 mersenneforum.org Is (1 + 1 + 1 + 1 + 1 + 1 + ...) less than (1 + 2 + 3 + 4 + 5 + 6 + ...) ?
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xilman
Bamboozled!

"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across

5×2,053 Posts Quote:
 Originally Posted by retina Good job. Now, anyone to suggest that they are uncomparable and all the above answers are meaningless? Why are infinities so confusing?
The devil lies in the details of what is meant by the term "..."

Let Ni represent the ith term in each series considered separately.

If ... is notational convenience for Ni = \aleph N_2, i > 6 the sums are incomparable.

Last fiddled with by xilman on 2020-04-01 at 17:15   2020-04-02, 00:08   #13
Dr Sardonicus

Feb 2017
Nowhere

66478 Posts Quote:
 Originally Posted by retina Now, anyone to suggest that they are uncomparable and all the above answers are meaningless?
Objection! Meaninglessness already pointed out in this post!
Quote:
 Of course, the series as written do not converge, so simply taken at face value the question is nonsense.   2020-04-02, 00:54   #14
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

2×32×11×29 Posts Quote:
 Originally Posted by xilman If ... is notational convenience for Ni = \aleph N_2, i > 6 the sums are incomparable.
Okay, sure, technically "..." could be anything, even demons flying out of one's nose. But that doesn't really follow the intention of the question.
Quote:
 Originally Posted by Dr Sardonicus Objection! Meaninglessness already pointed out in this post!
So you gave two answers in that post?   2020-04-02, 03:19   #15
LaurV
Romulan Interpreter

Jun 2011
Thailand

100010001101112 Posts Quote:
 Originally Posted by retina Okay, sure, technically "..." could be anything, even demons flying out of one's nose.
How about monkeys from butt? (you said "everything" )   2020-04-02, 03:50   #16
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

2×32×11×29 Posts Quote:
 Originally Posted by LaurV How about monkeys from butt? (you said "everything" )
Or a super-intelligent shade of the colour blue.   2020-04-02, 06:54 #17 kriesel   "TF79LL86GIMPS96gpu17" Mar 2017 US midwest 23·557 Posts L'Hopital's rule for determining the value of ratio of infinities as they approach a limit applies, and allows comparison of sum of series B / sum of series A even though both sums are infinite for infinite number of elements. (I think my calculus professor would approve.) The video linked in the original post makes clear the 3-dot symbology ... represents an infinite series' unwritten further elements, when referring to the series written S=1+2+3+4+5+6 ... as all the natural numbers out to infinity. S=-1/12 =1+2+3+4+5+6 ... does not work for me. Adding positive elements moves the sum further positive at every increment, all infinity of them. It amounts to saying that the number line is curved, intersecting itself at +inf and -1/12 (and perhaps there are other cases the men in the video could come up with too). What is there to bend a Euclidean geometry, one-dimensional, ideal, straight line back onto itself? https://theness.com/neurologicablog/...ion-have-mass/ https://arxiv.org/ftp/arxiv/papers/1309/1309.7889.pdf To have a noninteger sum of integers is a strange result. To have a negative sum of positive integers is a strange result. To have the absolute value of the sum smaller than the absolute value of any of the elements whose absolute values are all additive is a strange result. To have the absolute value of the sum smaller than the least element or the least difference of the absolute values of the summed terms are strange results. Pretty much any of these would be considered indications of error.   2020-04-02, 11:03 #18 Nick   Dec 2012 The Netherlands 2×7×103 Posts This game does not get anywhere useful with series. A similar game with polynomials and rational functions, however, is more fruitful. For polynomials f and g with real coefficients, define f to be greater than g if f(x)>g(x) for all sufficiently large x. To put this precisely, f>g iff there exists a real number c such that, for all real x≥c, f(x)>g(x). Convince yourself of the following: For constant polynomials, this definition agrees with our existing definition of "greater than" for real numbers. For any f,g exactly one of the following holds: f>g, f=g,, g>f. For any f,g,h, if f>g and g>h then f>h. For any f,g,h, if f>g then f+h>g+h. For any f,g, if f>0 and g>0 then fg>0. In technical terms, we summarize this by saying that the set $$\mathbb{R}[X]$$ of all polynomials with real coefficients forms an ordered ring with ordering defined in the above way. Given polynomials f,g with real coefficients, if g≠0 then we can form a fraction $$\frac{f}{g}$$. This is called a rational function over the reals. For rational functions $$\frac{f_1}{g_1}$$ and $$\frac{f_2}{g_2}$$, we define $$\frac{f_1}{g_1}$$ to be greater than $$\frac{f_2}{g_2}$$ if $$f_1g_2>f_2g_1$$ (using the ordering for polynomials defined earlier). (Just as with fractions of integers, it is possible to express each rational function in more than one way, but it is not difficult to show that this definition is independent of the representations we choose.) Convince yourself of the following: For rational functions with 1 as the denominator, this definition agrees with the earlier one. For any $$f_1,g_1,f_2,g_2$$, exactly one of the following holds: $$\frac{f_1}{g_1}>\frac{f_2}{g_2}$$, $$\frac{f_1}{g_1}=\frac{f_2}{g_2}$$, $$\frac{f_2}{g_2}>\frac{f_1}{g_1}$$. For any $$f_1,g_1,f_2,g_2,f_3,g_3$$, if $$\frac{f_1}{g_1}>\frac{f_2}{g_2}$$ and $$\frac{f_2}{g_2}>\frac{f_3}{g_3}$$ then $$\frac{f_1}{g_1}>\frac{f_3}{g_3}$$. For any $$f_1,g_1,f_2,g_2,f_3,g_3$$, if $$\frac{f_1}{g_1}>\frac{f_2}{g_2}$$ then $$\frac{f_1}{g_1}+\frac{f_3}{g_3}>\frac{f_2}{g_2}+\frac{f_3}{g_3}$$. For any $$f_1,g_1,f_2,g_2$$, if $$\frac{f_1}{g_1}>\frac{0}{1}$$ and $$\frac{f_2}{g_2}>\frac{0}{1}$$ then $$\frac{f_1}{g_1}\frac{f_2}{g_2}>\frac{0}{1}$$. In technical terms, we summarize this by saying that the set $$\mathbb{R}(X)$$ of all rational functions over the reals forms an ordered field with ordering defined in the above way. But if f is just the polynomial X then, for any real number c we have f>c. So this ordered field contains a copy of the real numbers as a bounded subset! It follows that this ordered field is not complete, and therefore that the usual rules of limits and calculus no longer apply here.   2020-04-02, 14:21 #19 Dr Sardonicus   Feb 2017 Nowhere 3×5×233 Posts In addition to ordered fields which are not complete, there are complete fields which are not ordered. The complex numbers are perhaps the best known example. But there are also complete "non-Archimedian" fields. Their topology is defined by a "valuation" which takes non-negative real values, and satisfies |x*y| = |x|*|y|, and |x + y| <= Max(|x|, |y|). The basic examples are the p-adic completions of the rational numbers, where p is a prime number. The p-adic valuation of a nonzero integer n is p-e, where pe is the exact power of p dividing n. The p-adic valuation of 0 is 0. The p-adic valuation of any non-multiple of p is 1, and integers divisible by high powers of p are "small." Thus, in the 2-adic rationals (and its completion), it is perfectly correct to say that the "infinite sum" of powers of 2 1 + 2 + 4 + 8 + ... = -1   2020-04-02, 16:06 #20 kriesel   "TF79LL86GIMPS96gpu17" Mar 2017 US midwest 23·557 Posts What the heck, as long as we're throwing out any pretense of following the sort of integer and rational number math the video seems to begin premised upon, let's compare the two series in the original post mod small n. I propose mod 1 to keep things simple. The sum of any series consisting of positive integers mod 1 is zero. Then the two series are equivalent. Mod 2 could also be a lot of fun, allowing the series to switch positions among >, < and = as the number of terms increases. Schroedinger's cat now has 3 states, and oscillates between them like neutrinos; perhaps awake, asleep, and dead. 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 A mod 2 alternates between 0 and 1 at each term; even+1=odd; odd+1=even. B mod 2 alternates between 0 and 1 at each two terms. Odd + even = odd; odd + odd = even. Last fiddled with by kriesel on 2020-04-02 at 16:11   2020-04-02, 17:09   #21
wpolly

Sep 2002
Vienna, Austria

3×73 Posts Quote:
 Originally Posted by retina How come?

See Terry Tao's blogpost regarding such regularized sums.   2020-04-02, 18:13   #22
retina
Undefined

"The unspeakable one"
Jun 2006
My evil lair

2·32·11·29 Posts Quote:
 Originally Posted by wpolly See Terry Tao's blogpost regarding such regularized sums.
Alright, so there is conflicting information.

The video I linked in the first post makes use of shifting the series to the right, and ignoring terms that are zero. Are they "doing it wrong"?   Thread Tools Show Printable Version Email this Page

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