Determining if series converges or diverges

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Homework Statement
Determine if the following series converges or diverges using any appropriate tests
Relevant Equations
limit comparison test
1633162415960.png

Is it valid to use limit comparison test to compute the following series?
If it is, would my reasoning be valid?

Thank you!
 
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What precisely is your reasoning?
 
An orthodox way is to investigate
|\frac{a_{n+1}}{a_n}|
is greater or less than 1.
 
Sunwoo Bae said:
Homework Statement:: Determine if the following series converges or diverges using any appropriate tests
Relevant Equations:: limit comparison test

View attachment 290042
Is it valid to use limit comparison test to compute the following series?
If it is, would my reasoning be valid?

Thank you!
What does the limit comparison test say exactly?
Isn't there a specific relation to be fulfilled between ##a_n## and ##b_n##?
Do your successions ##a## and ##b## fulfill such relation?
 
Sunwoo Bae said:
Homework Statement:: Determine if the following series converges or diverges using any appropriate tests
Relevant Equations:: limit comparison test

View attachment 290042
Is it valid to use limit comparison test to compute the following series?
If it is, would my reasoning be valid?

Thank you!

Is <br /> \frac{n^n}{n!} = \frac{n}{1} \frac{n}{2} \cdots \frac{n}{n-1} \frac{n}{n}<br /> greater than, or less than, 1 for large n? Given that, which of the following is true:
1. a_n &lt; b_n for large n.
2. a_n &gt; b_n for large n.

How does that affect the comparison test?
 
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Likes jim mcnamara and Delta2
I personally think the ratio test would be the easiest to use, here.
 
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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