Answered: "Find the interval of convergence of the following..."

Albarellak

Answered question

2021-02-22

Find the interval of convergence of the following series.

\frac{{(x - 4)}^{n}}{3^{n}}

Answer & Explanation

avortarF

Skilled2021-02-23Added 113 answers

Given the series:
$\sum_{n = 1}^{\infty} \frac{{(x - 4)}^{n}}{3^{n}}$
The nth term is given as:
$a_{n} = \frac{{(x - 4)}^{n}}{3^{n}}$
Applying the ratio test, we get:
$L = lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} |$
$\Rightarrow L = lim_{n \to \infty} \leq | \frac{\frac{{(x - 4)}^{n + 1}}{3^{n + 1}}}{\frac{{(x - 4)}^{n}}{3^{n}}} |$
$\Rightarrow L = lim_{n \to \infty} | \frac{{(x - 4)}^{n + 1}}{3^{n + 1}} \times \frac{3^{n}}{{(x - 4)}^{n}} |$
$\Rightarrow L = lim_{n \to \infty} | \frac{{(x - 4)}^{n + 1 - n}}{3^{n + 1 - n}} |$
$\Rightarrow L = lim_{n \to \infty} | \frac{(x - 4)}{3} |$
Now plugging the limit, we get:
$\Rightarrow L = | \frac{(x - 4)}{3} |$
For series to converge:
$| \frac{(x - 4)}{3} | < 1$
$| x - 4 | < 3$
Hence the radius of convergence is:
$R = 3$
The interval of convergence is:
$- 3 < x - 4 < 3$
$- 3 + 4 < x < 3 + 4$
$1 < x < 7$
Now checking for $x = 1$ , we get series:
$\sum_{n = 1}^{\infty} \frac{{(x - 4)}^{n}}{3^{n}} = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{3^{n}}$
The above series converges since $\sum | a_{n} | = \sum \frac{1}{3^{n}}$ converges by geometric series test.
Now checking for $x = 7$ , we get series:
$\sum_{n = 1}^{\infty} \frac{{(x - 4)}^{n}}{3^{n}} = \sum_{n = 1}^{\infty} {(3)}^{n}$

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