A simple nonlinear recurrence relation involving two sequences. Starting from a messy differential equation, I arrived at the following simple pair of update rules:
$$ a(t+1) = a(t) + \frac{1}{b(t)}$$ $$b(t+1) = b(t) + \frac{b(t)}{a(t)}$$ My question is: suppose we initialize it somewhere (say $a(0)=b(0)=1$). How will the answer grow with time?
Intuition: as time goes on, both $a(t),b(t)$ increase, but the increase of each inhibits the rate at which the other can grow. Simulations show that $a(t)$ seems to approach a finite limit while $b(t)$ appears to exponentially in $t$. 
 A: Yes, we can show that $\displaystyle a_n$ is convergent (say with limit $a$) and that $\displaystyle b_n$ is unbounded, with $\displaystyle b_n^{1/n} \to 1 + \frac{1}{a}$ 
I take the sequence to start with $\displaystyle 1$, so assume $\displaystyle a_1 = b_1 = 1$.
It is easy to prove that $\displaystyle b_n \ge a_n \ge 1$ (by induction) and thus we have that $b_n \ge n$.
Thus we have that $\displaystyle a_{n+1} - a_n \le \frac{1}{n}$
Adding up gives us $\displaystyle a_{n} - 1 \le \sum_{k=1}^{n-1} \frac{1}{k}$
and so $\displaystyle a_n \lt 2 \log (n+1)$
Now $\displaystyle \frac{b_{n+1}}{b_n} = 1 + \frac{1}{a_n}$ and taking the telescoping product gives us
$\displaystyle b_n = \prod_{k=1}^{n-1} (1 + \frac{1}{a_k}) \tag{1}$
Now since $a_n \lt 2 \log(n+1)$ we have that 
$\displaystyle  b_n \gt \prod_{k=1}^{n-1} \left(1 + \frac{1}{2\log (k+1)}\right)$
Using the inequality $\displaystyle 1 + a \ge 2\sqrt{a}$ we get
$\displaystyle b_n \gt \frac{2^{(n-1)/2}}{\sqrt{\log (n-1)!}}$
Thus $\displaystyle  b_n \gt \frac{2^{(n-1)/2}}{n^2}$ for sufficiently large $\displaystyle  n$ (Stirling approximation gives us $\displaystyle  \log n! \sim n \log n$)
Thus
$\displaystyle  a_{n+1} - a_n \lt \frac{n^2}{2^{(n-1)/2}}$
for sufficiently large $n$.
Thus $\displaystyle  a_n \lt C + \sum_{k=1}^{\infty} \frac{k^2}{2^{(k-1)/2}}$
Since the latter series converges, $\displaystyle  a_n$ is bounded above.
We can also easily see that $a_{n+1} \gt a_n$. 
Thus $\displaystyle  a_n$ is convergent, to say $\displaystyle  a$.
Using (1) we see that
$\displaystyle  \frac{\log b_n}{n} = \frac{1}{n}\sum_{k=1}^{n-1} \log (1+\frac{1}{a_k})$
Since $\log (1 + \frac{1}{a_n})$ converges, so does the arithmetic mean (and to the same limit).
Thus
$\displaystyle  \frac{\log b_n}{n} \to \log (1 + \frac{1}{a})$
Thus
$\displaystyle  b_n^{1/n} \to 1 + \frac{1}{a}$
