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
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)
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)
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
oddpuzzling, 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)
deleted by creator
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)