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

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

2021-02-26
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.