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JH
Numerade Educator

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Problem 51 Hard Difficulty

Given any series $ \sum a_n, $ we define a series $ \sum a_{n}^{+} $ whose terms are all the positive terms of $ \sum a_n, $ and a series $ \sum a_{n}^{-} $ whose terms are all the negative terms of $ \sum a_{n.} $ To be specific, we let

$ a_{n}^{+} = \frac {a_n + \mid a_n \mid }{2} $

$ a_{n}^{-} = \frac {a_n - \mid a_n \mid }{2} $

Notice that if $ a_n > 0 $, then $ a_{n}^{+} = a_n $ and $ a_{n}^{-} = 0 $ whereas if $ a_n < 0, $ then $ a_{n}^{-} = a_n $ and $ a_{n}^{+} = 0. $

(a) If $ \sum a_n $ is absolutely convergent, show that both of the series $ \sum a_{n}^{+} $ and $ \sum a_{n}^{-} $ are convergent.
(b) If $ \sum a_n $ is conditionally convergent, show that both of the series $ \sum a_{n}^{+} $ and $ \sum a_{n}^{-} $ are convergent.

Answer

a. we conclude by the Comparison Test that both $\sum a_{n}^{+}$ and $\sum a_{n}^{-}$ must be absolutely convergent. Or: Use Theorem 11.2 .8
b. $\sum a_{m}^{+}$ can't converge. Similarly, neither $\operatorname{can} \sum a_{m}^{-}$

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Video Transcript

well, given the sequence and we define the positive and negative terms by these two formulas here. So let's go ahead and show party here. So we're given that this sum is absolutely commercial and we'd like to use that to prove that these two series of the positive terms and one for the negative terms that these air also commercial. So let's go ahead and prove that the proofs will be very similar and part a of both of these facts here. So starting with the first series, just use the definition of an plus up here to rewrite our INSERM and I'LL do a step further I'LL go ahead and split up these fractions here now if I because A N is absolutely conversion, that implies that the sum of the absolute values and the sum of an both converge so I can go ahead and actually rewrite. This is the following. I could do an pull out the one half and distribute the sum inside and also pull out the one half. And as we just said, absolute convergence implies that the sum of the absolute values converges. And by comparison with this with this green, Siri's over here If you's comparison test. Oh, we write that out there. It's an important step that'LL also apply that this sum of a n also converges. So in general, absolute conversions implies some of absolute value and the sum of and both converge. So that means that these sums are real numbers. And if you divide them by two and add them, you still get a real number. So the whole sum convergence now in part, B R sees me in the second part of our way where we deal with the minus term. Oops. She just be a minus, not a minus one. There, Come on. Once again you can go from one to infinity and if I use the same steps is up in the previous line, you get the same thing except a minus sign. However, the fact is, if you subtract two conversion Siri's, you still get a real number. So since both of these sums converge, this is a real number. So conversions And so that resolves part. Ay, we showed both of these sums converge. Let's go on to part B. Now we'd like to show that if a and his conditionally conversion and I'll just remind you what that means. In a second, we'd like Tio Show that this implies that those sums that we just consider they don't convert But they'll divers in this case so conditionally convergent That means that a n ca merges. But if you take the absolute value on the inside, that will die, George. So that's definition of conditionally conversion. And if we write down those previous equations we had on the previous page for there are two sums that we're considering. So let's go ahead and write those back. Oh, now, using the fact that we have conditional convergence this first, some will converge here, so that is a real number. However, the second son diverges, so this will be infinity. And if you add infinity in other words, if you add a diversion to a conversion, Siri's the result will be infinite, not a real number. So this diverges. And similarly, if you do the one with the minus sign this time you'LL be, you would have a real number minus infinity. So this one will also diverge. But pull out the two one to infinity so this converges. But then this one diverges to infinity. So you have a real number minus infinity that'LL be minus infinity so that this some also diverges and that resolves party.