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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I personally think the ratio test would be the easiest to use, here.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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