1+2+3+4+… and −⅛

Question

Is there some deeper meaning to the following derivation (or rather one-parameter family of derivations) associating the divergent series $1+2+3+4+…$ with the value $-\frac 1 8$ (as opposed to the value $-\frac 1 {12}$ obtained by zeta-function regularization)? Or is it just a curiosity?

Formally put $x=1+2+3+4+\dotsb$. Writing $$x-1=(2+3+4)+(5+6+7)+\dotsb=9+18+27+\dotsb=9x$$ we get $8x=-1$. Or, writing $$x-1-2=(3+4+5+6+7)+(8+9+10+11+12)+\dotsb=25+50+75+\dotsb=25x$$ we get $24x=-3$. Or, writing $$x-1-2-3=(4+\dotsb+10)+(11+\dotsb+17)+\dotsb=49+98+147+\dotsb=49x$$ we get $48x=-6$. Etc.

It is not surprising that values other than $-\frac 1 {12}$ can be obtained as “values” of this divergent series. What surprises me is that all of these methods of grouping terms give the same answer. It makes me wonder whether there is a larger story here.

Answer 1

Yes there is, in $p$-adics.

You are probably familiar with the relation

$$8T(n)+1=(2n+1)^2.$$

Now for any $p$ except $2$ (which has to be excluded because of the non-unit coefficients in the above relation) we can identify a subsequence of whole numbers $n$ such that the squared quantity on the right converges $p$-adically to zero. Then $T(n)$ follows suit, converging to $-1/8$.

For instance, we may put in $p=3$, in which case $-1/8$ is rendered as the $3$-adic integer $\overline{01}$. Then using base $3$ arithmetic we develop the sequence \begin{align*} \newcommand\pdots{{\ldots}} & T(\pdots1)=\pdots01 \\ & T(\pdots11)=\pdots0101 \\ & T(\pdots111)=\pdots010101 \end{align*}

where the base on the left side is set up to converge to $-1/2$ (for which the corresponding square is zero) and the right side then converges quadratically to $-1/8$ in the subsequence. The quadratic convergence to $-1/8$ is unique to that target value bevause of the critical value of the corresponding square.

This quadratic convergence leads to ordinary integer triangular numbers being "attracted" to the $3$-adic representation $\overline{01}=-1/8$. Below is a table wherein the columns represent possible two-digit endings for any $81$ consecutive triangular numbers in base $3$; the rows represent possible values for the preceding two digits and the entries describe how many triangular numbers out of the block of $81$ will end with the resulting four-digit pattern. Combinations not shown correspond to no triangular numbers represented in base 3.

The table shows that there us an excess of triangular numbers ending with $...01$ in base $3$ ($27/81$ versus $18/81$ for the other possible two-digit endings) and among those triangular numbers ending with $...01$, the four-digit ending $...0101$ is further overrepresented. The overrepresentation of patterns matching $\overline{01}$ grows when we cobsider longer strings of terminal digits in base $3$. Similar attraction is seen to $\overline{03}$ in $5$-adics, $\overline{06}$ in $7$-adics, and so on.

Triangular numbers are not the only ones with this property. We can set up similar patterns with any polygonal number pattern, for instance octagonal numbers quadratically converging to $-1/3$ and favoring that fraction in $p$-adic subsequences where the prime $p\ne3$.

Answer 2

You are not using all partial sums, but only a restricted choice of them. So you are not really looking at the limit of all partial sums. An analogue would be deciding to compute $1 - 1 + 1 - 1 + 1 - 1 + \cdots$ by only focusing on the even-indexed partial sums (every other partial sum).

For your general calculation, assuming $x$ makes sense, pick $N \geq 1$ and subtract $\sum_{j=1}^N j$ from $x$ and then look only at the partial sums of what is left using multiples of $2N+1$: the $(2N+1)$-th partial sum, the $(2N+1)2$-th partial sum, and so on. Collecting terms $2N+1$ at a time, centered at the multiples of $2N+1$ (with $N$ terms preceding and $N$ terms succeeding those multiples), we get $$ x - \sum_{j=1}^N j = \sum_{k \geq 1} \left(\sum_{i=-N}^N ((2N+1)k + i)\right) = \sum_{k \geq 1} (2N+1)^2k = (2N+1)^2x, $$ so $$ x = (2N+1)^2x + \frac{N(N+1)}{2} = (4N(N+1) + 1)x + \frac{N(N+1)}{2}. $$ Subtract $x$ from both sides and solve for $x$: $$ x = -\frac{N(N+1)/2}{4N(N+1)} = -\frac{1}{8}. $$

The series $1 + 2 + 3 + \cdots$ is not convergent even in the $p$-adics: for no prime $p$ are the terms of the partial sums tending $p$-adically to $0$.