If $(1+\sqrt{2})^n=a_{n}+b_{n}\sqrt{2}\;,(\forall n\in \mathbb{N}).$Then $\lim_{n\rightarrow \infty}\frac{a_{n}}{b_{n}} = $

Question

Let $a_{n},b_{n}\in \mathbb{Q}$ such that $(1+\sqrt{2})^n=a_{n}+b_{n}\sqrt{2}\;,(\forall n\in \mathbb{N}).$Then $\displaystyle \lim_{n\rightarrow \infty}\frac{a_{n}}{b_{n}} = $

$\bf{My\; Try::}$ Using

$$(1+\sqrt{2})^n = 1+\binom{n}{1}(\sqrt{2})+\binom{n}{2}(\sqrt{2})^2+\cdots \cdots +\binom{n}{n}(\sqrt{2})^n$$

So $$a_{n} = \binom{n}{0}+\binom{n}{2}\cdot 2+\binom{n}{4}\cdot 2^2+\cdots \cdots $$

So $$b_{n} = \binom{n}{1}+\binom{n}{3}\cdot 2+\binom{n}{5}\cdot (2)^2+\cdots$$

Now How can i solve $$\lim_{n\rightarrow \infty}\frac{a_{n}}{b_{n}} = $$

Help required, Thanks

Answer 1

By the binomial theorem,

$$(1+\sqrt2)^n=a_n+b_n\sqrt2\implies(1-\sqrt2)^n=a_n-b_n\sqrt2.$$

Then

$$\frac{a_n}{b_n}=\sqrt2\frac{(1+\sqrt2)^n+(1-\sqrt2)^n}{(1+\sqrt2)^n-(1-\sqrt2)^n}=\sqrt2\frac{1+\left(\dfrac{1-\sqrt2}{1+\sqrt2}\right)^n}{1-\left(\dfrac{1-\sqrt2}{1+\sqrt2}\right)^n}.$$

This gives linearly converging rational approximations of $\sqrt2$. For instance, with $n=10$,

$$\sqrt2\approx\frac{6726}{4756}$$ with a relative error on the order of

$$2\left(\frac{1-\sqrt2}{1+\sqrt2}\right)^{10}\approx4.4\cdot10^{-8}.$$

Answer 2

Here is another path.

$$(a_n+b_n\sqrt{2})(1+\sqrt{2})=a_n+2b_n+(a_n+b_n)\sqrt{2}$$

thus if we let $c_n=\frac{a_n}{b_n}$ then

$$c_{n+1}=1+\frac{1}{c_n+1}$$

One can check by induction that the odd terms are increasing, the even decreasing and their difference goes to zero. Thus the limit must satisfy

$$L=1+\frac{1}{L+1}$$ and so $L=\sqrt{2}$.

Answer 3

Note first that the recursion

$$a_{n+1}+b_{n+1}\sqrt2=(1+\sqrt2)(a_n+b_n\sqrt2)=(a_n+2b_n)+(a_n+b_n)\sqrt2$$

implies $a_1,a_2,a_3,\ldots$ and $b_1,b_2,b_3,\ldots$ are both strictly increasing sequences of positive integers, since $a_1=b_1=1$ and $a_n,b_n\gt0$ implies $a_{n+1}=a_n+2b_n\gt a_n$ and $b_{n+1}=a_n+b_n\gt b_n$. In particular, $b_n\to\infty$ as $n\to\infty$.

Now, by the formal symmetry between $\sqrt2$ and $-\sqrt2$ and the general algebraic identity $(x+y)(x-y)=x^2-y^2$, we have

$$\begin{align} (1+\sqrt2)^n=a_n+b_n\sqrt2&\implies(1-\sqrt2)^n=a_n-b_n\sqrt2\\ &\implies(-1)^n=a_n^2-2b_n^2\\ &\implies{a_n\over b_n}=\sqrt{2\left(1+{(-1)^n\over b_n^2}\right)}\\ &\implies\lim_{n\to\infty}{a_n\over b_n}=\sqrt2 \end{align}$$

Answer 4

The following method may look heavy for such a simple example, but may be useful for less obvious ones. Since $$\begin{pmatrix}a_{n+1}\\b_{n+1}\end{pmatrix}=A\begin{pmatrix}a_n\\b_n\end{pmatrix}$$ where the eigenvalues of the matrix $A=\begin{pmatrix}1&2\\1&1\end{pmatrix}$ are $\lambda_\pm:=1\pm\sqrt2,$ we get the following closed form for the two sequences: $$a_n=\alpha\lambda_+^n+\beta\lambda_-^n,\quad b_n=\gamma\lambda_+^n+\delta\lambda_-^n$$ the four constants $\alpha,\beta,\gamma,\delta$ being determined by the four values $a_0=1,b_0=0,a_1=1,b_1=1.$ Solving the system gives $\alpha=\beta=\frac12,\gamma=-\delta=\frac1{2\sqrt2}$ hence $$\frac{a_n}{b_n}=\sqrt2\,\frac{\lambda_+^n+\lambda_-^n}{\lambda_+^n-\lambda_-^n}=\sqrt2\,\frac{1+(\lambda_-/\lambda_+)^n}{1-(\lambda_-/\lambda_+)^n}\to\sqrt2.$$