# The divergent series 1+1+…

In the present article I intend give plausible value for the divergent series 1+1+… according with Euler’s results and with relaxed properties of associativity, distributivity and so on.Most of the following results are rigorously nonsense, ” a facon de ecrire”

For example, we know that $1+2+3+4+5+\cdot\cdot\cdot=-\frac{1}{12}$ so

$1+1+1+1+1+1+\cdot\cdot\cdot \\ 0+1+2+3+4+5\cdot\cdot\cdot= \\1+2+3+4+5+6\cdot\cdot\cdot =-\frac{1}{12}$ (sum by columns…)

and so $s-\frac{1}{12}=-\frac{1}{12}$ that is s=0 so that zero is a plausibile value for s (actually it’s the most accreditated!).

$\alpha$ plausible value for s $\Rightarrow \alpha - n$, n positive integer , is plausible too: $1+1+1+1+1+1+\cdot\cdot\cdot= \alpha \Rightarrow 1+(1+1+1+\cdot\cdot\cdot)=\alpha, 1+s=\alpha \Rightarrow s=\alpha-1$ is plausibile too, and the same for $\alpha-2, \alpha-3\cdot\cdot\cdot,\alpha-n\cdot\cdot\cdot$

$\alpha$ plausible $\Rightarrow \cfrac{\alpha}{n}$, n positive integer, is plausible too:

$\alpha=1+1+1+1+1+1+\cdot\cdot\cdot\\ \quad =1+0+1+0+1+0+1+\cdot\cdot\cdot+\\ 0+1+0+1+0+1+0+1 \cdot\cdot\cdot = s+s$

that is $2s=\alpha\Rightarrow \frac{a}{2}$ is plausible too. And similarly for n integer positive.

$\alpha$ plausible $\Rightarrow n\alpha$ plausible too:

$3\alpha=\alpha+\alpha+\alpha =(1+1+\cdot\cdot\cdot)+(1+1+\cdot\cdot\cdot)+(1+1+\cdot\cdot\cdot)=\\1+0+0+1+0+0+1\cdot\cdot\cdot+\\0+1+0+0+1+0+0+\cdot\cdot\cdot\\0+0+1+0+0+1+0\cdot\cdot\cdot\\1+1+1+1+1+1+1+\cdot\cdot\cdot\\\Rightarrow 3\alpha$

is plausible too, and the same for all n.

Therefore starting with the initial value 0 we know that all negative rational numbers are plausible.

Let’s recall for example the value $-\frac{1}{2}$:

starting from the well known fact $1-1+1-1+1-\cdot\cdot\cdot=\cfrac{1}{2}$ we have:

$\frac{1}{2}=1-1+1-1+1-1+\cdot\cdot\cdot=\\ \quad\quad\quad\quad 1+1+1+1+1+1+1+1+1\cdot\cdot\cdot+\\ -2(1+\quad+1+\quad 1+\quad +1+\cdot\cdot\cdot)=s-2s=-s \Rightarrow s=-\frac{1}{2}$.

But we haven’t only rational negative numbers:

$-\frac {1}{12}=1+2+3+4+5+\cdot\cdot\cdot=\\ 1+1+1+1+1+1+\cdot\cdot\cdot+\\ 0+1+1+1+1+1+\cdot\cdot\cdot+\\ 0+0+1+1+1+1+\cdot\cdot\cdot+\\0+0+0+1+1+1+\cdot\cdot\cdot$      (sum the numbers in column !) $-\frac{1}{12}=s+s+s+s+s+\cdot\cdot\cdot= s(1+1+1+1\cdot\cdot\cdot)=s^{2}$ and then $s=\pm\sqrt{-\frac{1}{12}}$ is plausible too: a complex imaginary number!

$\cfrac {1}{s}=\cfrac {1}{1+1+\cdot\cdot\cdot}=\cfrac{1}{1+(1+1+\cdot\cdot\cdot)}=\cfrac {1}{1+s}=\cfrac {1(1-s)}{(1+s)(1-s)}=\cfrac {1-s}{1-s^{2}}=\cfrac {1-s}{1-(-\frac{1}{12})}\Rightarrow \cfrac{1}{s}=\frac {12}{13}(1-s)$ and solving $s=\frac{3\pm\sqrt{-30}}{6}$ are plausible values too.

This calculation gives also one continued fraction for s:

$\cfrac {1}{s}=\cfrac {12}{13}(1-s) \Rightarrow s=\cfrac{\frac{13}{12}}{1-s}=\cfrac{1}{\frac{12}{13}-\frac{12}{13}s}=\cfrac{1}{\frac{12}{13}-\frac{12}{13}(\cfrac{1}{\frac{12}{13}-\frac{12}{13}s})}=\cfrac{1}{\frac{12}{13}-\cfrac{1}{1-s}}=\cfrac{1}{\frac{12}{13}+\cfrac{1}{-1+s}}\Rightarrow s=[0;\frac{12}{13};-1;\frac{12}{13};-1;\cdot\cdot\cdot]$

$e^{s}=1+\cfrac {s}{1!}+\cfrac {s^{2}}{2!}+\cfrac{s^{3}}{3!}+\cdot\cdot\cdot =\\ (1+\cfrac {s^{2}}{2!}+\cfrac {s^{4}}{4!}+\cdot\cdot\cdot)+s(\cfrac{1}{1!}+\cfrac{s^{2}}{3!}+\cdot\cdot\cdot)=\\ (1+\cfrac{(-\frac{1}{12})}{2!}+\cfrac{(-\frac{1}{12})^{2}}{4!}+\cdot\cdot\cdot)+s(\cfrac{1}{1!}+\cfrac{(-\frac{1}{12})}{3!}+\cfrac{{(-\frac{1}{12})}^{2}}{5!}+\cdot\cdot\cdot)=\\(1-\cfrac{\sqrt{\frac{1}{12}}^{2}}{2!}+\cfrac{(\sqrt{\frac{1}{12}})^{4}}{4!}+\cdot\cdot\cdot)+\sqrt{12}s(\cfrac{\sqrt{\frac{1}{12}}}{1!}-\cfrac{(\sqrt{\frac{1}{12}})^{3}}{3!}+\cdot\cdot\cdot)=\\ cos\sqrt{\frac{1}{12}}+\sqrt{12}sen \sqrt{\frac{1}{12}}s\\\Rightarrow e^{1+1+\cdot\cdot\cdot}=cos\sqrt{\frac{1}{12}}+\sqrt{12}sen\sqrt{\frac{1}{12}}(1+1+\cdot\cdot\cdot)$

$log(s)=log(1+1+1+\cdot\cdot\cdot)=log(1+(1+1+\cdot\cdot\cdot))=log(1+s)=\\s-\frac{s^2}{2}+\frac{s^3}{3}-\frac{s^4}{4}+\cdot\cdot\cdot= (-\frac{s^2}{2}-\frac{s^4}{4}-\cdot\cdot\cdot)+s(1+\frac{s^2}{3}+\frac{s^4}{4}+\cdot\cdot\cdot)=\\ (\frac{1}{12})(\frac{(\frac{1}{12})}{1}-\frac{(\frac{1}{12})^2}{2}+\frac{(\frac{1}{12})^3}{3}-\cdot\cdot\cdot)+s(1-\frac{(\frac{1}{12})}{3}+\frac{(\frac{1}{12})^2}{5}+\cdot\cdot\cdot)=\\ \frac{1}{2}log(1+\frac{1}{12})+\sqrt{12}((\sqrt{\frac{1}{12}})-\frac{(\sqrt{\frac{1}{12})}^3}{3}+\frac{\sqrt{(\frac{1}{12})}^5}{5}+\cdot\cdot\cdot)s=\\ log\sqrt{\frac{13}{12}}+\sqrt{12}arctg\sqrt{\frac{1}{12}}s\\ \Rightarrow log(1+1+\cdot\cdot\cdot)=log\sqrt{\frac{13}{12}}+\sqrt{12}arctg(\sqrt{\frac{1}{12}}(1+1+\cdot\cdot\cdot)$

$s^{s}=s^{1+1+\cdot\cdot\cdot}=sss\cdot\cdot\cdot =(ss)(ss)(s s)\cdot\cdot\cdot=(-\frac{1}{12})(-\frac{1}{12})(\frac{1}{12})\cdot\cdot\cdot=(-\frac{1}{12})^{s}$

$\Rightarrow (1+1+\cdot\cdot\cdot)^{1+1+\cdot\cdot\cdot}=(-\frac{1}{12})^{1+1+\cdot\cdot\cdot}$

References:

Leonard Euler: Institutiones calculi differentialis

G.H.Hardy: Divergent Series, At the Clarendon Press 1949