I'm mesmerized by how intuitive you made the theorem seem. I always felt it was sort of like a "paradox", but your explanation made it look almost plain obvious. Great video!

It makes sense to a degree, the same way that ∞ - ∞ can equal whatever you want

Sometimes I get frustrated by how you always state the obvious (you do it very slowly, of course). Then I realise that I would never have come up with what is "obvious" in the middle of the video had you not told me about what is "obvious" previously. And then I just realise that that's how math works! You just state the "obvious" and you come up with more "obvious" statements. This proves how good of an educator you are, great job.

My strategies for the extra problems: To make the series oscillate, instead of having one target sum, make it two, for example 1 and 0. First take enough positive terms to get the partial sum above 1, then take negative terms to get it below 0, then repeat. This way the series will have infinitely many partial sums both above and below the interval [0,1]. To make the series diverge to infinity, use the same strategy, but make the target sums increase by 1 each time they are reached. I.e. first take positive terms to get above 1, then take negative terms to get below 0, then get above 2, then below 1, then above 3, then below 2, etc. Since the negative terms converge to 0, the sequence of partial sums will have an increasing lower bound that will go to infinity.

I like how this is your only video and it's an absolute banger.

This explanation video was on par with 3b1b, if not better. As a very loyal 3b1b viewer, I want to emphasize that it means a lot.

I remember learning about conditional convergent series when taking calculus class and it always felt "why do I care if a series is conditionally vs. absolutely convergent". Your video answered my long-standing question. Thank you!

fun fact: for the (-1)^n/(2n+1) sequence, the series exactly equals arctanh(p)/2 + pi/4 where p is the proportion of negative and positive terms, ranging from -1 (all negative terms) to 1 (all positive terms) For example, the ++- pattern converges to arctanh(1/3)/2 + pi/4 I would love to see a proof for why this is the case, because it seems too simple not to have an elegant reason for being that way.

A minor mistake in the argument at around 8:52 for about why the added terms must converge to 0. It is not because otherwise it must diverge to infinity, but because otherwise it must diverge to infinity OR by oscillating (e.g 1 -1 1 -1… series does not diverge to infinity.)

I think this one might be my fav one so far! Amazing job! the animation you have made it really to follow along and helped me understand the concept a lot!

It is amazing how it has already been two years... I remember the time I watched your first video when it was 2 months old.... time flies by.

A really impressively clear explanation. Thanks a lot for making this!

For the infinity rearrangement try this. Suppose that the up and down arrows are sorted in terms of decreasing length.What you do is add on the first down arrow. Then add enough up arrows such that the total sum of all the arrows is bigger than 1. Add the second down arrow, then add enough up arrows to make the total sum bigger than 2. Add the third down arrow then add enough up arrows to make the total sum bigger than 3. Add the 4th down arrow then add enough up arrows to make the total sum bigger than 4. And so on.

My idea for one oscillation algorithm: Sort each subset (up and down arrows) by size, as done in the video, use the arrows in order of decreasing size. Start at zero (use this as a new minimum), then use the first up arrow to create a new maximum. From here on, after reaching a new maximum, add down arrows until you exceed the previous global minimum. And every time you reach a new global minimum, switch to adding up arrows until you exceed the previous global maximum. This will have the series oscillate with increasing amplitude - because every time you go down, you go down to an all-time low, and every time you go up, you go up to an all-time high. Because the series of up and down arrows each diverge when viewed independently (see video), it will always be possible to add enough arrows to exceed the previous maximum/minimum.

This was an incredibly lucid explanation. I think I’m actually going to try to teach this to the students in my math-for-art-students course. Before seeing this video, I wouldn’t have dreamed of trying to explain something like this to them, but if I replicate your explanation, I think some (hopefully most) of them might actually get it!

You must have put a whole lot of work into this and it was so good, thanks a bunch for this!

How have i never came across you before? Excelent quality of content, far beyond your current 10k subscriber count. Keep it up!

amazing. I felt I watched a 5 minute video since your presentation was smooth and intuitive, great job!!

The production quality for the first video is insane! You totally matched mathologer's video quality on this one. (He did a video on the same topic a few years ago)

Wow i'm very impressed of the combination of such pretty visuals together with a very good explaination !

this video made me feel like I knew a lot about the subject by simply explaining the topic really well

This is good content. Thank You. The idea of absolute convergence is much more clear to me now.

Diverging to infinity: Consider the length of the largest red arrow, and place a number of green arrows that add up to more than twice that length. Now place the longest red arrow. Continue in the same pattern. Of course red and green can be switched here to make negative infinity

I don't know what about your channel is different, but you've helped me understand these paradoxical maths things better than anyone else! Both this and fractional derivatives, finally explained understandably

What an explanation!!! Absolutely the best Maths video I've seen this year!

Wow, that's a great video! Step by step easy to follow, and yet not glossing over any important detail.

Subscribed and watched every video. Great job with the visualizations and distilling the salient features and ideas of the proof into something not just manageable but intuitive, all without any handwaving. Keep going!

Superb video! To summarize, the trick is we have changed the underlying distribution of the negative numbers vs. that of the positive numbers. In other words, how often the often occurs with respect to the other, leading to the different result.

this got me in stitchs awesome vid!

19:40 My take on your problem: 1-So a series diverge by oscilation you will set 2 goal lines, one that need to be transpassed in each phase ( upward and downward ). 2-To diverge to infinity do the exactly same thing as diverging by oscilation, but move each target up or down by some amount on each cicle.

Imagine a dancer who takes one step forward followed by one step backward forever. Then rearrange their steps so instead they take two steps forward followed by one step backwards forever. Somehow in my example it is not surprising at all that the dancers end up in different places after you repeat their moves an arbitrary number of times, even though you can rearrange the infinite sequence of moves of one dancer into the other's.

Awesome presentation, hoping for more videos., Thank you !

Great job making counterintuitive and difficult-to-grasp concepts seem natural!

So so good! I'd love to see the other half of this, that ordering doesn't matter for absolutely convergent series

Oscillating sequence: add positive terms until the partial sum is greater than 1, then add negative terms until the partial sum is less than 0 Positive infinity: add positive terms until the partial sum is greater than 1, then add the first negative term, then keep adding positive terms until the partial sum is greater than the next integer followed by the next single negative term (negative infinity can be reached by starting with the negative terms)

A fantastic video, thank you!

❤This is my favorite theorem in the whole calculus, both for sounding so counterintuitive and with so intuitive a proof.

I just found this video a few seconds ago and i must respectfully give you a compliment by saying you have an outstanding sense of humor because i giggled and laughed and rewinded the intro several times. Are you also a comedian?

A very good explanation of what seemed impossible. Well done.

wow! this is mindblowing!

I added your video to my private list labeled '' treasure videos '' And that's enough as a compliment

I don't feel that it's particularly paradoxical to begin with. A (only slightly naive) definition of an infinite sum is the number that we approach as we take longer and longer partial sums. If you cram two times more positive terms into each sum as in the first example, it'll obviously make all partial sums bigger and since you changed the ordering for the whole infinite series, it hols for the infinite sum as well.

Amazing video, maybe the best maths video I saw on YT

That is such a vivid presentation

Stupendous work!

A great video, 3lue1brown level quality, really nice!

This was a really good video love your explanation

A beautiful entry. Thank you

I came up with one way to approach infinity. For each down arrow, pop enough up arrow until the magnitude is more than twice that of the down arrows. Since this meta-addition will result in more than the total magnitude of down arrows, the sum would approach positive infinity. Same can be applied to approach negative infinity.

This is an absolutely impressive video

That's actually quite important in quantum mechanics! When switching from Schrödinger's to Dirac's picture of QM, one substitutes the concept of "wavefunctions" with that of "state vectors" living in a Hilbert space that is (for most problems) infinite dimensional. In this case the usual representation of matrices and vectors is not very intuitive, since an infinite-dimensional matrix cannot even be written down. Nevertheless the maths behind it still works in the conventional ways, and the definition of scalar product between vectors of infinitely many components is coherent and most importantly finite. Turns out that the dot product of a state vector with itself is just the infinite sum of the square magnitudes of the Fourier coefficients needed to build the Fourier expansion of the wavefunction in the chosen basis. So yeah, a dot product between infinite dimensional vectors actually represents a converging series, and this fact has physical importance!

For divergence to infinity, get the arrows to build a stair case by placing up arrows until they’re over 2 and then down arrows until under 1, then up arrows until over 3 to 2 to 4 etc. All the arrows must be used and it diverges to infinity. For a divergence by oscillation, pick any two real numbers a,b with a>b (WLOG). Place up arrows until they’re sum is greater than ‘a’ and then place down arrows until the sum is less than ‘b’ and then repeat. The arrows are always enough to get from a to b and back again and oscillate between the two numbers.

This explanation is so beautiful and awesome

Very well explained 👏

The explanation that got me is that the serie is INF+ (-INF), when you arrange P then N - which can result in any number.

This was great. Thanks.

Elegant. The reason this seems counter intuitive is because we forget that two infinities are not always equal. If you pick one pos and one neg number each time, that is a different infinite series than taking two pos and one neg each time. They will not converge to the same number.

Very well articulated. Your channel is a great boon to your audience and burgeoning mathematicians the world over.

Where you showed π÷4=atan(1) was absolutely COMIC

Fun fact: using the same construct you can actually find a sub-sequence of the partial sum sequence adjacent to any sequence u(n) you want. If you call S(n) the partial (rearranged) sequence it mean that you can find a rearrangement and a sequence k_n such that S(k_n) - u(n) tends to zero. The sum will therefore approache every term of the sequence, getting better and better every time. Even more stronger, you can control how it tends to zero: if you fix a threshold ε>0, then you can make so that |S(k_n) - u(n)|

To make it diverge to infinity (or negative infinity), or to oscillate I feel like you could move the desired sum each time you cross it (but such that the addition/movement of the arrow is still sufficient to cross the line; for simplicity let's say you move it the maximum possible amount in your intended direction) in the direction of either positive or negative infinity, or in the case of oscillation, in the direction of your desired maximum or minimum (and have it be that whenever your sum reaches within some delta of the desired maximum or minimum, you instead choose to approach the other of the maximum or minimum instead). Since you cross it infinitely many times and because the series is only conditionally convergent, I believe that would mean that you move the line an infinite amount of times, and the sum of the movements of the desired sum would diverge because it's only conditionally convergent, meaning the movement of the desired sum would also diverge. The only aspect I'm not 100% sure on is you could possibly make a series where if you add only the absolute values of the terms which cross the threshold it would be absolutely convergent even if the total series is conditionally convergent, but I suspect that due to the nature of conditional convergence that that might not be the case, but I don't have the tools to prove that part.

The sums always added up to the same value for the number of terms shown or upto 1/23 in all the three cases for me. When taken in the original order, when the adjacent terms are interchanged, and also when all the positive terms are taken together and all the negative terms grouped together.

This has made me think of the axiom of choice and how careful one has to be to be confident of infinite proofs using just cardinality ignoring order, see Cantor and Gödel.

Change the number for when you switch, like every time you go up double the number, or every time you switch you switch between two values (or more). In fact you can probably do some fun things by switching the number randomly with different distributions. Like make it so it doesn't diverge to anything. What makes this weird is what happens if you pick a nonconstructable number? I suppose this would imply that there couldn't be a nice rule to pick the next set of arrows.

You can get infinite or finite divergent sums by changing the target line to some other function which has a strictly smaller derivative than the lower bounding function of the terms (or the negative of the negative terms). If Aln(n+1) is strictly less than the sum of all positive terms, then Bln(n+1) is a valid target line iff B

Very well done video!

Here's my ideas: For +∞: Sum all the negative numbers, then add the positive ones. Vice versa for -∞. (At least I think that would work). Divergent: First place the largest positive arrow (or number). Next place negative arrows (starting with the largest) such that they are longer than the positive one in total. Then place positive arrows such that they are longer then the previous negative ones together. Repeat for all the numbers.

The to infinity construction is fun. You have two types of steps: even steps and odd. At each even step you enumerate arrows to get to n - x where n is the step you're on and x is the next negative number. Then at each odd step you enumerate a negative number. The cleanest construction I can think of.

This is undoubtedly a nice video and contains a lot of new ideas but I think the real challenge is how can these visual thoughts be turned into a formal proof . This is the real challenge I believe 👍❤

Your explanation made me go from "wtf that's impossible" to "oh this is so obvious"

I know nobody will see this comment on a year old vid, but I feel clever so I'm posting it. My idea for making a sequence that approaches infinity goes as follows: With the arrows arranged by size you take enough new up arrows that in total they exceed the absolute value of the last down arrow (basically tip to tip the ups are longer than last down.) Add all those up arrows and then the next down arrow to the sequence, and just repeat that. Another way to say this is you just need to get a little higher each time you are going up. Unless I'm missing something this would approach infinity, just potentially very slowly.

For convergent series (sums where the terms get smaller and smaller, approaching zero), the rearrangement is a problem when: The nth term is moved m terms away from where it was, and m has no limit/bound to how large it can get. So, at 1:19, -1/15 should be in the 5th term, but gets rearranged to be in the 12th term. The sum shown is different because terms can be moved up to any distance from their starting positions, even an infinite distance. For non-convergent series, such as 1-1+1-1+1-1..., almost any rearrangement will result in the sum value changing, because the sum doesn't have a consistent clear value.

It's actually pretty simple. If the series converges absolutely, so do the positive and negative subseries. But in a series where all elements have the same sign, rearrangement can't change the sum. Conversely, in a conditionally convergent series, both the positive and the negative subsequence go to infinity. Hence, you can approach any value by just picking negative elements when the partial sum is above the target and positive values when below. You will always pass the target because you have an infinite budget and you will get closer and closer because the elements get arbitrarily small.

To make them non convergent on anything the positive and negative arrows must be trying to converge on different numbers where the positive ones must be trying to converge above where the negative ones are trying to converge. In order to make them blow up to infinity you add up enough positive ones to be at least some fixed amount greater than the next negative arrow and then add that one. Vice versa for negative infinity. Pretty fun puzzle tbh

my take: convergence to infinity: Separate the sequence into positive and negative numbers, sorted by value (like in the video). Use positive numbers until you get above 2, then use negative numbers until you get below 1. Then use positive numbers until you get above 3, then switch again to negative numbers until it falls below 2. Once again, use positive numbers to get above 4, and use negative numbers to get below 1. At any step you use the positive numbers to get add ‘+2’ to the sum and then you ‘take away’ 1 with negative numbers. The conditional convergence assumption should make this possible (since both the sequences of positive and negative numbers are divergent). Example where it doesnt converge: Same setup as before and as in the video: two sequences of positive and negative numbers, sorted. Use positive numbers to get above 1, then use negative numbers to get below 0, and repeat.

For a more general proof, replace the constant target number with an arbitrary infinite series: it literally doesn't matter what the series is as long as every term in the series is a real number. Starting with the first number in the target series, apply enough up or down arrows (one or three other; not both) to cross that value; then advance to the next value in the target series and repeat the process. The difference between your partial sum and the value of that target series will converge to zero, which can also be phrased as your infinite sum converting to the infinite series. Meaning that you can make the infinite sum become infinitely close to the behavior of the target series. So if the target series converges, so will your infinite sum; if it diverges, so will your infinite sum, and in exactly the same way: positive infinity, negative infinity, or forever oscillating. Or, if you have a more detailed means of measuring how the series behaves in the long run (e.g., hyperreal numbers or intervals), your partial sum can be made to match whatever the measurement of the target series is.

Oscillation solution: Pick two values, we'll say 1 and 2, then add positive numbers until you pass the greater value (2) then add negative numbers until you pass the lesser value (1). This will oscillate between 1 and 2. Infinity solution: Add a negative number from the list, then add positive numbers from the list until the current run of positive numbers is greater or equal to double the last negative value, the end result after these two steps will always be at least the absolute value of the sum of the negative values (infinity). Positive and negative may be swapped to get negative infinity.

Awesome!

What strikes me is that while you can prove that there will always be some re-arrangement of a conditionally convergent sequence that will generate any real number you choose, it doesn't mean that finding that precise rearrangement will be that easy in reality, as virtually all real numbers will be generated by a rearrangement of the sequence with no regular pattern, in which you simply need to know an infinite number of terms, rather than be able to infinitely generate them from a pattern (i.e. if you randomly choose a target number to converge on then it is almost certain that the rearrangement solution will not be of the form 'a times up followed by b times down, repeated infinitely many times' or anything of that sort).

Amazing quality content

I tried to solve your challenge and here’s what I came up with: To make a series converge to nothing, just add up-arrows until they exceed 1, then add down-arrows until they’re below 0, then add up-arrows again until they reach 1 and so on. (Obviously the numbers don’t have to be 1 and 0). To blow up to positive infinity, add up-arrows until you hit an integer (k), then add down-arrows until you’re below k-1/2, then add up-arrows until you reach k+1, then down-arrows until you’re below k+1/2 and so on. There’s probably a much simpler solution, but this works fine. Negative infinite is just the opposite strategy.

The crux of this is that conditionally convergent series are just discrete ways to write out infinity - infinity, which you know to be indeterminate, and can take on any value when given a concrete representation. It's not entirely obvious this is the case initially, but once you realize that the conditional convergence implies P+N = finite, but P-N (or P + abs(N)) is infinite, it becomes pretty clear imo For the end challenges, you could simply set a series of targets to reach, for +inf let's say we aim for 2, 1, 3, 2, 4, 3, 5, 4,... , for -inf simply the negation of all those terms, and for divergence we can do 1, -1, 1, -1,... . In all these cases, there's a total infinite upwards movement and downwards movement, so all the arrows will be used.

Absolutely amazing & beautiful Congratulations

nice intuitive explanation

Absolutely beautiful ❤️

19:25 The trick here is to make the "target line" dynamic. To blow it up to plus infinity you move the line up a finite distance every time you cross it downwards, to blow it down you move it down a finite distance every time you cross it upwards. To make it oscillate you move the line up when crossing it downwards and move it back to its original location when crossing the moved line upwards. You can also make the line jump a random (but finite) distance against the crossing direction every time you cross it to make a sum that has no limit and does not oscillate.

thank you for your perfect explications.

I lost brain cells very early but this video is amazing. Never expected something like this to happen.

Love this video

My strategies: Divergence: as soon as it crosses the value S_1, switch to a new value S_2, which it'll "travel" to until it reaches S_2, then when it reaches S_2, switch to a new value, and so on and so forth. Positive and negative infinity: I have no idea

Awesome sir

My approach for oscillating, choose S1, S2 in the reals so that S1 =/= S2. Without loss of generality we can choose S1 > S2. Then add up arrows until you pass S1, then add down arrows till you reach S2, then head back to S1, repeat forever. Again we can use the fact that the series is conditionally convergent to know we can always reach the other side. My approach for infinity, you could switch up and down for negative infinity. Start with a single up arrow, then add a down. From here on add the next N up arrows, where N is the number of term in the series so far, at this point you would add 2. Then a down arrow. Now you are adding the next 5 up terms, then a down term. Now 11 ups 1 down. 23 ups 1 down. 47 up 1 down. As you continue on, you will have more upward motion than downward. Unless the magnitude of the 1 down is larger than the N ups. However as the sum of positive terms approaches infinity (conditionally convergent) You will eventually add so many up terms no down term could compensate, as the down terms will approach 0.

This also provides a novel bijection from N^N to R (or R^k, as Lévy-Steinitz proves). Does anyone know about random rearrangements of conditionally convergent series?

begin with the first positive term (P). substract as many terms as needed until you reach -1*P. add as many terms as needed to reach a number P' such that P'>P substract as many terms as needed to reach -1*P'. add as many terms as needed to reach a number P'' such that P''>P' repeat as many times as you want. the result will diverge reaching infinite and minus infinity. if you want a series that approaches only infinity, use always only one negative term instead of enough terms to reach -1*P if you want a series that approaches only minus infinity, use always only one positive term instead of enough terms to reach P'>P if you want a series that oscilates between A and B with A>B add as many terms as you can without reaching A, substract as many terms as you can without reaching B, repeat until you get bored.

19:18 That makes sense. Instead of trying to try to approach the line y=S, you can make the arrows approach the line y=Sx. This is just a linear equation which at x=inf has the value y=inf. What x is in this case is the count of how many times the target value line is crossed, not how many arrows are taken into account. Otherwise it would not be possible to stay on the line (I think). In that same way, you can make the arrows approach y=sin(x), to have the series not approach anyting at all.

I've seen lots of videos about the Riemann Rearrangement Theorem, and I think this is a good introduction to someone with little experience. However, I think it's a shame you didn't bring up the definition that the sums of positive and negative terms must go to infinity. I think that shows the issue very clearly, because the total series is infinity minus infinity, which is undefined. So if you take infinity minus infinity, you can bracket and rearrange the terms to get whatever result you'd like, because the equation is ultimately not defined.

thanks for this, it was very interesting… subscribed after 5 minutes (watched the entire video)… good luck

If KZfaq allowed double likes I would have given them to you....that passing comment about L_2 spaces made my day; it makes so much sense - invariance of inner product under space transformations

how to rearrange the terms so that the infinite sum diverges into oscillation or infinity pick 2 values, S_0 and S_1, such that S_0

proof that arranging a certain sequence of numbers within a conditionally convergent sequence can reach infinity: Add up enough numbers to go in the direction of the infinity you want to reach to value V. Once you reach value V, add a negative number if you want to reach +infinity or a positive number if you want to reach -infinity. Then add up enough numbers to surpass value V up to new value V2, and repeat. This is guaranteed to get you to the infinity you wish to repeat because of the following: 1. Removing an item from an infinite set still renders that set size to be infinity because infinity-1=infinity 2. You will be guaranteed to surpass value V at least twice because the sum of the infinte series of positive/negative numbers of conditionally convergent sequences is an infinity. If you remove an element from the sequence, the sum of the sequence is still an infinity because infinity - x = infinity