20200401, 17:07  #12  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
5×2,053 Posts 
Quote:
Let N_{i} represent the i^{th} term in each series considered separately. If ... is notational convenience for N_{i} = \aleph _{N_2}, i > 6 the sums are incomparable. Last fiddled with by xilman on 20200401 at 17:15 

20200402, 00:08  #13  
Feb 2017
Nowhere
6647_{8} Posts 
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20200402, 00:54  #14  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2×3^{2}×11×29 Posts 
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20200402, 03:19  #15  
Romulan Interpreter
Jun 2011
Thailand
10001000110111_{2} Posts 
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20200402, 03:50  #16  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2×3^{2}×11×29 Posts 
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20200402, 06:54  #17 
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
2^{3}·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 3dot 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, onedimensional, ideal, straight line back onto itself? https://theness.com/neurologicablog/...ionhavemass/ 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. 
20200402, 11:03  #18 
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 xβ₯c, f(x)>g(x). Convince yourself of the following:
Given polynomials f,g with real coefficients, if gβ 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:
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. 
20200402, 14:21  #19 
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 "nonArchimedian" fields. Their topology is defined by a "valuation" which takes nonnegative real values, and satisfies x*y = x*y, and x + y <= Max(x, y). The basic examples are the padic completions of the rational numbers, where p is a prime number. The padic valuation of a nonzero integer n is p^{e}, where p^{e} is the exact power of p dividing n. The padic valuation of 0 is 0. The padic valuation of any nonmultiple of p is 1, and integers divisible by high powers of p are "small." Thus, in the 2adic rationals (and its completion), it is perfectly correct to say that the "infinite sum" of powers of 2 1 + 2 + 4 + 8 + ... = 1 
20200402, 16:06  #20 
"TF79LL86GIMPS96gpu17"
Mar 2017
US midwest
2^{3}·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 20200402 at 16:11 
20200402, 18:13  #22  
Undefined
"The unspeakable one"
Jun 2006
My evil lair
2·3^{2}·11·29 Posts 
Quote:
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"? 
