Region of Convergence for Series in x+2y: Description and Solution

ehrenfest
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[SOLVED] radius of convergence

Homework Statement


Let D be th region in the xy plane in which the series
\sum_{k=1}^{\infty}\frac{(x+2y)^k}{k}
converges. Describe D.

Homework Equations


The Attempt at a Solution


By the ratio test, we find the radius of converge of the series in x+ 2y to be 1. So, the series will converge when |x+2y| < 1. This region is a rectangle.

What is wrong with this?
 
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Be careful with the ratio test because it only tests for absolute convergence. However you are close. It should be x+2y &lt; 1 and x+2y \geq -1. This is not a rectangle. It's a "strip" through the plane.
 
I am an idiot.
 
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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