Question

Match the following series with the series below in which you can compare using the Limit Comparison Test. Then determine whether the series converge or diverge. A. ? n=1 to ? 1/n, B. ? n=1 to ? 1/n^2, C. ? n=1 to ? 1/n^3, and D. ? n=1 to ? (1/2)^(n-1) 1. ? n=1 to ? (n + 7)/n^2 Does this series converge or diverge? 2. ? n=1 to ? (2/(2^n + 5)) Does this series converge or diverge? 3. ? n=1 to ? 1/(n^2 + 4) Does this series converge or diverge? 4. ? n=1 to ? (n - 5)/(n^4 + 1) Does this series converge or diverge?

          Match the following series with the series below in which you can compare using the Limit Comparison Test. Then determine whether the series converge or diverge.
A. ? n=1 to ? 1/n, B. ? n=1 to ? 1/n^2, C. ? n=1 to ? 1/n^3, and D. ? n=1 to ? (1/2)^(n-1)
1. ? n=1 to ? (n + 7)/n^2 Does this series converge or diverge?
2. ? n=1 to ? (2/(2^n + 5)) Does this series converge or diverge?
3. ? n=1 to ? 1/(n^2 + 4) Does this series converge or diverge?
4. ? n=1 to ? (n - 5)/(n^4 + 1) Does this series converge or diverge?

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Match the following series with the series below in which you can compare using the Limit Comparison Test. Then determine whether the series converge or diverge.
A. ? n=1 to ? 1/n, B. ? n=1 to ? 1/n^2, C. ? n=1 to ? 1/n^3, and D. ? n=1 to ? (1/2)^(n-1)
1. ? n=1 to ? (n + 7)/n^2 Does this series converge or diverge?
2. ? n=1 to ? (2/(2^n + 5)) Does this series converge or diverge?
3. ? n=1 to ? 1/(n^2 + 4) Does this series converge or diverge?
4. ? n=1 to ? (n - 5)/(n^4 + 1) Does this series converge or diverge?

Added by Blanca H.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Supreeta N

08/03/2023

Step 1

For the series $\sum_{n=1}^{\infty} \frac{1}{n^2}$, we can compare it with the series $\sum_{n=1}^{\infty} \frac{1}{n}$ using the Limit Comparison Test. Let's find the limit: $$\lim_{n \to \infty} \frac{\frac{1}{n}}{\frac{1}{n^2}} = \lim_{n \to \infty} n = Show more…

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Match the following series with the series below in which you can compare using the Limit Comparison Test. Then determine whether the series converge or diverge. A. Σ n=1 to ∞ 1/n, B. Σ n=1 to ∞ 1/n^2, C. Σ n=1 to ∞ 1/n^3, and D. Σ n=1 to ∞ (1/2)^(n-1) 1. Σ n=1 to ∞ (n + 7)/n^2 Does this series converge or diverge? 2. Σ n=1 to ∞ (2/(2^n + 5)) Does this series converge or diverge? 3. Σ n=1 to ∞ 1/(n^2 + 4) Does this series converge or diverge? 4. Σ n=1 to ∞ (n - 5)/(n^4 + 1) Does this series converge or diverge?

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Transcript

00:02 Hi here for the given question.

00:03 We are given that we have series we need to check whether they are convergent or divergent.

00:08 We are using p -test series and we are like using comparison test and then using comparison test.

00:15 We are checking the p -series test.

00:17 So here for the first one, we have series summation and running from 1 to infinity and plus 7 divided by n square.

00:23 So this can be further written as summation and running from 1 to infinity 1 plus 7 upon n divided by n.

00:30 So here we have a n equals to 1 plus 7 upon n divided by n, which is less than b n which is equal to 1 upon n.

00:38 So here we know that b n is equal to summation and running from 1 to infinity 1 upon n.

00:44 So here as we can observe that here in our case, this is 1 upon n here.

00:49 We have p equals to 1 which is less than 1 less than or equal to 1.

00:54 So the series will be divergent using p -test.

00:57 So using p -test we have b n as divergent.

01:02 So here by comparison test, we can say that series a n is also divergent.

01:13 So here for the first part the given series is divergent.

01:16 Now here we are given second part.

01:18 We have summation and running from 1 to infinity 2 divided by 2 to the power n plus 5.

01:23 So here now further we know that a n is equal to 2 upon 2 to the power n plus 5.

01:29 This is less than b n which is equal to 1 upon 2 to the power n.

01:33 Now here we know that b n can be further written as 1 by 2 to the power n.

01:38 Now here if we use the value using r -test, then here it is in a geometric series and r is equal to 1 by 2 which is less than 1.

01:48 So here we have b n as a convergent series because if the value of ratio is less than 1, then the series will be convergent.

01:55 So here b n is convergent.

01:58 Therefore by comparison test, we can say that here in our case a n is also convergent...

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