# 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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