Can you make sure I'm doing this right?

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Homework Statement



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Homework Equations



If |z|>1, the series ∑z-n converges to 1/(z-1)

If |z|<1, the series ∑zn converges to 1/(1-z)

The Attempt at a Solution



screen-capture-7-3.png




? Am I getting there?
 
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Jamin2112 said:

Homework Equations



If |z|>1, the series ∑z-n converges to 1/(z-1)

I don't think that equation is right... Shouldn't it be for |z|>1,
\sum_{n=0}^\infty \frac{1}{z^n} = \frac{1}{1-\frac{1}{z}} = \frac{z}{z-1} \; \; ?<br />

Or am I missing something?
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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