Select the FIRST correct reason why the given series...

Select the FIRST correct reason why the given series diverges. A. Diverges because the terms don't have limit zero B. Divergent geometric series C. Divergent p-series D. Integral test E. Comparison with a divergent p-series F. Diverges by limit comparison test G. Diverges by alternating series test 1. ?_{n=1}^{?} 1/?n 2. ?_{n=1}^{?} cos(n??)/ln(3) 3. ?_{n=1}^{?} ln(n)/n 4. ?_{n=1}^{?} (7n + 7)/(-1)^n 5. ?_{n=1}^{?} (-1)^n (2n)!/(n!)^2 6. ?_{n=1}^{?} (n + 1)(9^2 + 1)^n/9^{2n} Note: You are allowed only six attempts on this problem.

Transcript

0:00 Hi there.

00:01 In this question, we have several series which are already divergent.

00:05 The first series is summation n equal to 1 to infinity 1 divided by root n.

00:09 The second series is summation in equal to 1 to infinity, 0x n by n3.

00:14 The third series is summation in equal to 1 to infinity, 0 .2 infinity 7n plus 7 divided by minus 1 power n.

00:20 And now the fifth one is summation in equal to 1 to infinity 7n plus 7 divided by minus 1 power n.

00:26 And now the fifth one is, summation n equal to 1 to infinity minus 1 power n multiplied by 2n factorial divided by n factorial square.

00:38 The next one is summation n equal to 1 to infinity.

00:41 N plus 1 multiplied by 9 square plus 1 whole power n divided by 9 power 2n.

00:47 Now we are given with several reasons for this divergence of the above series and we have to choose the first one which is the reason for the given series.

00:58 For the divergence of the given series.

01:01 The first reason is that divergence because the terms don't have limit zero.

01:05 The second one is that divergent geometric series.

01:08 The third one is that divergent p series.

01:10 The fourth one is that integral test.

01:13 Now fifth one is comparison with the diversion p series.

01:17 Next one is diverges by limit comparison test.

01:21 And last one is diverges by alternating series test.

01:24 So let's see how we'll do this.

01:28 So the first series is summation, equal to 1 to infinity 1 divided by root 10.

01:34 So here we'll check the reasons 1 by 1.

01:38 So the first reason is the divergence because the terms don't have limit 0.

01:43 Here we have the series or the sequence is 1 divided by root 1 and clearly limit 10 tends to infinity 1 divided by root 1 is equal to 0.

01:53 So a is not the point.

01:56 So a is not the option.

01:58 Now we'll move on to the second 1 b.

02:01 That is it is a divergent geometric series.

02:05 This is a geometric.

02:08 This is not a geometric series because a geometric series is of the form.

02:13 Summation n is equal to 1 to infinity, a r power n.

02:18 But this particular series is not in that form.

02:22 So we cannot see.

02:23 We cannot use that reason also.

02:25 So that is also not true.

02:27 Now the c option is divergent p series.

02:30 So here we can write a series as summation n equal to 1 to infinity 1 divided by n power 1 divided by 2.

02:40 And we know that this is of the form summation n equal to 1 to infinity 1 divided by n power p, where p is less than 1.

02:48 So we know that in a p series if p is less than 1.

02:51 That is if we have the series summation n equal to 1 to infinity 1 divided by n power p, n power p and p less than 1 implies that the series summation the series summation n equal to 1 to infinity 1 divided by n by n power p diverges okay so for the first one option c is the correct answer so for the first series option c is the correct answer that is divergent p series now we'll move on to the second one.

03:29 So the second series is given by summation n is equal to 1 to infinity, cosine of n pi divided by ln3.

03:41 So here also we'll check the reasons one by one.

03:44 So the first one is that we can see that the term cosine of n pi divided by ln3 is actually minus 1 power n divided by ln3 because we have that cosine n pi, cosine n pi is equal to minus 1 if n is owed and it is 1 if n is even.

04:13 So we can write this as like this and as n tends to infinity, we can see that limit n tends to infinity minus 1 power n divided by l and 3 will not attain the limit will not attain any limit that does not exist.

04:32 This particular limit does not exist.

04:37 So we can say that the first statement is true here.

04:41 So option a is correct.

04:43 Option a is correct.

04:49 That is diverges because the terms don't have limit zero.

04:52 Now we'll move on to the third series.

04:56 The third series is given as summation.

05:01 N is equal to 1 to infinity, l and n divided by n.

05:06 So here we'll check the options one by one.

05:08 For the first option, we have limit n tends to infinity lnn divided by n is equal to 0...

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