Find the interval of convergence of the power series. ∑n=1∞4n(x−1)nn

Given Data The power series is an=∑n=1∞4n(x−1)nn Using the ratio test to evaluate the interval of converges of the power series, limn→∞an+1an=limn→∞(4n+1(x−1)n+1n+1)(4n(x−1)nn) =limn→∞(4n+14n(x−1)n+1(x−1)nn+1n) =4(x−1)limn→∞(nn+1) =4(x−1)limn→∞(11+1n) =4(x−1)(11+1∞) =4(x−1)(11+0) =4|x−1| The value of 4|x−1|&lt;1 in the interval 0≤x≤1 Hence the interval of converges of the power series is 0≤x≤1.

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kuCAu

Answered question

2021-01-02

Find the interval of convergence of the power series.

\sum_{n = 1}^{\infty} \frac{4^{n} (x - 1)^{n}}{n}

Answer & Explanation

Talisha

Skilled2021-01-03Added 93 answers

Given Data
The power series is $a_{n} = \sum_{n = 1}^{\infty} \frac{4^{n} (x - 1)^{n}}{n}$
Using the ratio test to evaluate the interval of converges of the power series,
$lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} = lim_{n \to \infty} \frac{(\frac{4^{n + 1} (x - 1)^{n + 1}}{n + 1})}{(\frac{4^{n} (x - 1)^{n}}{n})}$
$= lim_{n \to \infty} (\frac{\frac{4^{n + 1}}{4^{n}} \frac{(x - 1)^{n + 1}}{(x - 1)^{n}}}{\frac{n + 1}{n}})$
$= 4 (x - 1) lim_{n \to \infty} (\frac{n}{n + 1})$
$= 4 (x - 1) lim_{n \to \infty} (\frac{1}{1 + \frac{1}{n}})$
$= 4 (x - 1) (\frac{1}{1 + \frac{1}{\infty}})$
$= 4 (x - 1) (\frac{1}{1 + 0})$
$= 4 | x - 1 |$
The value of $4 | x - 1 | < 1$ in the interval $0 \leq x \leq 1$
Hence the interval of converges of the power series is $0 \leq x \leq 1$ .

Jeffrey Jordon

Expert2021-12-27Added 2605 answers

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