# If $a_n+b_n\sqrt{3}=(2+\sqrt{3})^n$, then what's $\lim_{n\rightarrow\infty}\frac{a_n}{b_n}$?

Let $a_n$ and $b_n$ be integers defined in the following way: $$a_n+b_n\sqrt{3}=(2+\sqrt{3})^n.$$ Compute $$\lim_{n\rightarrow\infty}\frac{a_n}{b_n}.$$

I tried expanding using binomial theorem: $$(2+\sqrt{3})^n=\binom{n}{0}2+\binom{n}{1}2^2\sqrt{3}+\binom{n}{2}2^3\cdot 3+\ldots+\binom{n}{n-1}2\sqrt{3}^{n-1}+\binom{n}{n}\sqrt{3}^{n}$$ and I think we have that: $$a_n=\binom{n}{0}2+\binom{n}{2}2^3\cdot 3+\ldots+\binom{n}{n}\sqrt{3}^{n}$$

$$b_n=\binom{n}{1}2^2\sqrt{3}+\ldots+\binom{n}{n-1}2\sqrt{3}^{n-1}$$

But, frankly, I do not know what to do next.

• The first thing you should have done is work out examples. You can easily figure out what the answer should be if you look at numerical data (and know some basic decimal expansions). This would not help you directly solve the problem, but that's not an excuse for not being able to at least figure out on your own a candidate for the answer so you can ask a better question.
– KCd
Apr 9, 2017 at 16:36

For fun, you can also do this with linear algebra. Let $V = \mathbb{Q}(\sqrt{3})$ as a $\mathbb{Q}$-vector space with basis $\{1,\sqrt{3}\}$. The "multiplication-by-$(2+\sqrt{3})$" map looks like $\begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix},$ so $$\begin{pmatrix} a_n \\ b_n \end{pmatrix} = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}^n \begin{pmatrix} 1 \\ 0 \end{pmatrix}.$$ In the limit, the largest eigenvalue of $\begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}$ dominates, so $\begin{pmatrix} a_n \\ b_n \end{pmatrix}$ will be close to a multiple of the eigenvector $\begin{pmatrix} \sqrt{3} \\ 1 \end{pmatrix}$; i.e. $\frac{a_n}{b_n} \rightarrow \sqrt{3}.$

• Conceptually a beautiful answer. Apr 9, 2017 at 16:48

Hint: First observe that $a_n -b_n\sqrt{3}=(2-\sqrt{3})^n$. Combine with the assumption, we get $a_n^2 - 3b_n^2 =1$ for all $n$. Then $\frac{a_n^2}{b_n^2}-3=\frac{1}{b_n^2}$. You should prove that $b_n \to \infty$. Hence, $\frac{a_n}{b_n} \to \sqrt{3}$ since $\frac{a_n}{b_n}>0$.

Note that

$$a_n - b_n\sqrt{3} = (2-\sqrt{3})^n$$

Hence,

$$2a_n = (a_n - b_n\sqrt{3}) + (a_n + b_n\sqrt{3}) \Rightarrow a_n = \dfrac{(2+\sqrt{3})^n + (2-\sqrt{3})^n}{2}$$

Expressing $b_n$ we obtain:

$$b_n = \dfrac{(2+\sqrt{3})^n - a_n}{\sqrt{3}} = \dfrac{(2+\sqrt{3})^n - (2-\sqrt{3})^n}{2\sqrt{3}}$$

Now it should be easy to compute the limit of $\dfrac{a_n}{b_n}$. Since $(2-\sqrt{3})^n \to 0$, we can conclude that

$$\lim\limits_{n\to+\infty}\dfrac{a_n}{b_n} = \lim\limits_{n\to+\infty} \dfrac{(2+\sqrt{3})^n}{(2+\sqrt{3})^n/\sqrt{3}} = \sqrt{3}$$