Change of series sign: [1 - ((pi*z)^2)/6 + ...]^{-1} = [1 + ((pi*z)^2)/6 + ...]

[Mondo]

Hello,

I have stumbled upon following transformation [imath][1 - \frac{(\pi z)^2}{6} + ...]^{-1} = [1 + \frac{(\pi z)^2}{6} + ...][/imath].
The question is simple - how was it done?

 

[Dr.Peterson]

Hello,

I have stumbled upon following transformation [imath][1 - \frac{(\pi z)^2}{6} + ...]^{-1} = [1 + \frac{(\pi z)^2}{6} + ...][/imath].
The question is simple - how was it done?

What's the context? That can make a big difference!

Or to put it another way ... What's in the dots? You can hide a lot there.

But possibly the answer is related to the expansion of [imath](1-x)^{-1}[/imath].

 

[Mondo]

It is as part of the [imath]\cot(\pi z)[/imath] expansion -> [math]\cot(\pi z) = \frac{\cos(\pi z)}{\sin(\pi z)} = \frac{1}{\pi z}[1-\frac{(\pi z)^2}{2} + ...] [1 - \frac{(\pi z)^2}{6} + ...] ^{-1} = \frac{1}{\pi z} [1-\frac{(\pi z)^2}{2} + ...][1 + \frac{(\pi z)^2}{6} + ...][/math]And I don't understand the last part, where minus becomes a plus.

 

[khansaheb]

Using Taylor's series, can you expand [imath]\frac{1}{1-x}[/imath], around x=0 ?