Testing a series for convergence/divergence

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SUMMARY

The discussion focuses on determining the convergence or divergence of the series \(\frac{2+(-1)^n}{n^2+7}\) using the special comparison test with the convergent series \(\frac{1}{n^2}\). Participants analyze the limit as \(n\) approaches infinity of the ratio of the two series, specifically \(\lim_{n \to \infty} \frac{2+(-1)^n}{n^2+7} \div \frac{1}{n^2}\). The conclusion drawn is that since \(\frac{2+(-1)^n}{n^2+7}\) is bounded by \(\frac{3}{n^2+7}\), which converges, the original series also converges.

PREREQUISITES
  • Understanding of the special comparison test in series convergence
  • Familiarity with limits and evaluating them as \(n\) approaches infinity
  • Knowledge of convergent series, specifically \(\frac{1}{n^2}\)
  • Basic algebraic manipulation of series and inequalities
NEXT STEPS
  • Study the special comparison test in detail for series convergence
  • Learn how to evaluate limits involving alternating series
  • Explore the concept of the squeeze theorem in series analysis
  • Investigate other convergence tests such as the ratio test and root test
USEFUL FOR

Students studying calculus, particularly those focused on series convergence, as well as educators teaching convergence tests in mathematical analysis.

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Homework Statement


Use special comparison test to find if \frac{2+(-1)^n}{n^2+7} is convergent or divergent.

Homework Equations


Special comparison test using the convergent series \frac{1}{n^2}

and taking the limit as n-> infinity of my initial series \frac{2+(-1)^n}{n^2+7} divided by my comparison series \frac{1}{n^2}

which comes out to be lim n--> infinity of \frac{2n^2+(-1)^n(n^2)}{n^2+7}.

The Attempt at a Solution



I guess I need help evaluating that limit because what I'm getting is undefined (alternating) and the back of the book says that it's defined.
 
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0 \leq \frac{2+(-1)^n}{n^2+7} \leq \frac{3}{n^2+7}. You can squeeze it as such. If the series on the right converges, then so does yours.
 

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