Solving a recurrence to a problem I was given a problem by a friend where I had to place blocks in a specific pattern such that each block was of area 1 but had length/width such that it fit exactly on the side of the block made until then. What I had to find out is height to width ratio after a long time. Im skipping the problem details because it's irrelevant.
I've translated the problem into math, and Ive landed with the following recurrence relation. 
If $h_k, w_k$ are the height and width of the $k$-th block I place, then the following recurrence emerges:
\begin{align}
w_{2k} &= w_{2k-1} + w_{2k-2}\\
h_{2k} &= \frac1{w_{2k}} \\
\text{and}\\
h_{2k+1} &= h_{2k} + h_{2k-1} \\
w_{2k+1} &= \frac1{h_{2k+1}}\\
\end{align}
with $h_0, h_1, w_0, w_1 = 1$
And I want to explore the convergence and hence the limit of $h_k/w_k$.
I've managed to remove $w_k$ from the equations and arrived at 
$$h_{2k+1} = h_{2k} + h_{2k-1}$$
and 
$$h_{2k} = \frac1{\frac1{h_{2k-1}} + \frac1{h_{2k-2}}}$$
where I have to find $$\lim_{k\to\infty} h_{2k}^2$$
Any help or suggestion is appreciated. Links or references as to how such recurrences are solved is also welcome.
 A: We may unfold the previous term of the same parity in your recurrence, and arrive at:
$$h_{2k+1} = h_{2k} + h_{2k-1} = h_{2k} + h_{2k-2} + h_{2k-3} = h_{2k} + h_{2k-2} +\dots+ h_2$$
$$h_{2k} = \frac1{\frac1{h_{2k-1}} + \frac1{h_{2k-3}}+\dots+{1\over h_1}}$$
Quite obviously, a non-trivial limit is impossible: odd terms diverge to infinity, even terms converge to zero.
Should it be otherwise (say, $h_{2k}\underset{k\to\infty}\longrightarrow c>0$), it would follow that for $k$ large enough, $h_{2k+1}\approx c\cdot k$, hence $h_{2k}\approx{c\over{1\over2k}+{1\over2k-2}+\dots}\approx{2c\over\ln k}$, which tends to $0$, which would make a contradiction. (Yes, all that hand-waving with "$\approx$" can be made rigorous, and easily so.)
We don't even have to bother looking for the explicit form of $h_k$, though that's also doable.
A considerably more interesting problem would be to find $\lim\limits_{k\to\infty} \left(\color{red}k\cdot h_{2k}^2\right)$ (related to the Wallis infinite product), but that's another story.
A: I am not sure if there is a general method of solving such problems but you should almost always try to find a non-recurrent formula for your equations. After writing out a few terms of the height sequence, you may notice that $h_{2k}=\frac{(2k-1)!!}{(2k)!!}$ and $h_{2k+1}=\frac{(2k+1)!!}{(2k)!!}$, where n!! is the double factorial of n. This is indeed true and can be quite easily proven using induction.
Now you want to find $\lim_{n\to\infty}h^2_{2n}$. We can rewrite $h_{2n}=\frac{(2n)!}{(2^n n!)^2}$ and then use Stirling's approximation $n! \sim n^n \cdot e^{-n} \cdot \sqrt{2 \pi n}$:
$$\frac{(2n)!}{(2^n n!)^2} \sim \frac{(2n)^{2n} \cdot e^{-2n} \cdot \sqrt{4 \pi n}}{(2^n \cdot n^n \cdot e^{-n} \cdot \sqrt{2 \pi n})^2} = \frac{1}{\sqrt{\pi n}}$$
Therefore we see that $\lim_{n\to\infty}h^2_{2n} = \lim_{n\to\infty} \frac{1}{\pi n} = 0$.
