Convergence of an infinite series

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SUMMARY

The discussion focuses on determining the convergence of the series involving the natural logarithm of n, specifically the series represented as \(\sum \frac{1}{\log(n)}\). Participants concluded that the comparison test is the most suitable method for this analysis, particularly by comparing it to the convergent p-series \(\sum \frac{1}{n^{3/2}}\), where p > 1. The key takeaway is that since \(\log(n) < n^{1/2}\) asymptotically, the original series converges as well.

PREREQUISITES
  • Understanding of series convergence tests, specifically the comparison test.
  • Familiarity with p-series and their convergence criteria.
  • Knowledge of asymptotic notation and properties of logarithmic functions.
  • Basic calculus concepts, including limits and series summation.
NEXT STEPS
  • Study the comparison test for series convergence in more detail.
  • Learn about p-series and their convergence conditions.
  • Explore asymptotic analysis and its applications in series.
  • Review examples of series involving logarithmic functions and their convergence properties.
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Students studying calculus, particularly those focusing on series convergence, as well as educators looking for examples of convergence tests in action.

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Homework Statement


http://img840.imageshack.us/img840/3609/unleddn.png

note that by log(n), i really mean NATURAL log of n

Homework Equations


it's convergent, but I can't figure out which test to use

The Attempt at a Solution


there is no term to the nth power, so ratio test is useless; root test is useless too; comparison test would seem to be the best option, but I can't figure out how to compare when I have a natural log in the expression... limit comparison test takes me nowhere either.
 
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I think that the log(n) is just there to throw you off. [tex]\log(n) < n^{1/2}[/tex] asymptotically right. Do you have a theorem showing that [tex]\sum \frac{1}{n^{3/2}}[/tex] is convergent?
 
yes, p-series are convergen if p>1;

now i get it, I just need to compare it with the series you just shown, since with the (n+1) the original series will always be smaller than the convergent series you shown, therefore convergent too.
 

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