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Witchcraft (mander.xyz)

submitted 2 months ago* (last edited 2 months ago) by [email protected] to c/[email protected]

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[–] [email protected] 41 points 2 months ago (7 children)

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?

[–] BugleFingers 11 points 2 months ago (5 children)

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

[–] [email protected] 12 points 2 months ago* (last edited 2 months ago) (4 children)

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/n^2^) which converges to π^2^/3

So I can see why it converges, just not where to.

[–] Bender_on_Fire 6 points 2 months ago

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.

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