Determine whether the series is convergent or divergent.

Sabricd
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Hello,

I have to determine whether the series converges or diverges.

It is \Sigma (-1)^n * cos(Pi/n) where n=1 and goes to infinity.

First I took the absolute value of the function and got the limit from n to infinity of cos(pi/n) and as a result I got 1 because cos(0)=1. However my textbook says it's divergent. Could you please help me understand why it is divergent?

Thank you!
 
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Nevermind! I figured out why. I guess I confused the alternating series test with some other one. :)

Thanks!
 
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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