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$\begingroup$

I'm trying to expand the following as a power series in s but I'm having trouble.

$\frac{\mu(1-s)-(\mu-\beta s)p}{\beta(1-s)-(\mu-\beta s)p}$

and I've rewritten it in the form $1/(1-x)$ where $x = \frac{(\mu - \beta)+s(\beta - \mu)}{(\mu - \mu p)+s(\beta p - \mu)}$

I'm unsure how to proceed.

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    Hint: This is $$\frac{a-bs}{c-ds}$$ with $$(a,b,c,d)=(\mu(1-p),\mu-\beta p,\beta-\mu p,\beta(1-p))$$ which, provided that $|ds|<|c|$, is $$(a-bs)\sum_{n=0}^\infty c^{-n-1}d^ns^n$$ Can you finish this? – 2017-02-26
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    Thank you! I've finished it now. – 2017-02-26
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    Then you might post what you did as an answer. – 2017-02-26
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    Actually I made mistake. I'm confused because if you expand the bracket into the sum, you are left with an $s^n$ term and an $s^(n+1)$ term, but I only want an $s^n$ term – 2017-02-26
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    Yes -- which is why you should show what you did. – 2017-02-26
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    $(a−bs)\sum_{n=0}^{\infty} c^{-n-1}d^{n}s^{n} = \sum_{n=0}^{\infty} (ac^{-n-1}d^{n}s^{n} + bc^{-n-1}d^{n}s^{n+1}) = \sum_{n=0}^{\infty} s^{n}(ac^{-n-1}d^{n}+sbc^{-n-1}d^{n}) $ – 2017-02-26
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    Right, but note that $$\sum_{n=0}^\infty bc^{-n-1}d^ns^{n+1}=\sum_{n=1}^\infty bc^{-n}d^{n-1}s^n$$ and rejoice... – 2017-02-26

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