I like this series division method. Still wonder about the zeroes.

[Kharakov]

I can't figure out what that's getting at.

Before I explained the "infinite series division" method I was using, I mentioned that there is a set of specific sequences of non decreasing natural numbers that you could use to get zero sums (that I call "harmonic 0s") when dividing the harmonic series by the series formed by the non-decreasing natural number sequences.

Here's an example of a harmonic 0: (1 + 1/2 + 1/3...) / (1+1+1...) = 1 -1/2 -1/6 -1/12.... = 0

I asked if there were any other non-decreasing natural number sequences, NOT related to the simplex number sequences (1+1+1...)^n that generate harmonic 0s.

2+2+2... or 1/2+ 1/2 +1/2... give you harmonic 0s as well, as do other a*(1+1+1...)^n. The question excludes zeroes by series comparison between the harmonic series and the non decreasing natural number sequence series (because the one series "approaches" infinity much faster). These 0s are based on invalid reasoning about division by infinite series.

Division method:

Show

Simplex convolution series division (Inverse Cauchy Product????), results are r_whatever:
\(\begin{tabular}{ l | c c c c c } & a & b & c & d & e ... \\ \div & A & B & C & D & E ... \\ \hline r_0 & r_0 A & r_0 B & r_0 C & r_0 D & r_0 E \\ r_1 & r_1 A & r_1 B & r_1 C & r_1 D & \\ r_2 & r_2 A & r_2 B & r_2 C & & \\ r_3 & r_3 A & r_3 B & & & \\ r_4 & r_4 A & & & & \\ \end{tabular} r_0=\frac{a}{A} r_1=\frac{b-B r_0}{A} r_2=\frac{c- C r_0 - B r_1}{A} .... \)

results of division are in left hand columns:

\( \begin{tabular}{ r | r r r r r r} & 1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & ... \\ \div & 1 & 1 & 1 & 1 & 1& ... \\ \hline 1 & 1 & 1 & 1 & 1 & 1 \\ -\frac {1}{2} & -\frac {1}{2} & -\frac {1}{2} & -\frac {1}{2} & -\frac {1}{2} & \\ -\frac {1}{6} & -\frac {1}{6} & -\frac {1}{6} &-\frac {1}{6} & & \\ -\frac {1}{12} & -\frac {1}{12} & -\frac {1}{12} & & & \\ -\frac {1}{20} & -\frac {1}{20} & & & & \\ \end{tabular} \) ................ \(\begin{tabular}{ r | r r r r r r } & 1 & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5}& ... \\ \div & 1 & 2 & 3 & 4 & 5 & ... \\ \hline 1 & 1 & 2 & 3 & 4 & 5 \\ -\frac {3}{2} & -\frac {3}{2} & -3 & -\frac {9}{2} & -6 & \\ +\frac {1}{3} & \frac {1}{3} & \frac {2}{3} & 1 & & \\ +\frac {1}{12} & \frac {1}{12} & \frac {1}{6} & & & \\ +\frac {1}{30} & \frac {1}{30} & & & & \\ \end{tabular} \)

So it's 1-1 for the first division (by 1+1+1+...) and 1/2-1/2 for the second (by 1+2+3...).

Here are a couple of the series:

Show

Harmonic/ 1+1+1+1... = ..................................... 1 - (1/2 + 1/6 + 1/12 + 1/20....) series is reciprocal triangulars/2
Harmonic/ 1+2+3+4... = 1-3/2 +... = ............... -1/2 + (1/3 +1/12 + 1/30 +1/60...)=0 series is reciprocal tetras/3
Harmonic/ 1+3+6+10... = 1-5/2+11/6 -............= 1/3 - (1/4 + 1/20+ 1/60 +1/140...)= 0 series is reciprocal pentatopes/4
Harmonic/ 1+4+10+20...=1-7/2+13/3-25/12...= -1/4 + (1/5 + 1/30+ 1/105 +1/280...) = 0 series is reciprocal 6 simplex/5

The square convolution I mention for (1+1+1...)^2 isn't technically correct- I don't know if this is something that Cauchy proved, which is why the simplex product is called the Cauchy product, despite the fact that this leads people away from valuable information: that the product is directly related to simplexes and simplex number sequences.

(1+1+1...)^2 = does = 1+2+3+4.... because you're multiplying the first term in the series by (1+1+1+1...), then the second term, then the third, so you have:

\(\begin{tabular}{ l | c c c c c } & 1 & 1 & 1 & 1 & 1 ... \\ \times & 1 & 1 & 1 & 1 & 1 ... \\ \hline & 1 & 1 & 1 & 1 & 1... \\ + & & 1 & 1 & 1 & 1 ...\\ +& & & 1 & 1 & 1 ..\\ +& & & & 1 & 1... \\ +& & & & & 1 ...\\ \hline & 1 & 2 & 3 & 4 & 5... \\ \end{tabular} \) |||||||||| \( \begin{tabular}{ l | c c c c c } &1 & 1 & 1 & 1 & 1 ... \\ \times & 2 & 1 & 1 & 1 & 1 ... \\ \hline & 2 & 2 & 2 & 2 & 2... \\ + & & 1 & 1 & 1 & 1 ...\\ +& & & 1 & 1 & 1 ..\\ +& & & & 1 & 1... \\ +& & & & & 1 ...\\ \hline & 2 & 3 & 4 & 5 & 6... \\ \end{tabular} \)|||||||||| \( \begin{tabular}{ l | c c c c c } &2 & 1 & 1 & 1 & 1 ... \\ \times & 1 & 1 & 1 & 1 & 1 ... \\ \hline & 2 & 1 & 1 & 1 & 1... \\ + & &2 & 1 & 1 & 1 ...\\ +& & & 2 & 1 & 1 ..\\ +& & & & 2 & 1... \\ +& & & & & 2 ...\\ \hline & 2 & 3 & 4 & 5 & 6... \\ \end{tabular} \) ||||||| \( \begin{tabular}{ l | r r r r r r} &1 & \frac{1}{2} & \frac{3}{8} & \frac{5}{16} &\frac{35}{128} & ... \\ \times &1 & \frac{1}{2} & \frac{3}{8} & \frac{5}{16} &\frac{35}{128} & ... \\ \hline & 1 & \frac{1}{2} & \frac{3}{8} & \frac{5}{16} &\frac{35}{128} & ... \\ + & &\frac{1}{2} & \frac{1}{4} & \frac{3}{16} &\frac{5}{32} & ... \\ +& & & \frac{3}{8} & \frac{3}{16} &\frac{9}{64} & ... \\ +& & & & \frac{5}{16} &\frac{5}{32} & ... \\ +& & & & &\frac{35}{128} & ... \\ \hline & 1 & 1 & 1 & 1 & 1 & ... \\ \end{tabular} \)


Some properties of the specific harmonic 0s:

Dividing a smaller harmonic 0 by a larger harmonic 0 results in 0.

Dividing a larger harmonic 0 by a smaller harmonic 0 results in a specific infinity related to relative position of harmonic 0s.

First four harmonic 0s:

Show

Harmonic/ 1+1+1+1... = 1 - (1/2 + 1/6 + 1/12 + 1/20....) = Harmonic / (1+1+1...) = 0 = H0_1
Harmonic/ 1+2+3+4... = -1/2 + (1/3 +1/12 + 1/30 +1/60...)= Harmonic / (1+1+1...)^2 = 0 = H0_2
Harmonic/ 1+3+6+10... = 1/3 - (1/4 + 1/20+ 1/60 +1/140...)= Harmonic / (1+1+1...)^3 = 0 =H0_3
Harmonic/ 1+4+10+20...= -1/4 + (1/5 + 1/30+ 1/105 +1/280...) = Harmonic / (1+1+1...)^4 = 0 = H0_4

You have to keep series positions, like before, so the collapsed series about do not work... use the full series expansions:

H0_1= 1 - (1/2 + 1/6 + 1/12 + 1/20....)
H0_2= 1-3/2 + (1/3 +1/12 + 1/30 +1/60...)
H0_3= 1-5/2+11/6- (1/4 + 1/20+ 1/60 +1/140...)
H0_4= 1-7/2+13/3-25/12 + (1/5 + 1/30+ 1/105 +1/280...)

H0_1/ H0_2 = 0
H0_2/ H0_1 = 1+1+1....
H0_2/ H0_4 = 0

H0_4/H0_2 = (1+1+1...)^2 =1+2+3+4... (the division is below)

In the following division, the diagonals sum to the original series (the numerator), like mentioned at the top of this post. The results are in the first column. The second column is the first column times 1, which is the first value in the denominator (divisor).
\( \begin{tabular}{ r | r r r r r r} & 1 &-\frac{3}{2} & \frac{1}{3} & \frac{1}{12}& \frac{1}{30}& ... \\ \div & 1 & - \frac{7}{2} & \frac{13}{3} & -\frac{25}{12} & \frac{1}{5} & ... \\ \hline 1 & 1 & - \frac{7}{2} & \frac{13}{3} & -\frac{25}{12} & \frac{1}{5} & ... \\ 2 & 2 & -7 & \frac {26}{3} & - \frac{25}{6} & \\ 3 & 3 & -\frac {21}{2} & 13 & & \\ 4 & 4 & -14 & & & \\ 5 & 5 & & & & \\ \end{tabular} \)


Some questions:

It doesn't seem that Ramanujan summation works, if you are strict.

What about the physical case with 1+2+3+4... = -1/12? Are there other numbers associated with the physic's properties associated with that series?

Last, and first question: Are there any non simplex (not a*(1+1+1...)^n) series that result in harmonic 0s when you divide the harmonic series by the non simplex related series?

A voice in my head said "Ho Ho", and I thought "Ho" and laughed. Am I sane?

\( \begin{tabular}{ l | c c c c c } &1 & 0 & 0 & 0 & 0 ... \\ \div & 1 & -1 &0 &0 &0 ... \\ \hline 1& 1 & -1 & 0 & 0 & 0... \\ 1& 1 &-1 & 0 & 0 & 0 ...\\ 1& 1& -1&0 & 0 &0 ..\\ 1& 1& -1&0 & 0 & 0... \\ \end{tabular} \) (1-1)^2 = 1 -2 +1\( \begin{tabular}{ r | r r r r r } &1 & -2 & 1 \times & 1 & -1 & \hline 1 & 1 & -1 & 0 -1 & -1 & 1 & & 0 & \end{tabular} \).... \( \begin{tabular}{ r | r r r r r } &1 & 0 & 0 & 0 & 0 ... \\ \div & 1 & -2 &1 &0 &0 ... \\ \hline 1& 1 & -2 & 1 & 0 & 0... \\ 2& 2 &-4 & 2 & 0 & 0 ...\\ 3 & 3& -6&3 & 0 &0 ..\\ 4& 4& -8&4 & 0 & 0... \\ 5& 5& -10&4 & 0 & 0... \\ \end{tabular} \)


Generates various infinite series. (1+1+1...) = 1/(1-1). (1+1+1)^2 = 1/(1-1)^2 = 1/(1 -2 +1) = 1+2+3+4....

1/ (1-2+3-4...) =1 +2 +1 =4... so the summation of that series looks good. The Grandi series looks good as well. 1/(1+2+3+4...) =1 -2 +1, so while analytic continuation gives you -1/12... it doesn't look like the Ramanujan summation keeps things in the proper columns.