Solving the Radius of Convergence of a Periodic Power Series

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Homework Statement



Consider the power series

Σanxn = 1+2x+3x2+x3+2x4+3x5+x6+…

in which the coefficients an=1,2,3,1,2,3,1,... are periodic of period p=3. Find the radius of convergence.

Homework Equations





The Attempt at a Solution


My attempt at a solution was to first state that the series was geometric as r=x
In order for this series to converge, the absolute value of r must be less than 1.
|r|<1
Therefore |x|<1.
Based on this, the interval of convergence is (-1,1) and the radius is r=1.

Based on my solution, I did not take into account an... If there is another solution I would need to take an into account to solve the radius of convergence please let me know.
 
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What if x is negative? Then you'd have an alternating series. You have to consider where this series converges.
 
But the series isn't a geometric series. So you don't know |x|<1 yet. You can split it up into some series that are geometric, though.
 
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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