Convergence of a Challenging Series with an Additional Factor

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Homework Statement



\sum from n=2 to \infty of n/((n2-5)*(ln n)2)

Homework Equations





The Attempt at a Solution


I've tried Limit Comparison but I always get a limit of 0 which will not work. Ratio test doesn't help. I don't think a direct comparison can be made but that seems to be the only other option...
 
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Hi simba924!

To simplify the problem a bit, show that

\frac{n}{n^2-5}\le\frac{2}{n}

for large n. Then apply the http://en.wikipedia.org/wiki/Integral_test_for_convergence" .
 
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Ahh thanks. Can you explain that in a little more detail though, I don't really get it.
 
Hey I still need some help with this one. I'm pretty sure that \sum 2/n is not convergent because it is in form integer/np where p=1

Can anyone help?
 
simba924 said:
Hey I still need some help with this one. I'm pretty sure that \sum 2/n is not convergent because it is in form integer/np where p=1

Can anyone help?

Remember, you still have the additional factor 1/(ln n)^2 (without which it would diverge). To do the integral test you have to check that

\int_2^{\infty}\frac{1}{x(\log(x))^2}

exists (you can find an explicit antiderivative, Hint: substitution).
 
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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