Can you change a sum by rearranging its numbers? --- The Riemann Series Theorem

Comments · 614

• JO

  @joaofrancisco8864 · 4 years ago

  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!

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• FR

  @FrostDirt · 4 years ago

  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.

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• TL

  @tlanohoecr · 5 years ago

  My strategies for the extra problems:<br><br>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].<br><br>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.

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• NO

  @noahnaugler7611 · 4 years ago

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

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• KA

  @kalebmark2908 · 5 years ago

  I like how this is your only video and it&apos;s an absolute banger.

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• ME

  @menturinai1387 · 4 years ago

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

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• DE

  @debblez · 4 years ago (edited)

  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)<br>For example, the ++- pattern converges to arctanh(1/3)/2 + pi/4<br>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.

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• SY

  @systemsbyvedant · 4 years ago

  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.

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• EK

  @ekut1922 · 3 years ago

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

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• AN

  @aneeshsrinivas9088 · 3 years ago

  For the infinity rearrangement try this.<br>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. <br>Add the 4th down arrow then add enough up arrows to make the total sum bigger than 4. And so on.

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• MA

  @MartinPoulter · 5 years ago

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

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• KU

  @kumozenya · 4 years ago

  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!

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