Find a closed form solution for $B_n = \frac 3{B_{n-1}}$ I am trying to find a closed form solution for this recurrence relation, but it is a repeating one and I can't seem to figure it out. Any ideas?
 A: Hint:
Let $n$ be even, then
$$
B_n=\frac {3}{B_{n-1}}=\frac {3}{\frac{3}{B_{n-2}}}=B_{n-2}
$$
So the sequence is constant in the even terms. What happens with the odd terms?

As I said in the comments, I'd call 
$$
B_n=\begin{cases}B_1, & n\text{ odd.}\\
\frac 3 {B_1}, & n \text{ even.}
\end{cases}
$$
A closed form solution (there's no recurrence relation). However, I'm not sure what your definition of closed form solution is.
A: Let $B_1$ be the first term.
If $n$ is odd, $B_n=B_1$.
If $n$ is even, $B_n=\frac{3}{B_1}.$
A: $$
B_1 \longmapsto \frac 3 {B_1} \longmapsto \left( \frac 3 {\left( \frac 3 {B_1} \right)} = B_1 \right) \longmapsto \frac 3 {B_1} \mapsto B_1 \longmapsto \frac 3 {B_1} \longmapsto B_1 \longmapsto \cdots
$$
That's the solution.
You could say it's $\begin{cases} B_1 & \text{when $n$ is odd,} \\[2pt] 3/B_1 & \text{when $n$ is even.} \end{cases}$
However, maybe the expression "closed form" is what you're worried about.  That's a bit vaguely defined but people seem to know what it is when they see it.  How about something like this:
$$
\frac{B_1 + (3/B_1)} 2 + (-1)^n \frac {(3/B_1)-B_1} 2.
$$
