Convergence of Series with Alternating Terms

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Homework Statement


E from 1 to infinity [2+(-1)^n]/[n(n^1/2)]


Homework Equations



We only have had learned comparison tests, power series, and divergence tests.

The Attempt at a Solution



I decided to split the function into its numerator multiplied by its denominator:
2+(-1)^n * 1/[n(n^1/2)]
Then I perform the divergence test for the numerator in which the limit doesn't exist, meaning it diverges and the p-series test for the denominator whose ratio is 3/2 which means it converges. I then made the assumption that a diverging function multiplied by a converging function would diverge. But I don't know if I am allowed to do this.
 
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notice that
1 <= [2+(-1)^n] <= 3
 
mvpshaq32 said:
I decided to split the function into its numerator multiplied by its denominator...

I then made the assumption that a diverging function multiplied by a converging function would diverge. But I don't know if I am allowed to do this.
No, you can't do this. You need to look at the term as a whole.

You probably find the numerator confusing. A good tactic is to replace something that looks complicated with a simple quantity that either bounds it from above or from below, depending on what you want to prove. Then show the simplified series converges or diverges. Use lanedance's hint.
 
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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