4mo ago

Witchcraft

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tbf, the 2nd sum is exactly the first one just multiplied by 1/2. though i get that the progression is natural, even, and odd.
the last one is definitely ~~odd~~ puzzling, but i cannot intuitively get the first one. how does summing the inverse of triangular number equal 2?
- I believe starting with 1/1 which equals 1, you are then adding infinitely (fractions) on top of the 1. So 1, then 1 1/2, ect, so the next full integer to be hit (infinitely down the line) would be 2.
  I don't do high level math so I hope this explanation is correct or intelligible, this is just how I understand it intuitively
  
  But the first few values are:
  1 + 1/3 + 1/6 + 1/10 + 1/15 + 1/21 + 1/28...
  I really don't see any pattern there showing why it converges to 2 exactly
  Edit:
  After thinking some more, you could write the sum as:
  (Sum from n=1 to infinity of): 2/(n * (n + 1))
  That sum is smaller than the sum of:
  2 * (1/n²) which converges to π²/3
  So I can see why it converges, just not where to.
  
  I didn't see the pattern either and had to look it up. Apparently, you can rewrite 1 + 1/(1+2) + 1/(1+2+3)+... as 2(1 - 1/2 + 1/2 - 1/3 +...+1/n - 1/(n + 1)) = 2(1 - 1/(n + 1))
  From there, the limit of 2 is obvious, but I guess you just have to build up intuition with infinite sums to see the reformulation.
  
  So the amount you are adding is getting smaller with each iteration, 1/4 is smaller than 1/2, however you are still adding 1/4 on top of the 1/2, and those two are combined, closer to "1" than either of them independently correct? (1/2 +1/4 =1/3. 1/3>1/2)
  So if the number gets bigger forever than at some point it will eventually hit "1", since we already started with "1" the next "1" will be "2"
  I hope I'm explaining it well enough, it's similar to how 3.33(repeating)x3...=10 (though technically for different reasons)
  
  Those add to 1.75, just keep adding (infinitely)
- It's not saying the proof is obvious, just that the result is plausible on its face.
- 1/(n * (n+1)) = 1/n - 1/(n+1)
  1/(1 * 2) + 1/(2 * 3) + 1/(3 * 4) + ... = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... = 1
  
  Written a bit more explicitly (although I kinda handwaved away the final term--the point is that you end up with one unpaired term which goes to zero)
  
  edit: I was honestly confused about how exactly this related to the question, but seeing the comment from @yetAnotherUser@discuss.tchncs.de (not visible from Hexbear) which showed that the first sum in the image is equivalent to
  the sum from n = 1 to ∞ of 2/(n (n + 1))
  made things clear (just take the above, put 2 in the numerator, and you get a result of 2)

26 comments