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I have $b_n = \frac{a_n}{a_{n-1}}$ where $a_n$ is the number of fish caught per year and is calculated by an average of the two previous years catches.

I am trying to show that $(b_n)_{n=1}^{\infty}$ converges to a limit.

So far:

I have set $(b_n)_{n=2}^{\infty}$ to the orbit of $b_1$ under $f(x)$ where $$f(x) = \frac{1}{2}\left(1 + \frac{1}{x}\right)$$

And with this, considering $f^2(x)$, I have proved that $$\lim_{n\rightarrow\infty}f^{2n+1}(x) =\lim_{n\rightarrow\infty}f^{2n} = x_0 = 1$$ using the contraction mapping theorem.

Now however, I am not sure how to adapt this to show that $(b_n)_{n=1}^{\infty}$ also converges to the same limit of $1$.

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You have shown that both $b_{2n}\to 1$ and $b_{2n+1}\to 1$. This implies that $b_n\to 1$ (as easily seen by directly applying the $\epsilon$-definition of limit).

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