QuestionFind the interval of convergence for the series $\sum_{n=1}^{\infty} \frac{(x-1)^{n}}{3^{n}}$ .

Studdy Solution

STEP 1

Assumptions1. The given series is $\sum_{1}^{\infty} \frac{(x-1)^{n}}{3^{n}}$ . We are asked to find the interval of convergence of the series

STEP 2

The interval of convergence of a series is found by applying the ratio test. The ratio test states that if the limit as $n$ approaches infinity of the absolute value of the ratio of the $(n+1)$ th term to the $n$ th term of a series is less than1, the series converges absolutely. If the limit equals1, the test is inconclusive. If the limit is greater than1, the series diverges.

STEP 3

Apply the ratio test to the given series. The ratio of the $(n+1)$ th term to the $n$ th term is $\frac{\frac{(x-1)^{n+1}}{3^{n+1}}}{\frac{(x-1)^{n}}{3^{n}}}$

STEP 4

implify the ratio.
$\frac{(x-1)^{n+1}}{3^{n+1}} \cdot \frac{3^{n}}{(x-1)^{n}} = \frac{(x-1)}{3}$

STEP 5

The ratio test requires us to take the limit as $n$ approaches infinity. However, our simplified ratio does not depend on $n$ , so we can skip this step.

STEP 6

The series converges absolutely if the absolute value of the ratio is less than1. Set up the inequality and solve for $x$ .
$\left|\frac{(x-1)}{3}\right| <1$

STEP 7

olve the inequality.
$-1 < \frac{(x-1)}{3} <1$

STEP 8

Multiply each part of the inequality by3.
$-3 < x-1 <3$

STEP 9

Add to each part of the inequality.
$-2 < x <4$

STEP 10

We have found the interval of convergence to be $(-2,4)$ . However, the ratio test only tells us that the series converges absolutely for $x$ in this interval. We must still check the endpoints $x = -2$ and $x =4$ separately.

STEP 11

Substitute $x = -$ into the series.
$\sum_{}^{\infty} \frac{(--)^{n}}{3^{n}} = \sum_{}^{\infty} \frac{(-3)^{n}}{3^{n}}$

STEP 12

implify the series.
$\sum_{}^{\infty} (-)^{n}$

STEP 13

This is a geometric series with common ratio $-$ , which does not converge. Therefore, the series diverges at $x = -2$ .

STEP 14

Substitute $x =4$ into the series.
$\sum_{}^{\infty} \frac{(4-)^{n}}{3^{n}} = \sum_{}^{\infty} \frac{3^{n}}{3^{n}}$

STEP 15

implify the series.
$\sum_{}^{\infty}$

STEP 16

This is a series with constant term, which does not converge. Therefore, the series diverges at $x =4$ .

STEP 17

The series $\sum_{}^{\infty} \frac{(x-)^{n}}{3^{n}}$ converges absolutely for $x$ in the interval $(-2,4)$ and diverges at the endpoints $x = -2$ and $x =4$ . Therefore, the interval of convergence is $(-2,4)$ .
The interval of convergence of the series is $(-2,4)$ .

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