solution to the recurrence relation $a_n=\frac{n}{a_{n-1}}$

Question

Is there a recurrence solution to $a_n=\frac{n}{a_{n-1}}$? I'm wondering if it could be done in the form of an alternating series partial to $n$ or as a trigonometric function.

Answer 1

Hint:

Set $b_n = \ln a_n$. Then $b_n = - b_{n-1} + \ln n$, and we can write $$\begin{align} b_n &= - b_{n-1} + \ln n = b_{n-2} - \ln (n-1) + \ln n\\ &= -b_{n-3} + \ln(n-2) - \ln (n-1) + \ln n \\ &\vdots \\ &= b_1 + \sum_{k=1}^n (-1)^{n-k} \ln k \end{align}$$

Can you continue?

Answer 2

For $n=2k+1$ odd you have:

$$a_{2k+1} = \frac{c . (2k+1)!!}{ (2k)!!}$$

For $n=2k$ even you have:

$$a_{2k} = \frac{(2k)!!}{c . (2k-1)!!}$$

where $c = a_1$.

This pattern can be seen by writing down the first 5-6 terms.
It should be easy to prove it by induction.