Series Convergence: \sum_{n=2}^{\infty}\frac{1+n+n^2}{\sqrt{1+n^2+n^6}}

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Determine whether the series \sum_{n=2}^{\infty}a_n converges absolutely,converges conditionally or diverges.If

a_n=\frac{1+n+n^2}{\sqrt{1+n^2+n^6}}


For several n I get a_n>\frac{1}{n} so I decided that this series is divergent.Right?
 
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That would do it. But did you prove that inequality? Just looking at 'several n' basically means you are guessing.
 
In fact the result can be much weaker, we don't need to prove the inequality for all n, or even that a_n is more than 1/n, just equal. Divide the terms through by n^2 and see what the n-th term as n --> infinity is.
 
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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