Key to "Use the Root Test to determine the convergence..."

babeeb0oL

Answered question

2021-02-21

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

\sum_{n = 1}^{\infty} (\frac{3 n + 2}{n + 3})^{n}

Answer & Explanation

doplovif

Skilled2021-02-22Added 71 answers

The given series is $\sum_{n = 1}^{\infty} (\frac{3 n + 2}{n + 3})^{n}$
Ratio test:
The series $\sum_{n = 1}^{\infty} a_{n}$ converges if $lim_{n \to \infty} \sqrt[n]{| a_{n} |} < 1$
The series $\sum_{n = 1}^{\infty} a_{n}$ diverges if $lim_{n \to \infty} \sqrt[n]{| a_{n} |} > 1$
In the given series $\sum_{n = 1}^{\infty} (\frac{3 n + 2}{n + 3})^{n}, a_{n} = (\frac{3 n + 2}{n + 3})^{n}$
Find the limit $lim_{n \to \infty} \sqrt[n]{| a_{n} |}$
$lim_{n \to \infty} \sqrt[n]{| a_{n} |} = lim_{n \to \infty} | (\frac{3 n + 2}{n + 3})^{n} |^{\frac{1}{n}}$
$= lim_{n \to \infty} (\frac{3 n + 2}{n + 3})$
$= lim_{n \to \infty} (\frac{3 + \frac{2}{n}}{1 + \frac{3}{n}})$
$= \frac{3 + \frac{2}{\infty}}{1 + \frac{3}{\infty}}$
$= \frac{3 + 0}{1 + 0}$
$= 3$
$> 1$
By the Ratio test, the series $\sum_{n = 1}^{\infty} (\frac{3 n + 2}{n + 3})^{n}$ diverges.

Jeffrey Jordon

Expert2021-12-27Added 2605 answers

Answer is given below (on video)

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