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 [url]https://www.youtube.com/watch?v=wI6XTVZXww[/url] [size=1]I'm not sure if making a comparison like the above is valid. Perhaps I misunderstand comparisons of infinite sequences?[/size] 
[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 [url]https://www.youtube.com/watch?v=wI6XTVZXww[/url] [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 [tex]\zeta(s)\;=\;\sum_{n=1}^{\infty}n^{s}\text{, when }\Re(s)\;>\;1[/tex] 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=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 [tex]\zeta(s)\;=\;\sum_{n=1}^{\infty}n^{s}\text{, when }\Re(s)\;>\;1[/tex] 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 
[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: 
[QUOTE=Dr Sardonicus;540962]OTOH, we have
1 + A = A :missingteeth:[/QUOTE]And B  A  1 = B 
[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 + ... 
[QUOTE=wpolly;541040]Sadly, in the regime of regularized sums, we no longer have 0 + 1 + 2 + 3 + ... = 1 + 2 + 3 + ...[/QUOTE]How come?

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 = 11+11... 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 handwave in the video is literal and telling. A series is bounded at the low end of the list, like a semiinfinite 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 = (1undef) 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' = 1undef + 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 (halfcycle): 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/hmcmathematicscalculusonlinetutorials/singlevariablecalculus/lhopitalsrule/[/URL] 
[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. Now we have two answers. A > B, and B > A Anyone want to speculate now that A = B? 
[QUOTE=retina;541490]Anyone want to speculate now that A = B?[/QUOTE]
Sure! They look like exactly the same size of infinity to me. 
[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] 
All times are UTC. The time now is 16:33. 
Powered by vBulletin® Version 3.8.11
Copyright ©2000  2020, Jelsoft Enterprises Ltd.