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.
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2Use the fact that $a_{n+1} = (1 + \frac{1}{n}) a_{n-1}$. – anomaly Jun 04 '16 at 19:33
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1To put anomaly's point in another way: Doing one step of the recursion gives you an inconvenient reciprocal, but doing it again flips it back again. That suggest that one should think in terms of $a_{n}$ in relation to $a_{n-2}$ rather than $a_{n-1}$. – Semiclassical Jun 04 '16 at 19:47
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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.
peter.petrov
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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?
Clement C.
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