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# Suppose the series $\sum a_n$ is conditionally convergent. (a) Prove that the series $\sum n^2 a_n$ is divergent.(b) Conditional convergence of $\sum a_n$ is not enough to determine whether $\sum na_n$ is convergent. show this by giving an example of conditionally convergent series such that $\sum na_n$ converges and an example where $\sum na_n$ diverges.

## a. This contradiction shows that $\sum n^{2} a_{m}$ diverges. b. In both cases, $\sum n a_{n}$ diverges by the Test for Divergence.

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##### Top Calculus 2 / BC Educators    ##### Kristen K.

University of Michigan - Ann Arbor

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we were given that the syriza vahan is conditionally conversion. Yeah, so in part A, we'd like to show and squared And the sum of this diverges so away to prove this is to go by contradiction. So here, Lis just point that out there. So let's go ahead and suppose the opposite. That would be the sum of and square a n converges Now by the test for diversions. This would imply that the end's term here goes to zero in the limit. But we can rewrite this in a way that will be convenient for us. So let's go ahead and rewrite the and square as a fraction. And if this lemon goes to zero, then I could go ahead and take absolute value. Here is well, so let me just go ahead and throw the absolute value there. Now we go to the limit comparison test. So here's another Theron and we have a fraction here. We know that the Siri's one over and square converges. You can use the pizzas to see this P equals two. It's bigger than one. And that implies that too. Some convergence. So in this case, limit comparison test would say so does the Siri's here. However, this cannot happen because in the given information we have that this syriza's conditionally conversion that it means part of the definition that if you take the absolute value of a n, the Siri's will diverge. So is it recap. We suppose that it converges because we're using proof by contradiction. And under this assumption in the proof, we end up showing that the Siri's an absolute value, converges. But that contradicts. They've given information, so that completes the proof. A party. Okay, now let's go to the next page for part B. So just remember still the given information I'll put that here they and some this was conditionally conversion. So I'll just abbreviate that. And now we want to show that this is not enough enough information to determine if the Siri's and and commercials So we'll show this by giving an example one in which it converges, and then one of which diversions. So let's look at an example here. Par I. So in this case, let's go ahead and define the sum of a M the following way. Now the Siri's is conditionally conversion. You can see that it conversions by the alternating Siri's test. But if you take absolute value, the negative one just becomes a one. And this diverges because you can use the pee test here. P equals one. Yeah, so that guarantees that this example were using, in part one here that it is indeed conditionally conversion. We have to check that because that is the assumption here in the problem. Now I look at the Siri's and a N that just becomes negative, one to the end and this diverges and to see that you can use the test for divergence. So here's one example in which the sum of nn divergence now will provide an example in which it converges this time well defined. The sum of an let's go ahead and do it negative one to the end over and let's do and natural log and in the denominator. So this is conditionally comm urgent. You can use alternating Siri's test that all apply convergence. But if you look at the absolute value, this diverges and to see that you could use the integral test so that once again guarantees that we're under this hypothesis up here, conditionally convergent. And then now go ahead and look at the sun stand there that becomes negative one to the end, over natural log of end in the Siri's those converge by the alternating serious tests. So we gave two examples, one in which the sum of many and diverse and that another example in which it converges and that solves proud part B of this problem.

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##### Top Calculus 2 / BC Educators    ##### Kristen K.

University of Michigan - Ann Arbor

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