Asked by Ashley
Hello!
I'm taking a Calculus course and I came up with this problem, which I'm not sure whether below method is the best one to solve this.
Your thoughts on this is highly appreciated.
Many thanks!
Question:
Using a suitable test determine whether the infinite series,
Sigma n=1 - infinity [ (e^(n+1))/((ln(n))^n) ] converges or diverges
The method I followed:
I considered the the nth term of the series a_n = [ (e^(n+1))/((ln(n))^n) ]
Then I considered the Root test to determine the convergence of divergence of this series, which derived;
lim n-->infinity |a_n|^(1/n) = lim n-->infinity [(e^(1/n))/(ln(n))]
Hence,
lim n-->infinity
|a_n|^(1/n) = 1/infinity -->0 < 1
Hence, by Root test, the above series converges.
Is this a correct method or is(are) their any other more suitable method(s) for determining the convergence/divergence of this series?
I'm taking a Calculus course and I came up with this problem, which I'm not sure whether below method is the best one to solve this.
Your thoughts on this is highly appreciated.
Many thanks!
Question:
Using a suitable test determine whether the infinite series,
Sigma n=1 - infinity [ (e^(n+1))/((ln(n))^n) ] converges or diverges
The method I followed:
I considered the the nth term of the series a_n = [ (e^(n+1))/((ln(n))^n) ]
Then I considered the Root test to determine the convergence of divergence of this series, which derived;
lim n-->infinity |a_n|^(1/n) = lim n-->infinity [(e^(1/n))/(ln(n))]
Hence,
lim n-->infinity
|a_n|^(1/n) = 1/infinity -->0 < 1
Hence, by Root test, the above series converges.
Is this a correct method or is(are) their any other more suitable method(s) for determining the convergence/divergence of this series?
Answers
Answered by
oobleck
take a look here at stackexchange . com
questions/670345/comparison-test-of-sum1-lnnn
questions/670345/comparison-test-of-sum1-lnnn
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