Calculate Limit of Series: Find $\displaystyle\lim_{k\to\infty}\frac{n_k}{d_k}$

[sennyk]

I want to calculate the following:

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

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.

Please any help is appreciated.

 

[nicksauce]

The terms of n_k and d_k are members of the Fibonacci series.
http://en.wikipedia.org/wiki/Fibonacci_series

There are a few proofs of the convergence of the ratio of consecutive terms on that page.

 

[2^Oscar]

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

 

[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.

 

[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.

We can start with rearranging the second equation:
n_{k-1}=d_{k}-d_{k-1}\to{n_{k}=d_{k+1}-d_{k}(*)

Inserting these relations into the first, we get:
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.