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-   -   Is (1 + 1 + 1 + 1 + 1 + 1 + ...) less than (1 + 2 + 3 + 4 + 5 + 6 + ...) ? (https://www.mersenneforum.org/showthread.php?t=25399)

 retina 2020-03-26 17:57

Is (1 + 1 + 1 + 1 + 1 + 1 + ...) less than (1 + 2 + 3 + 4 + 5 + 6 + ...) ?

Is (1 + 1 + 1 + 1 + 1 + 1 + ...) less than (1 + 2 + 3 + 4 + 5 + 6 + ...) ?

It would seem not.

Because 1 + 2 + 3 + 4 + 5 + 6 + ... = -1/12

[size=1]I'm not sure if making a comparison like the above is valid. Perhaps I misunderstand comparisons of infinite sequences?[/size]

 Dr Sardonicus 2020-03-26 19:06

[QUOTE=retina;540948]Is (1 + 1 + 1 + 1 + 1 + 1 + ...) less than (1 + 2 + 3 + 4 + 5 + 6 + ...) ?

It would seem not.

Because 1 + 2 + 3 + 4 + 5 + 6 + ... = -1/12

[size=1]I'm not sure if making a comparison like the above is valid. Perhaps I misunderstand comparisons of infinite sequences?[/size][/QUOTE]Of course, the series as written do not converge, so simply taken at face value the question is nonsense.

One can [i]assign[/i] values to the sums by misusing formulas. The value -1/12 assigned to 1 + 2 + 3 + ... is a case in point. We have

$$\zeta(s)\;=\;\sum_{n=1}^{\infty}n^{-s}\text{, when }\Re(s)\;>\;1$$

The zeta function is defined at s = 0 and at s = -1 (though is [i]not[/i] given by the above series at those points), taking the values -1/2 and -1/12, respectively.

Cheerfully disregarding the invalidity of the formula, mindlessly plugging in s = 0 gives

1 + 1 + 1 + ... ad infinitum = -1/2

and plugging s = -1 into the formula gives

1 + 2 + 3 + 4 + 5 + 6 + ... ad infinitum = -1/12.

And -1/2 < -1/12.

:-D

 retina 2020-03-26 19:27

[QUOTE=Dr Sardonicus;540956]Of course, the series as written do not converge, so simply taken at face value the question is nonsense.

One can [i]assign[/i] values to the sums by misusing formulas. The value -1/12 assigned to 1 + 2 + 3 + ... is a case in point. We have

$$\zeta(s)\;=\;\sum_{n=1}^{\infty}n^{-s}\text{, when }\Re(s)\;>\;1$$

The zeta function is defined at s = 0 and at s = -1 (though is [i]not[/i] given by the above series at those points), taking the values -1/2 and -1/12, respectively.

Cheerfully disregarding the invalidity of the formula, mindlessly plugging in s = 0 gives

1 + 1 + 1 + ... ad infinitum = -1/2

and plugging s = -1 into the formula gives

1 + 2 + 3 + 4 + 5 + 6 + ... ad infinitum = -1/12.

And -1/2 < -1/12.

:-D[/QUOTE]Very good.

If we assign:

A = 1 + 1 + 1 + 1 + 1 + 1 + ...
B = 1 + 2 + 3 + 4 + 5 + 6 + ...

Then B - A = B, since pairwise subtraction gives 0 + 1 + 2 + 3 + ... = 1 + 2 + 3 + ...

Therefore A = 1 + 1 + 1 + 1 + 1 + 1 + ... = 0

 Dr Sardonicus 2020-03-26 19:44

[QUOTE=retina;540957]Very good.

If we assign:

A = 1 + 1 + 1 + 1 + 1 + 1 + ...
B = 1 + 2 + 3 + 4 + 5 + 6 + ...

Then B - A = B, since pairwise subtraction gives 0 + 1 + 2 + 3 + ... = 1 + 2 + 3 + ...

Therefore A = 1 + 1 + 1 + 1 + 1 + 1 + ... = 0[/QUOTE]
OTOH, we have

1 + A = A

:missingteeth:

 retina 2020-03-26 19:50

[QUOTE=Dr Sardonicus;540962]OTOH, we have

1 + A = A

:missingteeth:[/QUOTE]And B - A - 1 = B

 wpolly 2020-03-27 13:09

[QUOTE=retina;540957]Very good.

If we assign:

A = 1 + 1 + 1 + 1 + 1 + 1 + ...
B = 1 + 2 + 3 + 4 + 5 + 6 + ...

Then B - A = B, since pairwise subtraction gives 0 + 1 + 2 + 3 + ... = 1 + 2 + 3 + ...

Therefore A = 1 + 1 + 1 + 1 + 1 + 1 + ... = 0[/QUOTE]

Sadly, in the regime of regularized sums, we no longer have 0 + 1 + 2 + 3 + ... = 1 + 2 + 3 + ...

 retina 2020-03-27 13:31

[QUOTE=wpolly;541040]Sadly, in the regime of regularized sums, we no longer have 0 + 1 + 2 + 3 + ... = 1 + 2 + 3 + ...[/QUOTE]How come?

 kriesel 2020-04-01 15:41

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.
[URL]https://math.hmc.edu/calculus/hmc-mathematics-calculus-online-tutorials/single-variable-calculus/lhopitals-rule/[/URL]

 retina 2020-04-01 15:48

[QUOTE=kriesel;541489]<snip>
Series B is not only greater than series A, it is INFINITELY greater. Numerator and denominator are both positive.[/QUOTE]Good job.

A > B, and
B > A

Anyone want to speculate now that A = B?

 VBCurtis 2020-04-01 16:01

[QUOTE=retina;541490]Anyone want to speculate now that A = B?[/QUOTE]

Sure! They look like exactly the same size of infinity to me.

 retina 2020-04-01 16:07

[QUOTE=VBCurtis;541493]Sure! They look like exactly the same size of infinity to me.[/QUOTE]Good job.

Now, anyone to suggest that they are uncomparable and all the above answers are meaningless?

[size=1]Why are infinities so confusing?[/size]

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