How to show that this specific limit exists in this proof?

Question

Given the sequence $T_n$ where $T_1 = 0, T_2 = 1, T_3 = 1$ and $T_n = T_{n-1} + T_{n-2} + T_{n-3}$ for $n >= 3$. Find what the ratio of consecutive terms, $\frac{T_{n+1}}{T_n}$ is converging to.

I already have figured out what the ratio is converging to: $$\lim_{n\to\infty} \frac{T_{n+1}}{T_n}$$} Now, we first assume that $\lim_{n\to\infty} \frac{T_{n+1}}{T_n}$ exists and converges at a finite value. Let's say that: $$A = \lim_{n\to\infty} \frac{T_{n+1}}{T_n} = \lim_{n\to\infty} \frac{T_{n} + T_{n-1} + T_{n-2}}{T_n}$$ Due to properties of limits, we can individually find each limit $$\lim_{n\to\infty} \frac{T_{n} + T_{n-1} + T_{n-2}}{T_n} = \lim_{n\to\infty}\frac{T_{n}}{T_n} + \lim_{n\to\infty}\frac{T_{n-1}}{T_n} + \lim_{n\to\infty}\frac{T_{n-2}}{T_n}$$ {Now let's solve each individual limit.} $$\begin{align} \lim_{n\to\infty}\frac{T_{n}}{T_n} &= 1\\ \lim_{n\to\infty}\frac{T_{n-1}}{T_n} &= \frac{1}{A}\\ \lim_{n\to\infty}\frac{T_{n-2}}{T_n} = \lim_{n\to\infty}\frac{ T_{n-2} T_{n-1} }{ T_n T_{n-1}} = \lim_{n\to\infty}\frac{T_{n-2}}{T_{n-1}} \cdot \frac{T_{n-1}}{T_{n}} &= \frac{1}{A^2} \end{align} $$ Substitute all our identities into our original equation: $$\begin{align}\lim_{n\to\infty} \frac{T_{n+1}}{T_n} = A &= 1 + \frac{1}{A} + \frac{1}{A^2}\\ A^3 &= A^2 + A + 1\\ A &= \frac{1}{3}(1 + \sqrt[3]{19 -3\sqrt{33}} + \sqrt[3]{19 + 3\sqrt{33}})\\ A &\approx 1.839286755\ldots \end{align} $$ How do I prove my underlying assumption that $\lim_{n\to\infty} \frac{T_{n+1}}{T_n}$ converges and exists? Also as another side question: Why does $\lim_{n\to\infty}\frac{T_{n-1}}{T_n} = \frac{1}{A}$? Same with $\lim_{n\to\infty}\frac{T_{n-2}}{T_{n-1}} = \frac{1}{A}$?

Link to answer of the original question: I don't understand what this question is asking and how to show it/prove it$

Answer 1

Let $f(x)=x^3-x^2-x-1.$ The equation $f(x)=0$ has $3$ complex solutions $x_1,x_2,x_3$ with $x_1\in \Bbb R$ and $1<x_1<2$ and where $x_2,x_3$ are non-real complex conjugates of each other.

And $1=x_1x_2x_3=x_1|x_2|^2$ so $x_1>1>|x_2|=|x_3|.$

(i). If $A,B,C$ are constants and $V_n=Ax_1^n+Bx_2^n+Cx_3^n$ for $n\in \Bbb N$ then $V_{n+3}=V_{n+2}+V_{n+1}+V_n$ for all $n\in \Bbb N.$

(ii). Given any values $T_1,T_2,T_3,$ there exist a unique $(A,B,C)$ such that $Ax_1^j+Bx_2^j+Cx_3^j=T_j$ for $j\in \{1,2,3\}.$ Then by induction on $n\geq 3$, if $T_n$ satisfies the recursion formula, we have $T_n=Ax_1^n+Bx_2^n+Cx_3^n$ for all $n\in \Bbb N.$

(iii). Therefore by (i) and (ii), with $T_1=0, T_2=1,T_3=1,$ there exists $(A,B,C)$ such that $T_n=Ax_1^n+Bx_2^n +Cx_3^n$ for all n\in \Bbb N.

And $A\ne 0.$ Because $|x_2|=|x_3|<1$ so $|Bx_2^n+Cx_3^n|\to 0$ as $x\to \infty,$ but $|T_n|=|Ax_1^n+Bx_2^n+Cx_3^n|\to \infty$ as $n\to \infty.$

So $\frac {T_{n+1}}{T_n}=$ $x_1\cdot \frac { 1+BA^{-1}(x_2/x_1)^{n+1} +CA^{-1}(x_3/x_1)^{n+1} } { 1+ BA^{-1}(x_2/x_1)^n + CA^{-1}(x_3/x_1)^n }.$

Since $|x_2/x_1|=|x_3/x_1|<1,$ therefore $\lim_{n\to \infty}\frac {T_{n+1}}{T_n}=x_1\approx 1.839287, $ the real-number solution of $x^3=x^2+x+1.$

Answer 2

$$T_n = T_{n-1} + T_{n-2} + T_{n-3}$$

$$T_1 = 0, T_2 = 1, T_3 = 1$$

Note that the terms are growing without bound because $a_{n+1}>a_n +1$ for $ n\ge 2 $

$$0,1,1,2,4,7,13,24,....$$

Thus you are dealing with a strictly increasing sequence which goes to infinity.

Answer 3

HINT

Looking for solution in the form of $$T_n = e^{an},$$ one can obtain the characteristic equation $$z^3-z^2-z-1 = 0$$ with the known roots.

That leads to the solution in the form of $$T_n = C_0A^n + C_1B^ne^{n\varphi}+C_2B_ne^{-n\varphi},$$ wherein $$T_0=0,\quad T_1=T_2=1.$$