Determining Convergence of Series with Logarithmic Terms

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SUMMARY

The series \(\sum \frac{1}{n^2 \ln n}\) converges based on the limit test, as \(\lim_{n \to \infty} \frac{1}{n^2 \ln n} = 0\). The nth term test for divergence indicates that if \(\lim a_n \neq 0\), the series diverges; however, this does not apply here. For the series \(\sum \frac{1}{n (\ln n)^p}\), convergence occurs when \(p > 1\). The comparison test is also suggested as a useful method for determining convergence.

PREREQUISITES
  • Understanding of series convergence tests, including the limit test and nth term test.
  • Familiarity with logarithmic functions and their properties.
  • Knowledge of the comparison test for series.
  • Basic calculus concepts, particularly limits and asymptotic behavior.
NEXT STEPS
  • Study the comparison test for series convergence in detail.
  • Learn about the integral test for convergence of series.
  • Explore the implications of the nth term test for series divergence.
  • Investigate the behavior of logarithmic functions in series, particularly in relation to convergence criteria.
USEFUL FOR

Mathematicians, students studying calculus or real analysis, and anyone interested in understanding series convergence, particularly those involving logarithmic terms.

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Homework Statement



Is the series convergent

1) [tex]\sum[/tex]1/(n^2 * ln n)

and 2) which value of p does the series converge. [tex]\sum[/tex] 1/(n*(ln n)^p)


The Attempt at a Solution



1)

I cannot see how the root method ([tex]\sqrt[n]{Cn}[/tex]) would work, or the ratio test would work (cn+1)/cn

Unless you use the limit test and lim 1/(n^2 * ln n) = 1/infinity = 0 therefore the series converges. I didnt think that question would be that easy though.

2) similar to the first wouldn't it just be if P > 1.
 
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ziggie125 said:

Homework Statement



Is the series convergent

1) [tex]\sum[/tex]1/(n^2 * ln n)

and 2) which value of p does the series converge. [tex]\sum[/tex] 1/(n*(ln n)^p)


The Attempt at a Solution



1)

I cannot see how the root method ([tex]\sqrt[n]{Cn}[/tex]) would work, or the ratio test would work (cn+1)/cn

Unless you use the limit test and lim 1/(n^2 * ln n) = 1/infinity = 0
The test you are thinking of, I believe, is the nth term test for divergence. If lim an != 0, the series diverges. You cannot use this test to conclude a series converges.

A test that might be useful in this problem is the comparison test.
ziggie125 said:
therefore the series converges. I didnt think that question would be that easy though.

2) similar to the first wouldn't it just be if P > 1.
 

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