Find x for Summation Series Convergence

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SUMMARY

The discussion centers on determining the convergence of the series ∞ Σ (x^n/ln(n+5)) from n=1. The ratio test indicates that the series converges for |x| < 1. However, it is crucial to note that the endpoints x = 1 and x = -1 must be excluded from the solution, as convergence does not occur at these values. The final conclusion is that the series converges strictly for -1 < x < 1.

PREREQUISITES
  • Understanding of series convergence tests, specifically the ratio test.
  • Familiarity with logarithmic functions and their properties.
  • Knowledge of limits and behavior of sequences as n approaches infinity.
  • Basic algebraic manipulation skills for handling inequalities.
NEXT STEPS
  • Review the Ratio Test for series convergence in more detail.
  • Study the behavior of logarithmic functions in convergence tests.
  • Learn about endpoint convergence for series and how to test it.
  • Explore additional convergence tests such as the Root Test and Comparison Test.
USEFUL FOR

This discussion is beneficial for students studying calculus, particularly those focusing on series and convergence, as well as educators teaching these concepts in mathematics courses.

cloveryeah
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Homework Statement



Σ ( x^n/ln(n+5) )
n=1

find the value of x that the above series converges

Homework Equations

The Attempt at a Solution


i cal. it by ratio test
and i found that |x|<1

but when i input (-1,1) into my webwork...it said it's wrong[/B]
 
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cloveryeah said:

Homework Statement



Σ ( x^n/ln(n+5) )
n=1

find the value of x that the above series converges

Homework Equations

The Attempt at a Solution


i cal. it by ratio test
and i found that |x|<1

but when i input (-1,1) into my webwork...it said it's wrong[/B]

Your answer may be overly conservative. You need to look at whether the series converges or diverges at x = +1 and at x = -1.
 
Hi Cloveryeah,
Great work, just something you've missed
In order for the series to converge,
Whenver n→∞ (x^n/ln(n+5)) = 0
You said that |x| < 1, you can't try x= 1 or x = -1, it's strictly less not even equal
 

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