Sequence of fractions which converge to √2

Question

I am a math aficionado and would like to know how to prove that the following sequence of fractions converge to √2 .

1/1; 3/2; 7/5; 17/12; ...

In general a(n)=a(n-1)+a(n-2) and a(n+1)=a(n-1)+2a(n-2)

Thanks in advance for your help! Matt

Answer 1

I assume your general term is $\frac{a_n}{b_n}$ where $a_n=2a_{n-1}+a_{n-2}$ and $b_n=2b_{n-1}+b_{n-2}$ (you seem to have put the 2s in the wrong place in your question).

You have a "linear recurrence relation", which is easily solved. If the roots of the associated quadratic $x^2=2x+1$ are $\alpha,\beta$, then the general solution is $a_n=A\alpha^n+B\beta^n$. Since $b_n$ satisfies the same recurrence relation, we have $b_n=A'\alpha^n+B'\beta^n$. You find the constants $A,B,A',B'$ by looking at the first two terms of each sequence.

In this case we have $\alpha=\sqrt2+1,\beta=1-\sqrt2$. Note that $|\beta|<1$, so for large $n$, we have $a_n\approx A\alpha^n,b_n\approx A'\alpha^n$. Indeed, more formally, it is easy to check that $\lim_{n\to\infty}\frac{a_n}{b_n}=\frac{A}{A'}$. If you work out $A,A'$ in this case you find that $\frac{A}{A'}=\sqrt2$.