Quick question about Ratio Test for Series Convergence

ColtonCM
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Homework Statement


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This is the question I have (from a worksheet that is a practice for a quiz). Its a conceptual question (I guess). I understand how to solve ratio test problems.

"Is this test only sufficient, or is it an exact criterion for convergence?"

Homework Equations



Recall the ratio-test: If {an}n∈N is a positive sequence and there is a number a < 1 such that eventually an+1 ≤ a then the series is convergent. If, eventually, an+1 ≥ 1 then the series is divergent.

The Attempt at a Solution



I would assume that it would be considered "only sufficient," since if the result yields a ratio of one, convergence cannot be determined, thus it is not an absolute criterion.

Would this line of reasoning be correct?

Thanks,

Colton
 
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that sounds right. it's sufficient but not necessary because there are other ways to determine whether or not a series converges eg ##\sum_{n} \frac{1}{n^{2}}## is known to converge but the ratio test doesn't give any information about it.
 
Sounds good, thanks!
 
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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