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

retina · 2020-03-26, 17:57

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
https://www.youtube.com/watch?v=w-I6XTVZXww

I'm not sure if making a comparison like the above is valid. Perhaps I misunderstand comparisons of infinite sequences?

Dr Sardonicus · 2020-03-26, 19:06

Quote:

Originally Posted by retina

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
https://www.youtube.com/watch?v=w-I6XTVZXww

I'm not sure if making a comparison like the above is valid. Perhaps I misunderstand comparisons of infinite sequences?

Of course, the series as written do not converge, so simply taken at face value the question is nonsense.

One can assign 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 not 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:

Originally Posted by Dr Sardonicus

Of course, the series as written do not converge, so simply taken at face value the question is nonsense.

One can assign 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 not 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

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:

Originally Posted by retina

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

OTOH, we have

1 + A = A

retina · 2020-03-26, 19:50

Quote:

Originally Posted by Dr Sardonicus

OTOH, we have

1 + A = A

And B - A - 1 = B

wpolly · 2020-03-27, 13:09

Quote:

Originally Posted by retina

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

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:

Originally Posted by wpolly

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

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.
https://math.hmc.edu/calculus/hmc-ma...hopitals-rule/

retina · 2020-04-01, 15:48

Quote:

Originally Posted by kriesel

<snip>
Series B is not only greater than series A, it is INFINITELY greater. Numerator and denominator are both positive.

Good job.

Now we have two answers.

A > B, and
B > A

Anyone want to speculate now that A = B?

VBCurtis · 2020-04-01, 16:01

Quote:

Originally Posted by retina

Anyone want to speculate now that A = B?

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

retina · 2020-04-01, 16:07

Quote:

Originally Posted by VBCurtis

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

Good job.

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

Why are infinities so confusing?

2020-03-26, 17:57	#1
retina Undefined "The unspeakable one" Jun 2006 My evil lair 5·7·191 Posts	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 https://www.youtube.com/watch?v=w-I6XTVZXww I'm not sure if making a comparison like the above is valid. Perhaps I misunderstand comparisons of infinite sequences?

2020-04-01, 15:41	#8
kriesel "TF79LL86GIMPS96gpu17" Mar 2017 US midwest 16306₈ Posts	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