Figuring out whether a series converges or not

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Homework Statement


I am supposed to show whether the series converges or diverges. I guess I need to use the ratio test … at least my calculus professor told us that we should use that test whenever we had complicated terms like n! and such.


Homework Equations


lim_n->inf a_(n+1)/a_n = L
If L<1 the series converges, if L>1 it diverges and if L=0 ... who knows? :p


The Attempt at a Solution


I tried to do the problem, but I am not sure if I got it right or not. My attempt at a solution is attached, along with the problem, in the image.


Thanks in advance.
 

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Sure. The ratio test gives you 0. It converges.
 
eventob said:

Homework Equations


lim_n->inf a_(n+1)/a_n = L
If L<1 the series converges, if L>1 it diverges and if L=0 ... who knows? :p
0 < 1, so if L = 0 in the Ratio Test, the series converges.
 
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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