Mastering the Series Ratio Test: A Comprehensive Guide

nameVoid
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%5Cfrac%7B%20cos(%20%5Cfrac%7Bn%20%5Cpi%7D%7B3%7D)%7D%7Bn!%7D%20%5Cmbox%7B%20convergent%7D%0D%0A.jpg

 
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nameVoid said:
%5Cfrac%7B%20cos(%20%5Cfrac%7Bn%20%5Cpi%7D%7B3%7D)%7D%7Bn!%7D%20%5Cmbox%7B%20convergent%7D%0D%0A.jpg

What's your question?
 
is it correct to apply the ratio test on An>=an and then use the results in the comparison test where bn=An for bn convergence and bn>an
 
nameVoid said:
is it correct to apply the ratio test on An>=an and then use the results in the comparison test where bn=An for bn convergence and bn>an
I have no idea what you're saying here. You have An, an, and bn. How do these relate to your original series?

The first test you did was a comparison test, comparing with \sum 1/n!. When you use the comparison test, you should already know whether the series you're comparing to converges or diverges. If you are comparing \sum a_n to a convergent series, you have to show that an <= the corresponding term in your convergent series. If you are comparing \sum a_n to a divergent series, you have to show that an >= the corresponding term in the divergent series.

Also, the comparison test and the ratio test have to be used on series without negative terms. Your cosine series has negative terms.
 
You could use the ratio test on the original problem; remember when using the ratio test you are taking the limit of the abs value of your function. You could also use the squeeze theorem along with the comparison test.
 
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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