3. Use the ratio test to find the radius of convergence and...

3. Use the ratio test to find the radius of convergence and interval of convergence for the series. (Make sure to pay careful attention to where the ratio test is inconclusive.) sum_{n=1}^{infty} frac{(x - 3)^n}{n} 4. Find the radius of convergence and the interval of convergence for the series. sum_{n=0}^{infty} frac{(- 3x)^n}{sqrt{n + 1}} 5. Find the radius of convergence and the interval of convergence for the series. sum_{n=0}^{infty} frac{n(x + 2)^n}{3^{n+1}}

Transcript

00:01 Hello everyone here in this question.

00:04 As for the first condition, we have the information as sigma n equals to 1 to infinity multiplied with x minus 3 whole power n divided by n where where n is a n is equals to x minus 3 whole power n divided by n and and a n plus 1 is equals to x minus 3, x minus 3, whole power n plus 1 divided by n plus 1.

00:45 So now the radius of the convergence using the ratio test is, by using the ratio test, the radius of the convergence is 1 divided by r equals to limit n goes to infinity, multiplied with an n plus 1 divided by a n so by inserting these values into the formula we will get the equation as limit n to infinity multiplied with x minus 3 whole power n plus 1 divided by n plus 1 multiplied with n divided by x minus 3 whole power n so here once divided by r will be equals to module x minus 3.

01:41 So hence series of convergence is if module x minus 3 is less than 1 then r will be equals to l.

01:56 So now the interval of convergence is interval of convergence is minus 1 less than x minus 3 less than l so the equation will be so the equation will be minus 1 plus 3 less than x less than 1 plus 3 which will be equals to 2 less than x less than x less than 4 so if x is equal to 2 then the series will be sigma n equals to 1 to infinity multiplied with minus 1 whole power n divided by n is convergent m is convergent by the alternative test so if x is f is if x is equals to 4 then the series of sigma n equals to 1 to infinity multiplied with 1 divided by n is divergent.

03:30 So hence the interval of convergence is equals to 2 .4.

03:46 That is 2 is less than or equals to x less than or equals to 4.

03:56 So this is the answer for the first condition.

04:00 And coming to the second condition, here the given information is sigma n equals to 0 to infinity multiplied with minus 3x whole power n divided by square root n plus 1...

Question

Please give Ace some feedback