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Given $ I_n = \int_0^{\pi/2} \sin^n\theta d\theta $, the recursive formula can be found as

$$ I_n = \frac{n-1}{n}I_{n-2} $$

The lecture notes later state that

$$ \frac{I_{2n-1}}{I_{2n+1}} = 1 + \frac{1}{2n} $$

When I try to calculate the ratio based on the recursive formula, I get other terms that cannot be cancelled, e.g.:

$$ \frac{I_{2n-1}}{I_{2n+1}} = \frac{\frac{2n-2}{2n-1}I_{2n-3}}{\frac{2n}{2n+1}I_{2n-1}} = \frac{(n-2)(2n+1)I_{2n-5}}{n(2n-3)I_{2n-3}} $$

Please, could you help to get the ratio?

tudou
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1 Answers1

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Initial Hint: By the first equation of $$ I_n=\frac{n-1}{n}I_{n-2} $$ you know that plugging in $n=2n-3$ you can get that $$ I_{2n-3} = \frac{2n-4}{2n-3}I_{2n-5}. $$ Try plugging this in and seeing if that gets you anywhere

Even easier version I realized after the fact: When you first do $$ \frac{I_{2n-1}}{I_{2n+1}} $$

just plug in the version of $$ I_{2n+1} = \frac{2n}{2n+1}I_{2n-1} $$ only, instead of substituting for both $I_{2n-1}$ and $I_{2n+1}$

wjmccann
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