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

(ii) \(\sum_{n=1}^\infty a_n\) diverges if L>1

(iii) For L=1, the test fails.

Here, \(a_n=(-\frac{1}{\ln n})^n,\forall n\geq2\)

\(\lim_{n\to\infty}\sqrt[n]{|a_n|}=\lim_{n\to\infty}\sqrt[n]{|(-\frac{1}{\ln n})^n|}\)

\(=\lim_{n\to\infty}((\frac{1}{\ln n})^n)^{\frac{1}{n}}\)

\(=\lim_{n\to\infty}\frac{1}{\ln n}\)

\(=0\)

\(\lim_{n\to\infty}\sqrt[n]{|a_n|}<1\)</span>

\(\Rightarrow L<1\)</span>

Hence, the series converges.

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

(ii) \(\sum_{n=1}^\infty a_n\) diverges if L>1

(iii) For L=1, the test fails.

Here, \(a_n=(-\frac{1}{\ln n})^n,\forall n\geq2\)

\(\lim_{n\to\infty}\sqrt[n]{|a_n|}=\lim_{n\to\infty}\sqrt[n]{|(-\frac{1}{\ln n})^n|}\)

\(=\lim_{n\to\infty}((\frac{1}{\ln n})^n)^{\frac{1}{n}}\)

\(=\lim_{n\to\infty}\frac{1}{\ln n}\)

\(=0\)

\(\lim_{n\to\infty}\sqrt[n]{|a_n|}<1\)</span>

\(\Rightarrow L<1\)</span>

Hence, the series converges.