Help finding the behaviour of this series?

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SUMMARY

The series in question is 0 + ln(2)/4 + ln(3)/9 + ln(4)/16 + ... + ln(n)/n^2. To determine its convergence or divergence, the comparison test is employed, which requires identifying a known series B(n) with established behavior. The discussion emphasizes that logarithmic functions, such as ln(x), grow slower than polynomial functions for large x, suggesting that a polynomial series can be used for comparison. Specifically, it is noted that for large x, all x^a (where a > 0) is less than ln(x), guiding the selection of an appropriate convergent polynomial series.

PREREQUISITES
  • Understanding of series convergence tests, particularly the comparison test.
  • Familiarity with logarithmic functions and their growth rates compared to polynomial functions.
  • Basic knowledge of graphing functions to visualize behavior of ln(x) against polynomials.
  • Experience with mathematical notation and series representation.
NEXT STEPS
  • Study the comparison test for series convergence in detail.
  • Learn about the behavior of logarithmic functions relative to polynomial functions.
  • Graph ln(x) alongside various polynomial functions to observe their growth rates.
  • Investigate specific convergent series that can be used for comparison, such as p-series.
USEFUL FOR

Mathematicians, students studying calculus or real analysis, and anyone involved in series convergence analysis will benefit from this discussion.

Dell
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given the following series and i am asked if it converges or diverges in its boundaries

0 + ln(2)/4 + ln(3)/9 + ln(4)/16 + ... + ln(n)/n^2

to find out if series converge/diverge, i have been using comparisons, finding a similar series, B(n) whose behaviour i know or can easily find, then saying :
# if B(n) > A(n), and B(n) converges, then A(n) converges
# if B(n) < A(n), and B(n) diverges, then A(n) diverges
# if A(n)/B(n)=K (K= not 0, not ∞) then A(n) and B(n) behave the same

but i cannot find any series that answers to any of these 3 condidtions, any ideas? the series must either have a known or relatively simple to find behaviour.
 
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Finding the convergence of series involving logs can be tough because it can be hard to see how log behaves in comparison to polynomials. To get a good sense of this, try graphing in the same window y=ln(x) along with some polynomials such as y=x, y=x^(1/2), y=x^(1/5), etc. What you should generally find is that all x^a (with a > 0) is less than ln(x) for large x. This should help you find a convergent polynomial series greater than yours.
 

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