Give a recursive definition of the set: $C:= \{ \frac{1}{3^2}, \frac{1}{6^2}, \frac {1}{9^2}, \frac{1}{12^2}, \frac{1}{15^2} \ldots \}$

Question

Question: Give a recursive definition of the set: $$C:= \left\{ \frac{1}{3^2}, \frac{1}{6^2}, \frac {1}{9^2}, \frac{1}{12^2}, \frac{1}{15^2}, \ldots \right\}$$

So I see this as the set $\left\{ \frac{1}{(3k)^2}\right\}$ for integers $k$ with $k \geq 1 $.

Attempt: First the base case will be that $\frac{1}{3^2}$ is in the set.

Then if $x = \frac{1}{(3k)^2}$ is in the set, then

$$k = \sqrt{\frac{1}{9x}}$$

And so I am imagining that for $k+1$ we say that if $x$ is in the set then $$\frac{1}{(3 (\sqrt{1/(9x)} + 1))^2}$$ is also in the set. I see that this works for $x=1/9$ since plugging this value in gives us $\frac {1}{6^2}$.

I'm not at all confident this is the correct solution or if there is an easier way of seeing this problem. Any hints insights appreciated.

Answer 1

Let $a_k$ be the $k$th element of the sequence. Then $$\frac{a_{k+1}}{a_k} = \frac{1/(3(k+1))^2}{1/(3k)^2} = \frac{1}{((3k+1)/3k)^2} = \frac{1}{\left(1+\frac{1}{3k}\right)^2},$$ and you can derive a recurrence formula for $a_{k+1}$ from that.