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Numerade Educator



Problem 52 Hard Difficulty

Prove that if $ \sum_n $ is a conditionally convergent series and $ r $ is any real number, then there is a rearrangement $ \sum a_n $ whose sum is $ r. $ [Hints: Use the notation of Exercise 51. Take just enough positive terms $ a_{n}^{+} $ so that their sum is greater than $ r $. Then add just enough negative terms $ a_{n}^{-} $ so that the cumulative sum is less than $ r $. Continue in this manner and use Theorem 11.2.6.]


Thus, for all $n>0,0<\left|a_{\sigma(n)}-M\right|<\left|a_{k}\right| .$ So the following is true:
$\sum_{n=1}^{\infty} a_{\sigma(n)}=M .$ (similar for $M \leq 0 )$


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Video Transcript

in this problem, we have, ah, conditionally converge in Siri's and ours. Any real number. Now we'd like to show that we can rearrange the sequence. And so let's say right now we have something like this. And then now we're going to rearrange that into a new sequence here. So where that these bees are equal to the Jays, but the order has changed. So, for example, you can say something like would have something like of that form or more generally even or hear Sigma's once one function. Okay, so let's make some observations here. So conditionally conversion implies conversion. I can. And this will imply if we go to the diversions, test that if you take the limit of the ends, those have to be zero now. Also, let's look back to exercise fifty one, where we define additional sequences, positive part of an also the negative part of an and notice that since this limit goes zero, that would imply that both of these limits also have to go to zero. Hey, because a Zen ghosts infinity a and goes zero. So these guys vanished. But so do these and one more remark from number fifty one. If you go to part B of that problem and assuming that the original is conditionally conversion, we could conclude at the positive part Diversions to infinity And that the negative part whatever. Just to minus infinity. Okay, let's go on to the next page here. So now the next step is to we want to rearrange the A m. So we'LL go ahead and call this new arrangement this rearrangement being so here's how we'LL do so. So if we have the following If you take some of these sums of these bien terms any B ends, you want any of the ends I mean and if you include one more term, it's called us, be care. Then we can conclude that this being close one had to be negative and we could even go a step further. Now let me go and verify prove this last case. So we have one toe and b k bigger than our and then let's go ahead and add the negative sum from one toe and plus one of the beings crumble sides. And now, since being plus one is a negative number, this is a positive number, and so that's why we can just to note this by being plus one. So here in this case, we want to know how should we choose Bien plus one? Well, we see that it has to be a negative number because of this, so that you could help us here. So we'LL go ahead and take Bien plus one to be the smallest remaining Sze Herm in the sequence of negative terms. And as we notice already that beat the as this is an increasing so we could rearrange He's an increasing order and as we observed on the last page, he's got a zero and this is useful here because if we look at this term right here, we just showed that the upper bound is going to zero in this case and it looks like that will imply that this partial some is converging toe are. But in order to conclude that well, look at another case similar to this first one here. So let's go on to the next page. So this is the other case where you some end of the bees and you get less than our but then you include one more term and it's bigger than our sown here. This must mean that being plus one is positive, and it also means this is a positive number because of this. So if you want, you could even put absolute values there. But it's not necessary. This will be Weston being close one again. I don't need the absolute value because it's positive. But as we noticed on the last page, we should mention how to define the n plus one. And this time we'LL take it to be the the largest remaining term of the sequence of positive parts. And this's now because we're we ordered this in decreasing order, so they're getting smaller each time. So by the diversions Testa's we noticed on the first page this will go to zero. So what we just showed is the following, regardless of which case were in either case, one on the previous page. Your case too. We just showed that this number is can always be set less than some number that tense zero. So I get more formally let me write this on the next page. Given any positive number, Absalon bigger than zero. Since our sequence is converging to zero, we have let me take a step back here. There exists are about this, by the way, we chose the bees. In each case, they have to get smaller and smaller. So eventually we could make it less than any Absalon. But also taking and larger than the same. And here, by our previous work, shows this that the air between the actual some are in the partial, some less than being plus one. But that's already lesson Absalom. So therefore, we just proved that this some converges two are and that resolves the problem.