Real Analysis | Rearrangements of absolutely convergent series.

[@AhmedBachir]

I like this guy

[@goodplacetostop2973]

11:17

[@ArpanDasgupta-q4n]

One thing that wasn't obvious to me initially was when exactly we were using the fact that f is injective and surjective. We are using this fact between the second and third lines at 11:16. Because f is injective, we know M > N, so m >= M > N. Because f is surjective, all of the terms in the list a_1, ..., a_N appear in the list b_1, ..., b_M, proving that the third line at 11:16 follows from the second. Also, ? = M.

[@lilylikesmarkies]

Great video as always. Perhaps an interesting result to present in this area would be Riemann's rearrangement theorem on conditionally convergent series

[@爸爸到底-s9x]

Since f is a bijection, the set {1,..., M} contains the set {1,..., N}. Hence the blue part = |b_1 + ... + b_f(1) + ... + b_f(N) - a_1 + ... + a_N| = what then follows in the video. I personally think the fact that f is bijective makes the identity hold is worth noting somewhere inside the proof.

[@carl3260]

Agree this is the key bit. But I think the set of b indices {1,...,M} contains {f(1),...,f(N)}, so the subtraction leaves only “b”s equal to “a_i”s, i>N (not nec. Including N+1, but def later so Cauchyable)

[@thesecondderivative8967]

10:27 I'm somewhat unsatisfied. We didn't use the fact that the series is absolutely convergent in our proof at all. What we proved is that we can find a bijective rearrangement such that our series converges for that rearrangement. Someone correct me if I am wrong. This is because we're letting our M be so big that all of the first N terms are contained in it. Otherwise, negative a_k terms would be in the absolute value which we haven't proven will not make the expression non-Cauchy-able. I tried to see if absolute convergence can fix that. I've been stuck on this problem for days. I can't solve it. It feels like you can take an arbitrary sum and create some bijection and assign the rearrangement tto that sum. It feels like we proved conditionally convergent sums can take any value supposed we have the index to be bijective to another index of an arbitrary sum. Someone please, if you find a fault in my reasoning, tell me. Please note that the crux of my argument is that the proof does not use the absolute convergence of S_n at all. If you would point to 7:16 as proof, note that the proof still works if you just assumed normal convergence.

[@MuffinsAPlenty]

|(b1+...+bm)−(a1+...+aN)| may not be less than ε/2 if the original series isn't absolutely convergent. One thing to keep in mind is that (b1+...+bm)−(a1+...+aN) may not be equal to a_(N+1) + a_(N+2) + ... + a_?, where every term from N+1 to ? is included. It could, for instance, be something like a_(N+1) + a_(N+3) + a_(N+5) + a_(N+7), because maybe a_(N+2) and a_(N+4) and a_(N+6) might be further along in the rearrangement. Therefore, |(b1+...+bm)−(a1+...+a_N)| may be LARGER than |a_(N+1) + a_(N+2) + ... + a_?|. Again, what if we have an alternating series, and (b1+...+bm)−(a1+...+aN) = a_(N+1) + a_(N+3) + a_(N+5) + a_(N+7). Then all four of these terms have the same sign, and the missing a_(N+2), a_(N+4), and a_(N+6) terms have opposite signs. So |a_(N+1) + a_(N+3) + a_(N+5) + a_(N+7)| > |a_(N+1) + a_(N+2) + a_(N+3) + a_(N+4) + a_(N+5) + a_(N+6) + a_(N+7)| So we don't have any control on |(b1+...+bm)−(a1+...+aN)|. We can't say for sure that it's less than ε/2. We can, for sure, say that |a_(N+1) + a_(N+3) + a_(N+5) + a_(N+7)|

[@thesecondderivative8967]

@@MuffinsAPlenty Thank you very much. I figured the gaps would make the series smaller than what was needed for the Cauchyness of the difference inside the absolute value but I failed to notice that a_N+k, k being natural, and consisting of all the gaps, might be negative and might skew the absolute value expression making it smaller instead of bigger like I initially thought. You're right; the only way to control the endpoint and be sure that it is bigger is so that the absolute value of each individual term even with the gaps filled in would still be less than e/2. The flaw in my reasoning was that I thought |a_N+1 (missing a few a_N+k terms)+...+ a_?| < |a_N+1 +...+a_?| ,but for some sequences, that might not be the case. However, which ever is bigger would be smaller than |a_N+1| +...+|a_?| which is less than e/2. Your counterexample to my argument help put things into perspective. I'm immensely grateful. I'm somewhat embarrassed I spent so much time on this though.

[@think_logic]

But where exactly does this proof break down if you try it with a conditionally convergent series?

[@yoyostutoring]

For conditionally convergent series, you can always rearrange terms to get different results (in fact, any number you like). For example if you take the sum from 1 to infinity of (-1)^n/n is a conditionally convergent series, but by summing it normally, you'll get ln(2) and there is a rearrangement such that the same series adds a total to 0.5ln(2).

[@matthiasbergner8911]

Here is suggestion for a related result, I am not sure whether you already did a video on this : Take a series that converges, but does not converge absolutely. Then for any real number M there exists a rearrangement of that series such that the rearranged series converges to that chosen number M.

[@爸爸到底-s9x]

You can make a conditionally-but-not-absolutely convergent series diverges as well if you want.

[@monu_7712]

I like that 4:05

[@ethanbartiromo2888]

I feel like the logic doesn’t follow in my head. Let’s say f(1) = 21 and f(k) = k for 1 < k < 21 and f(21) = 22 and f(22) = 1 then if N_1 = 5 and N_2 = 4 then N = max(N_1, N_2) = 5 and max(f(1), f(2), f(3), f(4), f(5)) = 21 and therefore b_1 + … + b_m where m = 21 does not include a_1 and therefore with it would be a_6 + a_7 + … + a_21 - a_1 inside the absolute value, but let’s say 0 < a_6, a_7, …, a_21 < 1 and a_1 = 100 then obviously |a_6| + |a_7| + … + |a_21| < |a_6 + a_7 + … + a_21 - a_1| therefore the statement at [10:16] doesn’t hold… am I missing something?

[@uzferry5524]

I think you got your definition of rearrangement confused when converting between b's and a's. f(1)=21 means a_1 becomes b_21, right? so if M=21, b_1 + ... + b_21 will certainly include a_1 (from b_21), a_2, a_3, ... , a_19, a_20, a_22 (from b_1), what it's not going to have is a_21 (because b_22 gets cut off). therefore b_1 + ... + b_21 - (a1 + ... + a_5) will be a_6 + a_7 +.. + a_20 + a_22

[@jonaskoelker]

Here's a related, simpler theorem: let b_n be a rearrangement of a_n, i.e. b_f(n) = a_n for some bijection f: N -> N with g(f(n)) = n for all n in N [and so b_n = a_g(n)]. If the a_n converge then b_n converges to the same limit. Proof: Let epsilon > 0 given with M in N such that |a_n - a| < epsilon for all n > M. Consider all natural numbers before M, let H = {f(1), ..., f(M)} and let K = max(H). When k > K we have g(k) not in {1, ..., M} i.e. g(k) > M and thus |b_k - a| = |a_g(k) - a| < epsilon. But that says the b_n sequence converges: choose K for that particular epsilon. Since the convergence of a series is the converges of a sequence of prefix sums, we get the converge part of the video's theorem as a corollary of this theorem.

[@lily_littleangel]

10:50 Yikes... I don't think e/2 + e/2 < e.

[@H3XED_OwO]

"equation" < e/2 + e/2 implies "equation" < e that's what he meant

[@rajeshsalvi8249]

May I know the reference book u use for real analysis ?

[@prakashsharma8339]

Hey micheal been following your channel for long , was wondering if you could start membership.

[@ΧρήστοςΠαπαδημητρίου-μ6π]

Is the converse of the thumbnail true?

[@isaacmammel9186]

Depends. Do you consider the sum of two divergent series to be equal?

[@ΧρήστοςΠαπαδημητρίου-μ6π]

@@isaacmammel9186 excluding divergent series, if the series convergences for every possible way of rearrangements does it mean that it converges absolutely?

[@stephenbeck7222]

Yes it is true for real valued series. Another way of stating that a series converges for all rearrangements is that the series unconditionally convergent. In general, absolute convergence implies unconditional convergence (the theorem proved in this video). For real series, unconditional convergence also implies absolute convergence.

[@manun7105]

@@ΧρήστοςΠαπαδημητρίου-μ6π It can easily seen by the Riemann rearrangement theorem. If we has an unconditionnally convergent serie, in particular it converges (the bijection is the identity). But if It's not absolutely convergent, the riemann theorem would imply that for a given bijection, the sum is +oo, which contradicts the unconditionnally convergence.

[@stephenbeck7222]

Weird that a relatively small advanced math channel gets spam (robot?) comments.

[@mcalkis5771]

I lost you at 9:30, I don't understand how the next line follows.

[@suhailawm]

Super sir. Thanks a lot

[@orenfivel6247]

Is there a video that proofs: if (and maybe only if also) f:ℕ→ℕ is a bijection (i.e., f(n) is the rearrangement or the permutation of the natural #'s), then f(n)→∞ as n→∞? Is it true also we need to demand that for all finite n∈ ℕ, f(n) must be finite?

[@MuffinsAPlenty]

All natural numbers are finite, so yes, f(n) must be finite. And that fact is sort of the key to proving that f(n)→∞ as n→∞. Formally, you want that, for each positive integer M, there exists a positive integer N so that whenever n >= N, we have f(n) > M. So fix some positive integer M. What you do is you look at the set S = {m ∈ ℕ | f(m) = N, we will have f(n) > M.

[@mikecaetano]

Would the colloquial analogy to this proof run along the lines of something like, the change in your pocket adds up to the same amount whether you start counting quarters first or pennies first? So that rearrangements correspond with permutations and f(n) maps the terms of the series ordered one way to the terms of the series ordered another way?

[@isaacmammel9186]

Well that makes it obvious when you have a finite amount of coins. However, when you have an infinite amount of terms, things become more weird. If the sum of the terms converges, but does not converge absolutely, then with a suitable rearrangement, you can actually make the sum equal to whatever you want!

[@nicolasguarin1344]

@@isaacmammel9186 ¿do you have an example?

[@isaacmammel9186]

@@nicolasguarin1344 Sure! Let's take the series 1-1/2+1/3-1/4+1/5... This series converges by the Alternating Series Test (specifically, it converges to ln(2)), but if you take just the odd terms or just the even terms, this is not the case. 1+1/3+1/5+... diverges to infinity, and -1/2-1/4-1/6... diverges to negative infinity. Now, in my previous comment, I claimed that it could converge to whatever I wanted it to be. Let's choose negative pi as an example. How we would make this happen is to put enough negative terms at the beginning until the partial sum is less than negative pi, then put positive terms until it is more than negative pi, and repeat the process. This will obviously make the partial sums get closer and closer to negative pi as the series progresses, making the limit of this new series (that uses the same terms, just rearranged) equal to negative pi.

[@ativjoshi1049]

@@isaacmammel9186 aptly explained

[@ancyvarghese1134]

cauchy criteria is int sufficient

[@pbj4184]

I get rigor n' all but isn't this fact common sense?

[@cletushumphrey9163]

no

[@pbj4184]

@@cletushumphrey9163 Why not? You're just rearranging the terms if the function is bijective. And we all know addition over real numbers is associative. So the order of the terms doesn't matter when you're adding them. And so the total sum doesn't change if you have a bijective function. QED.

[@sabhrant5533]

@@pbj4184 A conditionally convergent series of reals can be rearranged to get any real number.

[@pbj4184]

@@sabhrant5533 Yeah but he assumed the series is absolutely convergent. So we don't have that issue either

[@keithchan379]

@@pbj4184 he is not showing series of absolute value of a converges to same value as series of absolute value of b. he is showing series of a converges to same value as series of b. Your claim is the first one, which is trivial. The theorem is about the second one, which is not true for any a and its rearrangement b, as @sabhrant has pointed out.

[@jimiwills]

Click ;P