Sum of Geometric Series with cosine?

dan38
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Homework Statement


With a series like:
pi^(n/2)*cos(n*pi)

How am I meant to approach this?
Do I use the Squeeze Theorem?


Homework Equations





The Attempt at a Solution

 
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I believe you are trying to find the sum as n->infinity?

If so, you should start by checking whether the series converges or diverges.
 
Hi dan38
As Infinitum suggested, you probably are looking for convergence/divergence of the serie
I suppose you are being confused by the cos 'trick'.
But look at is closely.. what possible values do you have for cos(n∏) ?
1, -1, 1, -1, ... that is, (-1)^n
Are you familiar with D'Alembert's rule to decide on convergence ?

Cheers...
 
ah I see
so then it would become

(pi)^0.5 * (pi)^n * (-1)^n

pi^0.5 * ( - pi )^n

-pi^1.5 * ( - pi )^(n-1)

Of the form required
where a = -pi^1.5 and r = -pi

since r > 1
then it is divergent?
 
Well you're notation is confusing, you are supposed to use absolute values and somehow you get to -∏,
but yes, you would get to ∏>1 and therefore it diverges

Cheers...
 
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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