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How do you test for convergence of #Sigma (ln(n))^-n# from #n=[2,oo)#?

Answer 1

Ezekiel Alexander

The series:

#sum_(n=2)^oo (ln(n))^(-n)#

is convergent.

The series has positive terms so we can use the root test and we have:

#L= lim_(n->oo) root(n)(a_n) = lim_(n->oo) root(n)((ln(n)^(-n))) = lim_(n->oo) 1/lnn = 0#

As #L < 1# the series is convergent.

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Answer 2

Ezra Adams

To test for the convergence of the series Σ(ln(n))^(-n) from n=2 to infinity, you can use the integral test. This test states that if f(n) is a positive, continuous, and decreasing function for n ≥ 1, then the series Σf(n) converges if and only if the improper integral ∫f(x) dx from 1 to infinity converges.

In this case, f(n) = (ln(n))^(-n). To apply the integral test, you need to check the convergence of the integral ∫(ln(x))^(-x) dx from 2 to infinity.

Unfortunately, this integral doesn't have a simple closed-form antiderivative, making direct integration difficult. However, you can still analyze the convergence using other methods, such as comparison test, limit comparison test, or the ratio test, which may be more appropriate for this series.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.