Radius and Interval of Convergence for (3^n x^n)/(n+1)^2 Series

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Have to find the radius of convergence and interval of convergence,
the series is (3^n x^n ) / (n+1)^2,
did the ratio test and found the radius of convergence to be the 1/3.
now for finding the interval of convergence I plug in -1/3 and 1/3 into x and find out if it converges or not

For 1/3, it converges due to p-series, 2>1.

But for -1/3 I know it converges but can see why? Any help here at this endpoint?
 
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You could use the alternating series test.

More simply, you can relate the series at -1/3 to the series at 1/3...
 
I don't quite understand, the alternating series only works when An+1< or = to An, and in that series it doesnt, because An is negative and An+1 is postivie?
Could you elaborate and how I would relate it to the series at 1/3?
 
Look at the alternating series test again, it's the absolute values of the terms that are decreasing (and going to zero) while the sign is alternating.


The series at 1/3 is the absolute values of the terms of the series at -1/3, i.e. you've already should that the series at -1/3 is absolutely convergent.
 
iiiiiiiiiii...
my book doesn't say absolute value... so idk. I see what you mean if that's true. Why doesn't my book say that it says just that its decreasing or equal too...
 
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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