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I have found that the residue of the function

$f(x) = \frac{1}{(1+x)^{n+1}}$

is

$R= 2\pi i\frac{(n+1) \cdot (n+2) \cdot ... \cdot (2n)}{n! (2i)^{2n+1}}$.

I am having trouble with showing that this can be written as

$R=\frac{1 \cdot 3 \cdot 5 \cdot... \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot ... \cdot (2n)} \pi.$

Eugenia Dickens
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1 Answers1

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Using the odd-factorial, $$ 1\cdot 3\cdot 5 \cdot\cdots\cdot (2n-1) = \frac{(2n)!}{2^n n!}, 2\cdot4\cdot\cdots\cdot 2n = 2^n n! $$ The second equation is equal to (discarding the $\pi$) $$\frac{1 \cdot 3 \cdot 5 \cdot... \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot ... \cdot (2n)} = \frac{(2n)!}{2^{2n} (n!)^2} $$ Notice that $$ \frac{(2n)!}{n!} = (n+1)(n+2)\cdot\cdots\cdot(2n) $$ Therefore, $$ \frac{(2n)!}{2^{2n} (n!)^2} = \frac{(n+1)(n+2)\cdot\cdots\cdot(2n)}{2^{2n} n!} $$ The rest should be easy.

Community
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Henricus V.
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  • Thank you! This was very helpful. I figured there was some kind of identity for the odd and even factorials but wasn't able to find them. – Eugenia Dickens Mar 29 '16 at 14:57