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Let Caz1 &n be & conditionally convergent scries. Provc that for any A € R there exists & bijective map k from N to N such [email protected](n) = A.(That is,...

Question

Let Caz1 &n be & conditionally convergent scries. Provc that for any A € R there exists & bijective map k from N to N such [email protected](n) = A.(That is, some rearrangement of the terms sums to A)

Let Caz1 &n be & conditionally convergent scries. Provc that for any A € R there exists & bijective map k from N to N such that @k(n) = A. (That is, some rearrangement of the terms sums to A)



Answers

Let $\sum a_{n}$ be a series that converges to a number $L$ and whose terms are all positive. We show that any rearrangement $\sum c_{n}$ of this series also converges to $L$. (a) Show that $S_{n}<L$ for any partial sum $S_{n}$ of $\sum c_{n}$ (b) Show that for any $K<L,$ there is a partial sum $S_{n}$ of $\sum c_{n}$ with $S_{n}>K$ (c) Use parts (a) and (b) to show $\sum c_{n}$ converges to $L$

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.

If the's sums are both conversion and we have dealing with positive terms, so this means and and be enter both positive. Is it true that this some also convergence? So let's try to see if this is true here. So fence, there's some converges by the diversions test. This is from section eleven point two. So let me write that on the side eleven point two. We have that be and must approach zero in the limit. Since B ends eventually are getting very close to zero. There exist and inside your end. So let's say a natural number and such that if we let the index little and be larger than that, then the fiend must be less than or equal to, Let's say, one. This is just a consequence of the limit being zero sense it zero. That means there's a point at which B n plus one B and plus two and so on. All these numbers will be less than or equal to one. So all I'm saying here, therefore, because of this fact, I can rewrite this some let's say there are using a starting point here unless you say the starting point is one. It doesn't really matter what the starting point is. You could replace that one with any other number, and this will still be true. So first, let me instead of writing is infinite. Some break this into two parts. So look at the sun for all the way up to Capitol. And then I'll look at the remaining part now in this some over here, our little and is bigger than capital and therefore, for these values and this sum over here in the green, I can go ahead and use the fact that being is less than or equal to one due to this fact. So that means that our Siri's over here is less than or equal to the sum from one to capitol and a NBN. But then over here I'm just replacing being with one, and I have the inequality here. So this inequality is what made me write this inequality here. And I just have the sum from capital and plus one to infinity of a end. All I'm doing here is a NBN less than or equal to a M times one equals and and finally I know that this some here let me use a different color this red some over here, which I'm now using blue. This some converges because we were given that there's some over here convergence and the sum from N plus one to infinity of a end is less than or equal to, actually strictly less than the sum from one toe infinity of AM so sense we're given at this one emerges then by Red, a comparison test. This also converges. So let's say, by comparison with the entire series of the A M so sense it converges and rolling, adding positive numbers. That just means that we just showed that the sum from N plus one to infinity of a M is less than infinity. However, this original sum over here in the green, which I'm not circling this is automatically less than infinity because it's only a finite sum. Therefore, we're adding two numbers together that are less than infinity, So the result will also be lesson infinity. So therefore, we have shown that the sum of the A m. Bien first of all, we know it has to be larger than zero because a end and beyond are positive. On the other hand, We just showed that it's less than infinity, so it has to converge to a real number because again we're only dealing with positive terms, and that's our final answer.


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