Complex Analysis: Radius of Convergence

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Homework Statement


Find the radius of convergence of the power series:
a) \sum z^{n!}
n=0 to infinity

b) \sum (n+2^{n})z^{n}
n=0 to infinity

Homework Equations


Radius = 1/(limsup n=>infinity |cn|^1/n)


The Attempt at a Solution


a) Is cn in this case just 1? And plugging it in, the radius is 1?

b) cn = n+2^{n}, so then limsup n=> infinity |n+2^{n}|^{1/n} => ?? I'm stuck at this point.

i'm also confused in general, is cn just a sequence of coefficients, and what is zn? And I have other formulas for figuring out the radius of convergence, such as the ratio test. I'm not sure when to use which methods. Thank you!
 
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I also don't understand why z^n isn't used in the calculation of the radius of convergence.
 
So I'm running into trouble for part b still, any help would be greatly appreciated. The answer key says 1/2, but I don't know how to derive that.
 
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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