Use the Root Test to determine the convergence or divergence of the series. sum_{n=2}^inftyfrac{(-1)^n}{(ln n)^n}

ruigE 2021-02-25 Answered
Use the Root Test to determine the convergence or divergence of the series.
\(\sum_{n=2}^\infty\frac{(-1)^n}{(\ln n)^n}\)

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Expert Answer

Cristiano Sears
Answered 2021-02-26 Author has 26499 answers

Given:
The series, \(\sum_{n=2}^\infty\frac{(-1)^n}{(\ln n)^n}\)
To determine the convergence or divergence of the series using the Root Test.
Let, \(\sum_{n=2}^\infty\frac{(-1)^n}{(\ln n)^n}\)
The Root Test:
Let \(\sum_{n=1}^\infty a_n\) be a sequence of real numbers such that,
\(\lim_{n\to\infty}\sqrt[n]{a_n}=L,a_n\geq0\forall n\)
Then, (i) \(\sum_{n=1}^\infty a_n\) converges if \(L<1\)
(ii) \(\sum_{n=1}^\infty a_n\) diverges if \(L>1\)
(iii) For \(L=1\), the test fails.
Here, \(a_n=(-\frac{1}{\ln n})^n,\forall n\geq2\)
\(\lim_{n\to\infty}\sqrt[n]{|a_n|}=\lim_{n\to\infty}\sqrt[n]{|(-\frac{1}{\ln n})^n|}\)
\(=\lim_{n\to\infty}((\frac{1}{\ln n})^n)^{\frac{1}{n}}\)
\(=\lim_{n\to\infty}\frac{1}{\ln n}\)
\(=0\)
\(\lim_{n\to\infty}\sqrt[n]{|a_n|}<1\)
\(\Rightarrow L<1\)
Hence, the series converges.

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Answered 2021-12-27 Author has 9769 answers

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