Absolute convergence proving that limit =1

Chaoticoli
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Homework Statement


limit as k--> infinity ABS VALUE((cos(pi*k+pi))/(cos(k*pi)))


Homework Equations





The Attempt at a Solution



Can someone prove to me why this limit is equal to 1? I have tried several other sources and I have not had any luck.
 
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Chaoticoli said:

Homework Statement


limit as k--> infinity ABS VALUE((cos(pi*k+pi))/(cos(k*pi)))

Homework Equations



The Attempt at a Solution



Can someone prove to me why this limit is equal to 1? I have tried several other sources and I have not had any luck.
If k is an integer, what is cos(πk) ?
 
It must alternate between -1 and 1.
 
Chaoticoli said:

Homework Statement


limit as k--> infinity ABS VALUE((cos(pi*k+pi))/(cos(k*pi)))

I have tried several other sources and I have not had any luck.

Tried other sources? Have you tried working it yourself? What have you done so far?
 
Chaoticoli said:
It must alternate between -1 and 1.

What is \displaystyle\ \left|\frac{\pm1}{\pm1}\right|\ ?
 
I understand now. Thanks :)
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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