Calculus 2, Series Convergence Questions?

MMhawk607
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I have some problems here with Series and Convergence...

Here are the problems and my guesses at it.

http://img822.imageshack.us/img822/9523/23341530.png

It won't tell me which one is wrong, but it just says one/all is wrong. Any help is appreciated.

Attempts at solving, I tried using ratio test for these but I can't seem to get it. I need more comprehension of these types of problems I know, but these problems are due by tonight.
Alternating series really give me trouble too.
 
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OH

It was because of the third problem

sin(3n)/n^2

Is it because

sin(3n) / (n^2) ≤ 1/(n^2)

Therefore the p test makes it absolutely convergent?

I had the other 4 right, I figured that one was A though, is this the reason?
 
MMhawk607 said:
OH

It was because of the third problem

sin(3n)/n^2

Is it because

sin(3n) / (n^2) ≤ 1/(n^2)

Therefore the p test makes it absolutely convergent?

I had the other 4 right, I figured that one was A though, is this the reason?
Hello MMhawk607. Welcome to PF !

Yes, you had #3 wrong .

It is absolutely convergent.
 
mmhawk607 said:
i have some problems here with series and convergence...

Here are the problems and my guesses at it.

http://img822.imageshack.us/img822/9523/23341530.png

it won't tell me which one is wrong, but it just says one/all is wrong. Any help is appreciated.

Attempts at solving, i tried using ratio test for these but i can't seem to get it. I need more comprehension of these types of problems i know, but these problems are due by tonight.
Alternating series really give me trouble too.
...                         .
 
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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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