Radius & interval of convergence of power series

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SUMMARY

The discussion focuses on determining the radius and interval of convergence for the power series ∞Ʃ (x^n)/ln(n+1) using the ratio test. The ratio test involves evaluating the limit lim | a(n+1)/a(n)| as n approaches infinity, leading to the expression lim | [x ln(n+1)] /ln(n+2) |. The conclusion drawn is that as n approaches infinity, ln(n)/ln(n+1) approaches 1, which can be proven using L'Hôpital's rule, confirming that the series converges for specific values of x.

PREREQUISITES
  • Understanding of power series and their convergence
  • Familiarity with the ratio test for series convergence
  • Knowledge of logarithmic functions and their properties
  • Ability to apply L'Hôpital's rule in calculus
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  • Study the application of the ratio test in depth
  • Learn about the convergence of power series in detail
  • Explore the properties of logarithmic functions
  • Review L'Hôpital's rule and its applications in limits
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Students studying calculus, particularly those focusing on series convergence, as well as educators teaching power series and convergence tests.

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Homework Statement


i was doing this exercise and came across this example.

Ʃ (x^n)/ln(n+1)
n=1

The Attempt at a Solution


i know you have to do the ratio test which is
lim | a(n+1)/a(n)|
n>∞

i got to

lim | [x ln(n+1)] /ln(n+2) |
n>∞

and have no idea how to continue? is there a way to get rid of the logs? since in this case as n>∞ the top and bottom just goes to ∞?
 
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ln(n)/ln(n+1) goes to 1 for n -> infty
You can prove that easily by proving that ln(n)/ln(n+1)-1 approaches zero.
 
Or use L'Hopital's rule: the derivative of ln(x+1) is 1/(x+1) and the derivative of ln(x) is 1/x so this becomes (1/x)(x+1/1)= (x+1)/x goes to 1 as x goes to infinity.

That shows that ln(x)/ln(x+1) goes to 1 as x goes to infinity which includes n going to infinity.
 

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