Testing a series for convergence

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SUMMARY

The discussion focuses on determining the convergence of the series \(\sum_{n=2}^\infty{\sqrt{n^3+3}-\sqrt{n^3-3}}\). The user attempts to apply various convergence tests, including the ratio test and the comparison test, but encounters difficulties in finding a suitable comparison series. The transformation using the conjugate leads to the series \(\sum_{n=2}^\infty{\frac{6}{\sqrt{n^3+3}+\sqrt{n^3-3}}}\), which is pivotal for further analysis. The user seeks assistance in identifying an appropriate series for comparison.

PREREQUISITES
  • Understanding of series convergence tests, specifically the comparison test, ratio test, and root test.
  • Familiarity with algebraic manipulation, particularly the use of conjugates in simplifying expressions.
  • Knowledge of limits and their application in evaluating series behavior as \(n\) approaches infinity.
  • Basic proficiency in LaTeX for mathematical notation and expression formatting.
NEXT STEPS
  • Study the application of the comparison test in detail, focusing on identifying suitable series for comparison.
  • Learn about the integral test for convergence and its practical applications in series analysis.
  • Explore advanced techniques for manipulating series, including the use of asymptotic analysis.
  • Practice solving similar series convergence problems to enhance problem-solving skills in calculus.
USEFUL FOR

Students and educators in calculus, mathematicians focusing on series analysis, and anyone involved in advanced mathematical problem-solving who seeks to deepen their understanding of series convergence techniques.

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EDIT: The latex doesn't seem to be working at all... not exactly sure why this is. I can't delete the post, so, uh... never mind, I guess?
EDIT v. 2.0: Yeah, so copying and pasting images from LatexIt is apparently beyond me... Thanks for the help, razored! Right, so, any help anyone can give will be well appreciated!

Homework Statement


Determine whether the following series converges or diverges. If possible, determine the sum of the series exactly. Justify your answer with the proper series test.
[tex]\sum_{n=2}^\infty{\sqrt{n^3+3}-\sqrt{n^3-3}}[/tex]

Homework Equations


Comparison test, integral test, root test, ratio test, etc.

The Attempt at a Solution


Multiplying by the conjugate
[tex]\sqrt{n^3+3}+\sqrt{n^3-3}[/tex]
produces the series
[tex]\sum_{n=2}^\infty{\frac{6}{\sqrt{n^3+3}+\sqrt{n^3-3}}}[/tex].

My first inclination was to try and find a series which is clearly larger, but I'm having trouble doing that. In previous problems like this that I've seen, there's been an n in the denominator, allowing me to eliminate the terms with square roots and produce a series which is always greater.

In this case, though, I can't, so I gave the ratio test a try, with painful results:
[tex]\lim_{n\rightarrow\infty}\frac{\sqrt{(n+1)^3+3}+\sqrt{(n+1)^3-3}}{\sqrt{n^3+3}+\sqrt{n^3-3}}[/tex].
I tried expanding the cubic terms, making it a bit ridiculous.
[tex]\lim_{n\rightarrow\infty}\frac{\sqrt{n^3+3n^2+3n+4}+\sqrt{n^3+3n^2+3n-2}}{\sqrt{n^3+3}+\sqrt{n^3-3}}[/tex]
 
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Your Work :

http://texify.com/img/%5CLARGE%5C%21%5Ctext%7BWork%20from%20poster%3A%20%7D%20%5C%5C%5Csum_%7Bn%3D2%7D%5E%5Cinfty%7B%5Csqrt%7Bn%5E3%2B3%7D-%5Csqrt%7Bn%5E3-3%7D%7D%20%5C%5C%5Csqrt%7Bn%5E3%2B3%7D%2B%5Csqrt%7Bn%5E3-3%7D%20%5C%5C%5Csum_%7Bn%3D2%7D%5E%5Cinfty%7B%5Cfrac%7B6%7D%7B%5Csqrt%7Bn%5E3%2B3%7D%2B%5Csqrt%7Bn%5E3-3%7D%7D%7D%5C%5C%5Clim_%7Bn%5Crightarrow%5Cinfty%7D%5Cfrac%7B%5Csqrt%7B%28n%2B1%29%5E3%2B3%7D%2B%5Cs%20%20qrt%7B%28n%2B1%29%5E3-3%7D%7D%7B%5Csqrt%7Bn%5E3%2B3%7D%2B%5Csqrt%7Bn%5E3-3%7D%7D%5C%5C%20%5Clim_%7Bn%5Crightarrow%5Cinfty%7D%5Cfrac%7B%5Csqrt%7Bn%5E3%2B3n%5E2%2B3n%2B4%20%20%7D%2B%5Csqrt%7Bn%5E3%2B3n%5E2%2B3n-2%7D%7D%7B%5Csqrt%7Bn%5E3%2B3%7D%2B%5Csqrt%7Bn%5E3-3%7D%7D.gif
 
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