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.
So I think the wording for this problem is not great, but I don't really know what it is asking and what to do to show it. Is it like a limit? A sum?
On
You have been given a linear homogeneous recurrence relation. Can you see that you have enough information to calculate $T_n$ for any $n$ by stepping up from below?. For example, $T_4=T_3+T_2+T_1=1+1+0=2$ You can program this into a spreadsheet easily with copy down and find a numeric approximation. You are expected to find the characteristic polynomial and then find its greatest root. That will be your answer. Read the linked article, or your text, to find out why.
$\lim_{n \to \infty} \frac{T_{n+1}}{T_n} = \lim_{n\to\infty}\frac{T_n + T_{n-1} + T_{n-2}}{T_n}$. If the sequence $\frac{T_{n+1}}{T_n}$ converges, set $\lim_{n \to \infty} \frac{T_{n+1}}{T_n} = L$.
Then:
$\lim_{n \to \infty} \frac{T_{n}}{T_{n-1}} = \frac{1}{L}$ and $\lim_{n \to \infty} \frac{T_{n-2}}{T_n} = \lim_{n \to \infty} \frac{T_{n-2}}{T_n} \frac{T_{n-1}}{T_{n-1}} = \frac{1}{L^2}$. This comes from basic properties of limits, such as the product of two converging sequences converges to the product of their limits. From this, we have:
$$L = 1 + \frac{1}{L} + \frac{1}{L^2} \Rightarrow L^3 = L^2 + L + 1$$ The only real solution to this polynomial is $L = \frac{1}{3} (1 + \sqrt[3]{19 - 3 \sqrt{33}} + \sqrt[3]{19 + 3 \sqrt{33}} )$ - which is the answer to your problem.