"Determine the radius of convergence and the interval..." was answered by our experts

Maiclubk

Answered question

2021-02-10

Determine the radius of convergence and the interval of convergence for each power series.

\sum_{n = 1}^{\infty} \frac{(- 2)^{n} x^{n}}{\sqrt[4]{n}}

AGRFTr

Skilled2021-02-11Added 95 answers

Given series is:

\sum_{n = 1}^{\infty} a_{n} x^{n} = \sum_{n = 1}^{\infty} \frac{(- 2)^{n} x^{n}}{\sqrt[4]{n}}

By Applying ratio test , we have

lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} | = lim_{n \to \infty} | \frac{(- 2)^{n + 1} x^{n + 1} \sqrt[4]{n}}{\sqrt[4]{n + 1} (- 2)^{n} x^{n}} |

= lim_{n \to \infty} | \frac{(- 2) x \sqrt[4]{n}}{\sqrt[4]{n} (1 + \frac{1}{n})^{\frac{1}{4}}} |

= lim_{n \to \infty} | \frac{(- 2) x}{(1 + \frac{1}{n})^{\frac{1}{4}}} |

= | - 2 x |

= 2 | x |

for radius of convergence

2 | x | = 1

\Rightarrow | x | = \frac{1}{2}

r = \frac{1}{2}

is the radius of convergence.
Now, when

x = \frac{1}{2}

, given series becomes

\sum_{n = 1}^{\infty} \frac{(- 2)^{n} (\frac{1}{2})^{n}}{\sqrt[4]{n}}

\sum_{n = 1}^{\infty} (- 1)^{n} a_{n} = \sum_{n = 1}^{\infty} \frac{(- 1)^{n}}{\sqrt[4]{n}}

lim_{n \to \infty} a_{n} = 0

and seq.

a_{n}

is decreasing, so by Leibnitz alternating test, this series is convergent at

x = \frac{1}{2}

Now, when

x = - \frac{1}{2}

, given series becomes

\sum_{n = 1}^{\infty} \frac{(- 2)^{n} (\frac{1}{2})^{n}}{\sqrt[4]{n}}

= \sum_{n = 1}^{\infty} \frac{1}{\sqrt[4]{n}}

By, p-test this series is convergent.
so, interval of convergence is

[- \frac{1}{2}, \frac{1}{2}]

Jeffrey Jordon

Expert2021-12-27Added 2605 answers

Answer is given below (on video)

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