Convergence or divergence (series)

bfusco
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Homework Statement


Ʃ[(-1)^n (cosn)^2]/√n

The Attempt at a Solution


i don't have the slightest clue where to start
 
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Since this is a series and there is an alternating sign in the consecutive partial sums, you should use the Alternating Series Test.
\sum^{\infty}_{n=?}(-1)^n \frac{(\cos n)^2}{√n}
You have not stated the initial value for n.

The first step: Let a_n=\frac{(\cos n)^2}{√n}. Find the limit and test if it's zero.
Second step: Is a_{n+1}\leq a_n?
If both of these conditions are satisfied, then the series converges.
 
Last edited:
The initial value for n doesn't matter. It's presumably not zero.
 
JG89 said:
The initial value for n doesn't matter. It's presumably not zero.

The initial value of n is normally ignored, but stating the latter forms part of the proper notation when writing the series with the summation symbol.
 
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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