Converge or Diverge: Solving ∞Ʃ (kth root of k)/k^3 using Comparison Test

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Homework Statement


Ok, so I was given this problem:

Ʃ (kth root of k)/k^3 = k^(1/k) / k^3
k=1

Homework Equations


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The Attempt at a Solution


I know that I have to use the comparison test, but am unsure how to apply it. Can somebody help me?
 
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You can write that into Ʃk1/k-3. Can you find a nice real number r that has the property that r>1/k-3 for all k and Ʃkr converges?
 
No I can't.
 
Can you tell us for which real numbers r, the series \sum_k k^r converges?
 
I know that the series converges, but do not know what to use for the second series. I need to prove its convergence using one of the tests.
 
catsfanj said:
I know that the series converges, but do not know what to use for the second series. I need to prove its convergence using one of the tests.

What series converges?? What second series?? :confused:

Can you answer my question about \sum k^r?? For which r does it converge?
 
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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