# Thread: Approximation of root 2.

1. ## Approximation of root 2.

Hello it's me again and this is the last algebra question for a while!

Start with $\displaystyle a_1 = b_1 = 1$, and define

$\displaystyle a_{n+1} = a_n + 2b_n, \:\ b_{n+1} = a_n + b_n \:\ (n \geq\ 20)$

Work out $\displaystyle \frac{a_n}{b_n}$ for $\displaystyle n = \{1,2,4,5,6,\}$. You should find that $\displaystyle \frac{a_n}{b_n}$ is getting closer to $\displaystyle \surd{2}$ as $\displaystyle n$ increases.

Prove by induction that $\displaystyle a_n^2 - 2b_n^2 = (-1)^n$,
and deduce that $\displaystyle \frac{a_n}{b_n} \rightarrow \surd{2}$ as $\displaystyle n \rightarrow \infty$.

I have no problem up to the point of proving the formula by induction but I have no clue as to how to use the formula to make the deduction. Any help would be appreciated.

2. Hello, slevvio!

Prove by induction that $\displaystyle a_n^2 - 2b_n^2 \:= \:(-1)^n$.

Deduce that $\displaystyle \frac{a_n}{b_n} \rightarrow \sqrt{2}$ as $\displaystyle n \rightarrow \infty$.
You got the inductive proof . . . Good for you!

We have: .$\displaystyle a_n^2 \;=\;2b_n^2 + (-1)^n$

Divide by $\displaystyle b_n^2\!:\;\;\frac{a_n^2}{b_n^2} \;=\;2 + \frac{(-1)^n}{b^n}$

. . Then: .$\displaystyle \left(\frac{a_n}{b_n}\right)^2 \;=\;2 + \frac{(-1)^n}{b_n}$

Take limits: .$\displaystyle \lim_{n\to\infty}\left(\frac{a_n}{b_n}\right)^2 \;=\;\lim_{n\to\infty}\left[2 + \frac{(-1)^n}{b_n}\right]$

As $\displaystyle n\to\infty, \;b_n \to \infty$ also.
Hence, that last fraction approaches 0.

. . and we have: .$\displaystyle \lim_{n\to\infty}\left(\frac{a_n}{b_n}\right)^2 \;=\;2 + 0 \;=\;2$

Therefore: .$\displaystyle \lim_{n\to\infty}\,\frac{a_n}{b_n} \;=\;\sqrt{2}$