Limit of a series help

1. Sep 8, 2009

sennyk

I want to calculate the following:

$$\displaystyle\lim_{k\to\infty}\frac{n_k}{d_k}$$
where,
$$n_0 = 2$$
$$d_0 = 1$$
$$n_k = 2n_k_-_1 +d_k_-_1$$
$$d_k = n_k_-_1 + d_k_-_1$$

For the life of me I have no idea how to do this. By the way, the answer is supposed to be
$$\frac{1 + \sqrt{5}}{2}$$

This is not a homework problem. I was doing an electrical engineering problem and to solve the problem this series was magically solved.

2. Sep 8, 2009

3. Sep 8, 2009

2^Oscar

I'm intrigued... as the answer it is supposed to be is the golden ratio.

4. Sep 8, 2009

arildno

Okay, you may proceed as follows:
$$\frac{n_{k}}{d_{k}}=\frac{2n_{k-1}+d_{k-1}}{n_{k-1}+d_{k-1}}=1+\frac{1}{1+\frac{d_{k-1}}{n_{k-1}}}$$

Assuming that a limit exists as k (and therefore k-1) trundles off into infinity, call that limit "x".

Thus, you get the equation,
which ought to be easily solved.
$$x=1+\frac{1}{1+\frac{1}{x}}$$

Note that this is simply another way of writing the continued fractions representation of the golden ratio.

Last edited: Sep 8, 2009
5. Sep 8, 2009

arildno

Another way to proceed, would be to first solve for the one sequence, then for the seconde, and finally solve for the limiting ratio.

$$n_{k-1}=d_{k}-d_{k-1}\to{n_{k}=d_{k+1}-d_{k}(*)$$
$$d_{k+1}-d_{k}=2(d_{k}-d_{k-1})+d_{k-1}\to{d}_{k+1}-3d_{k}+d_{k-1}=0$$
Noting from (*) that we have $d_{0}=1,d_{1}=3$, you should be able to solve for the d-sequence.