How to find $\lim\limits_{n\to +\infty}(3+\frac{1}{a_n})$, where $a_n = 3+\frac{1}{a_{n-1}}$?

Question

How to find $\lim\limits_{n\to +\infty}(3+\frac{1}{a_n})$, where $a_n = 3+\frac{1}{a_{n-1}}$?

I tried to find $(a_n-a_{n-1})$ but it didn't help.

Answer 1

A fun but too-advanced approach is to use linear algebra. Take $a_0=a$ and write:

$$a_n = \frac{P_n(a)}{Q_n(a)}$$

where $P_n,Q_n$ are polynomials. Then you get that: $$a_{n+1}=3+\frac{Q_n(a)}{P_n(a)}=\frac{3P_n(a)+Q_n(a)}{P_n(a)}$$

or $P_{n+1}(a)=3P_n(a)+Q_n(a), Q_{n+1}(a)=P_n(a).$

This means that $$\begin{pmatrix}P_{n+1}\\Q_{n+1}\end{pmatrix}=\begin{pmatrix}3&1\\1&0\end{pmatrix}^n\begin{pmatrix}a\\1\end{pmatrix}$$

Now, the matrix has eigenvalues $\lambda_1,\lambda_2=\frac{3\pm\sqrt{13}}2$ with eigenvectors $v_i=\begin{pmatrix}\lambda_i\\1\end{pmatrix}$.

If $a=\lambda_2$ then the limit is $\lambda_2$.

If $a\neq \lambda_2$, then, since $|\lambda_2|<1$, the value will converge to $\lambda_1$, assuming $P_n(a)\neq 0$ for all $n$.

To figure out if $P_n$ is ever zero, we write:

$$\begin{pmatrix}a\\1\end{pmatrix}=b_1v_1+b_2v_2$$

and solve for scalars $b_1,b_2$. We know that $b_1+b_2=1$ and $b_1\lambda_1+b_2\lambda_2=a.$

Then $a_n=0$ iff $b_1\lambda_1^n+b_2\lambda_2^n=0$ or $b_1\lambda_1^{2n}=(-1)^{n+1}b_2.$

Since $a\neq \lambda_2$, we know $b_1\neq 0$, you want: $$2n=\log_{\lambda_1}\left((-1)^{n+1}\frac{b_2}{b_1}\right)$$

That means you want $\left|\frac{b_2}{b_1}\right|$ to be an even power of $\lambda_1$, and certain sign conditions.

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Alternatively, since $Q_{n+1}=P_n$ you get $P_{n+1}=3P_n+P_{n-1}$ and you can use a linear recurrence to show that $P_{n}=c_1\lambda_1^n+c_2\lambda_2^n$ with $P_0=a, P_{1}=3a+1.$ Then, for $n>0$ you have:

$$a_n=\frac{P_n}{Q_n}=\frac{c_1\lambda_1^{n}+c_2\lambda_2^n}{c_1\lambda_1^{n-1}+c_2\lambda_2^{n-1}}=\frac{c_1\lambda_1+(-1)^{n-1}c_2\lambda_2^{2n-1}}{c_1+(-1)^nc_2\lambda_2^{2n-2}}$$

Since $|\lambda_2|<1$ this converges to $\lambda_1$ if $c_1\neq 0$. If $c_1=0$ the the limit is $\lambda_2.$

Then again, we deal with the question of for what values of $a$ do we get some $a_n=0$.

Also, $$a_n-\lambda_1 = c_2\frac{c\lambda_2^n-\lambda_1\lambda_2^{n-1}}{P_{n-1}}=c_2\lambda_2^{n-2}\frac{\lambda_2^2+1}{P_{n-1}}=c_2\lambda_2^{n-2}\frac{3\lambda_2+2}{P_{n-1}}$$

Now, $$\frac{\lambda_2^{n-2}}{P_{n-1}}=\frac{(-1)^{n-2}\lambda_1}{c_1\lambda_1^{2n-2}+(-1)^{n-1}c_2}$$

So you get: $$a_n-\lambda_1 = c_2\frac{(-1)^{n-2}(2\lambda_1-3)}{c_1\lambda_1^{2n-2}+(-1)^{n-1}c_2}$$

showing, if $c_1\neq 0$ thatn $a_n-\lambda_1 = O\left(\lambda_1^{-2n}\right)$

Answer 2

Here is a method which analyses the error term - which seems to me to be like what you are trying to do with $a_{n+1}-a_n$. But this uses the equation you have been given in a different way. I've put enough to give you a start. Other methods may be more efficient, so this is more for interest.

Suppose that $a^2-3a-1=0$ and set $a_n=a+e_n$ then $$a_{n+1}=a+e_{n+1}=3+\frac 1{a_n}=\frac{3a+3e_n+1}{a+e_n}$$ so that $$a^2+ae_n+ae_{n+1}+e_ne_{n+1}=3a+3e_n+1$$ whence $$(a+e_n)e_{n+1}=(3-a)e_n$$ and $$e_{n+1}=\frac {(3-a)}{(a+e_n)}e_n$$

Now note what happens when you take the root which is greater than $3$ and $e_n$ is small enough in absolute value. - you should be able to prove that in this case the error term tends to zero, and hence establish the limit.

Answer 3

If the limit exists, it is a solution of the equation $$x=3+\frac1x$$

Nevertheless, the existence of the limit must be proved, and it can depend on the value of $a_1$ which you have not specified.

Answer 4

Let's show that for $\forall a_0>0$, the sequence is converging ...

Proposition 1. $3< a_n < 4$, for $\forall n \geq 2$.

If $a_0 > 0 \Rightarrow \frac{1}{a_0}>0 \Rightarrow \color{red}{a_1}=3+\frac{1}{a_0}\color{red}{>3}$. So $a_1 >0 \Rightarrow \frac{1}{a_1}>0 \Rightarrow \color{red}{a_2}=3+\frac{1}{a_1}\color{red}{>3}$ and, by induction $$\color{red}{a_n > 3}, \forall n >0 \tag{1}$$ Also, from $(1)$ $a_n > 3 \Rightarrow 0< \frac{1}{a_n} < \frac{1}{3} < 1 \Rightarrow \color{red}{a_{n+1}}=3+\frac{1}{a_n}\color{red}{<4}$. But $(1)$ is true for $n > 0$, so (by induction again) $$\color{red}{a_n < 4}, \forall n>1 \tag{2}$$ Altogether $$3< a_n < 4,\forall n \geq 2$$

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Now, let's look at the function $$f(x)=3+\frac{1}{x}$$ which is an important one, because $a_{n+1}=f(a_n)$.

Proposition 2. if $3\leq x \leq 4 \Rightarrow 3 \leq f(x) \leq 4$

$3\leq x \leq 4 \Rightarrow \frac{1}{3} \geq \frac{1}{x} \geq \frac{1}{4} \Rightarrow 4 > 3+ \frac{1}{3} \geq 3+\frac{1}{x} \geq 3+\frac{1}{3} > 3$ which is $3 < f(x) < 4$

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Proposition 3 Function $f: [3,4] \rightarrow [3,4]$ is a contraction mapping.

From MVT $$|f(x)-f(y)|= |f'(\varepsilon)| |x-y|=\left|\frac{1}{\varepsilon^2}\right| |x-y|$$ where $\varepsilon \in (x,y) \subset [3,4]$, thus $$|f(x)-f(y)| \leq \frac{1}{9} |x-y|$$

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Altogether, we can ignore the first 2 elements (Proposition 1) of the sequence, when the initial element is greater than $0$, they don't affect the convergence of the sequence. And according to the Banach fixed-point theorem, the sequence has a limit. More reading on this subject here