquad \mathbf{v}_{+} \equiv {1 \choose 0}\,,\quad \mathbf{v}_{-} \equiv {0 \choose 1} $$

It turns out that $\ds{\mathbf{u} \propto {3 \choose 1}}$ such that

$\ds{\mathbf{u}\mathbf{u}^{\mathsf{T}}{1 \choose 1} = \pars{3 \quad 1}{3 \choose 1}{1 \choose 1} = \pars{\begin{array}{cc} \ds{9} & \ds{3} \\ \ds{3} & \ds{1} \end{array}}{1 \choose 1} = {\color{red}{12} \choose \color{red}{4}}}$ which yields

$$ \lim_{n \to \infty}a_{n + 1} = {\mathbf{v}_{+}^{\mathsf{T}}\mathbf{u}\,\mathbf{u}^{\mathsf{T}}{1 \choose 1} \over \mathbf{v}_{-}^{\mathsf{T}}\mathbf{u}\,\mathbf{u}^{\mathsf{T}}{1 \choose 1}} = {12 \over 4} = \bbx{\large 3} $$