# 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.