Does the series converge or diverge?

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Homework Statement


Determine if series converges or diverges.

Homework Equations


\displaystyle \sum^{∞}_{n=0} \frac{cos n* \pi}{5^n}

The Attempt at a Solution


I have tried using the nth term test but not sure how to take the limit of cosine.
 
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whatlifeforme said:

Homework Statement


Determine if series converges or diverges.

Homework Equations


\displaystyle \sum^{∞}_{n=0} \frac{cos n* \pi}{5^n}

The Attempt at a Solution


I have tried using the nth term test but not sure how to take the limit of cosine.

Why don't you try working out the value of that cosine term for the first few values of n and see what you get?

Another hint: absolute convergence.
 
First, cos(n\pi) is just (-1)^n so this is \frac{(-1)^n}{5^n} and the ratio test works nicely.
 
HallsofIvy said:
First, cos(n\pi) is just (-1)^n so this is \frac{(-1)^n}{5^n} and the ratio test works nicely.

What is ##\sum |a_n|##, can it be bounded in some way?

Then perhaps another test will clean the rest up for you.
 
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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