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ruigE

Answered question

2021-02-25

Use the Root Test to determine the convergence or divergence of the series.

\sum_{n = 2}^{\infty} \frac{(- 1)^{n}}{(\ln n)^{n}}

Answer & Explanation

Cristiano Sears

Skilled2021-02-26Added 96 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,
$lim_{n \to \infty} \sqrt[n]{a_{n}} = L, a_{n} \geq 0 \forall n$
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, $a_{n} = (- \frac{1}{\ln n})^{n}, \forall n \geq 2$
$lim_{n \to \infty} \sqrt[n]{| a_{n} |} = lim_{n \to \infty} \sqrt[n]{| (- \frac{1}{\ln n})^{n} |}$
$= lim_{n \to \infty} ((\frac{1}{\ln n})^{n})^{\frac{1}{n}}$
$= lim_{n \to \infty} \frac{1}{\ln n}$
$= 0$
$lim_{n \to \infty} \sqrt[n]{| a_{n} |} < 1$
$\Rightarrow L < 1$
Hence, the series converges.

Jeffrey Jordon

Expert2021-12-27Added 2605 answers

Answer is given below (on video)

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