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Old 2020-04-01, 15:41   #8
kriesel
 
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Mar 2017
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Let A = 1 + 1 + 1 + ...
and A(i) = 1, defined for i a positive integer
Let B = 1 + 2 +3 + ...
and B(i) = i, defined for i a positive integer
Sum(B)=(i^2+i)/2
sum(A) = i
Sum(B)/sum(a) = (i^2+i)/2 / i = (i+1)/2
i=inf
Sum(B)/sum(A) = inf/2

Sum (Ai/Bi) = 1/i: sum is 1 + log i ~ log i
Sum (Bi/Ai) = i; sum is ( i^2 + i ) /2
limit as i->inf of Sum(Bi/Ai) / Sum (Ai/Bi) = (i^2 + i) /2 /(1+log i) ~inf^2/log(inf)/2

The value of S1 = 1-1+1-1... is not 0.5, the average of an even number of terms.
It has no single value. It's 0,1,0,1,... for sums of 0 or more terms. It's a biased square wave. It has DC amplitude 0.5 and AC amplitude 0.5.
The hand-wave in the video is literal and telling.

A series is bounded at the low end of the list, like a semi-infinite line or finite line. Terms preceding the first element are not zero, they are nonexistent and therefore undefined. Zero is a value; they have no value.

Summing two copies of S2, with one shifted, introduces an undefined term into the sum, not a zero:
S2 = 1 - 2 + 3 - 4 ...
S2a = undef + 1 - 2 + 3 - 4 ...
S2+S2a = (1-undef) -1 +1 -1 +1 ...

One could play the same trick with S1 to cancel the AC component
S1 = 1 - 1 + 1 - 1 ...
S1'= undef + 1 - 1 + 1 ...
S1+S1' = 1-undef + 0 +0 +0...
and get the result 2 * S1 = 1 so S1 = 1/2 if ignoring the undef problem and the fact S1 and S1' are different series.

And shift S1 the other way ignoring one first term (half-cycle):
S1"= - 1 + 1 - 1 ...
s1 = 1 - 1 + 1 ...
2 S1 = 0; S1 = 0
But 2 S1 = 1 from earlier, so 1 = 0.

In the S2 portion of the video, there's a sleight of hand in discarding zero terms.
The series obtained is 0 4 0 8 0 12 ...
which is reduced in the video to the series 4 8 12 ...
which seems to rescale it on the "time" axis and omit some terms, drastically changing it into a staircase series. That changes the values of initial terms from 0 4 0 8 0 12 to 4 8 12 16 20 24, and the partial sums from 24 to 84 for equal number of initial terms. Making a series out of the nonzero elements of the series 0 4 0 8 0 12 ... creates a new different series.

Going back to the originals, series A and B, apply L'Hopital's rule.
g = sumA(i) = i; g' = 1
f = sumB(i) = (i^2+i)/2; f' = i + 1/2

f(inf)/g(inf) = lim f'(i) / g'(i) = (i + 1/2) / 1 = i + 1/2 = inf.
Series B is not only greater than series A, it is INFINITELY greater. Numerator and denominator are both positive.
https://math.hmc.edu/calculus/hmc-ma...hopitals-rule/

Last fiddled with by kriesel on 2020-04-01 at 15:51
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