Comparing Series Convergence: Limit or Direct Comparison Test?

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SUMMARY

The series \(\sum_{n=1}^{+\infty} \frac{3^{n-1}+1}{3^{n}}\) converges, as established through the Limit Comparison Test with the series \(\sum_{n=1}^{+\infty} \frac{1}{3^{n}}\). The comparison shows that \(\frac{3^{n-1}+1}{3^{n}} > \frac{1}{3}\), which is a convergent geometric series. Therefore, the original series converges based on the properties of convergent series.

PREREQUISITES
  • Understanding of series convergence tests, specifically the Limit Comparison Test and Direct Comparison Test.
  • Familiarity with geometric series and their convergence criteria.
  • Ability to manipulate series, including reindexing and simplifying terms.
  • Knowledge of asymptotic behavior of functions as \(n\) approaches infinity.
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  • Study the Limit Comparison Test in detail, including examples and edge cases.
  • Explore the Direct Comparison Test and its applications in series convergence.
  • Learn about geometric series and their convergence properties, focusing on the ratio test.
  • Practice reindexing series and applying convergence tests to various series types.
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Homework Statement


Does the following series converge or diverge (use either the Limit Comparison or the Direct Comparison Test):

[tex]\sum_{n=1}^{+\infty} \frac{3^{n-1}+1}{3^{n}}[/tex]


Homework Equations



In a previous problem that was
[tex]\sum_{n=1}^{+\infty} \frac{1}{3^{n-1}+1}[/tex]
I was able to reindex the series to make it
[tex]\sum_{n=0}^{+\infty} \frac{1}{3^{n}+1}[/tex]
From there, I took 1/(3n+1)<1/3n.

Therefore, since 1/3n converges, [tex]\sum_{n=1}^{+\infty} a_{n}[/tex] also converges.


The Attempt at a Solution



However, I don't know how to solve the series that I'm currently on.

Thanks,
Erik
 
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(3^(n-1)+1)/3^n>1/3, isn't it?
 

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