Proving Whether an Alternating Series is Divergent or Convergent

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Homework Statement




Determine an explicit function for this sequence and determine whether it is convergent.

an={1, 0, -1, 0, 1, 0, -1, 0, 1, ...}


The Attempt at a Solution



I came up with this function:

an = cos(nπ/2), and wrote that as sigma notation from n=0 to infinity. Is that good or should I use a quadratic?

I don't think it converges because it just oscillates and never changes, but how do I prove that?
 
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Do you have to prove it using epsilon's and stuff??
If so, what is the definition of convergence? What is the negation?
 
see answer 5
 
Last edited:
I don't have to use epsilons. But I have to provide an equation showing it. And is the function I made satisfactory, or should I express it as a quadratic?
 
This sequence has 3 different limit points -1,0 and 1, therefore it does not have a limit.
 
If you reffer to the corresponding series,then it diverges because the general term does not converge to 0.
 
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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