Rational Limit for non-rational square roots

I discovered the following relationship while playing around with concepts discussed in recent Veritasium video.

Let's consider the sequence \[𝑎_{𝑛+1}=(𝑥−1) 𝑏_{𝑛}+𝑏_{𝑛+1}\] \[𝑏_{𝑛+1}=𝑎_{𝑛}+𝑏_{𝑛} \] where x is any positive integer and $a_{0}$ and $b_{0}$ can be any non zero real numbers

Now any idea on how to prove the following result: $\lim_{n\to\infty} \frac{𝑎_{𝑛+1}}{𝑎_{𝑛}} = \lim_{n\to\infty} \frac{b_{𝑛+1}}{b_{𝑛}} =1 + \sqrt{x}$ $AND$ $\lim_{n\to\infty} \frac{𝑎_{𝑛}}{b_{𝑛}} = \sqrt{x}$

#Calculus

Easy Math Editor

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Comments

Let $c_n = a_n-b_n.$ Then the equations become $\begin{aligned} c_{n+1} &= (x-1)b_n \\ b_{n+1} &= c_n + 2b_n \\ \end{aligned}$ which becomes $b_{n+1} - 2b_n - (x-1)b_{n-1} = 0.$ This is a second-order linear recurrence, and the roots of the associated equation are $1 \pm \sqrt{x}.$ Since $x$ is a positive integer, and the limit can't be negative, we always choose the positive square root. Anyway, that should get you started.

Patrick Corn - 1 year, 1 month ago

Thanks!

Vikram Pandya - 1 year, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$