Series Convergence: What Can the Nth Term Test Tell Us?

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SUMMARY

The discussion centers on the application of the Nth Term Test for Divergence in evaluating series convergence. The user initially attempts the divergence test and then explores the ratio test, ultimately finding it inconclusive. They also consider the integral test, breaking it down into manageable parts. The key conclusion is that the Nth Term Test indicates divergence if the limit of the series does not approach zero.

PREREQUISITES
  • Understanding of series convergence tests, including the Nth Term Test for Divergence.
  • Familiarity with limits and indeterminate forms in calculus.
  • Knowledge of the ratio test and integral test for series.
  • Ability to perform u-substitution in integral calculus.
NEXT STEPS
  • Study the application of the Nth Term Test for Divergence in various series.
  • Learn about the conditions under which the ratio test is conclusive.
  • Explore the integral test in greater depth, focusing on improper integrals.
  • Review techniques for resolving indeterminate forms in limits.
USEFUL FOR

Students and educators in calculus, mathematicians analyzing series, and anyone seeking to deepen their understanding of convergence tests in mathematical analysis.

woopydalan
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Homework Statement
Determine whether each of the following series converges or not.
## \sum_{n=1}^{\infty} \frac {n+3}{\sqrt{5n^2+1}}##
Relevant Equations
Divergence test, ratio test, etc
I'm not sure which test is the best to use, so I just start with a divergence test

##\lim_{n \to \infty} \frac {n+3}{\sqrt{5n^2+1}}##
The +3 and +1 are negligible
##\lim_{n \to \infty} \frac {n}{\sqrt{5n^2}}##

So now I have ##\infty / \infty##. So it's not conclusive. Trying ratio test

##\lim_{n \to \infty} \lvert \frac {n+4}{\sqrt{5(n+1)^2+1}} \cdot \frac {\sqrt{5n^2+1}}{n+3} \rvert##
seems to yield 1, so inconclusive

Integral test
## \int_{1}^{\infty} \frac {x+3}{\sqrt{5x^2+1}} dx ##. I could separate
## \int_{1}^{\infty} \frac {x}{\sqrt{5x^2+1}} dx + \int_{1}^{\infty} \frac {3}{\sqrt{5x^2+1}} dx ##
First part of the sum would be u-sub, not sure if I even know how to do the second part of the sum
 
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Think again about ##\displaystyle{\lim_{n \to \infty}\dfrac{n}{\sqrt{5n^2}}}.##
 
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I see, 1/sqrt(5)
 
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woopydalan said:
I see, 1/sqrt(5)
Ok, so what does the nth Term Test for Divergence tell you then?
 
woopydalan said:
So now I have ∞/∞.
Which means you haven't gone far enough in evaluating the limit. If you get any of the indeterminate forms, such as ##\frac \infty \infty, \frac 0 0, \infty - \infty,## or a few others, there is still work to do.
 
Mark44 said:
Ok, so what does the nth Term Test for Divergence tell you then?
Diverges if it's not 0
 
woopydalan said:
Diverges if it's not 0
What I meant was, what does the Nth Term Test tell you about this series, something I think you have now figured out.
 

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