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.

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    $\begingroup$ Silly comment. Let $z = 1 - 2 + 3 - 4 + \cdots$. Your method gives $z-1 = (-2+3-4) + (5-6+7) + (-8+9-10) + \cdots = -3 + 6 - 9 + \cdots = -3z$ so that $4z=1$, i.e., $z=\frac{1}{4}$. This is the "correct" value (en.wikipedia.org/wiki/…) $\endgroup$ Commented Feb 25, 2023 at 0:30
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    $\begingroup$ @SamHopkins The alternating version is Abel-summable, so pretty much any summation technique should give the same result. And it doesn't seem to be part of a general phenomenon, because $t + 2t^2 + 3t^3 + \cdots$ doesn't appear to be summable using the technique in the question ($t=1$) and in your comment ($t=-1$). $\endgroup$
    – Gro-Tsen
    Commented Feb 25, 2023 at 0:46
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    $\begingroup$ I think the "larger story" is: the series diverges. So all of these grouping steps fail. Note: $x-1=9x$ has solution $x=+\infty$, which is more nearly correct than $x=-1/8$. $\endgroup$ Commented Feb 25, 2023 at 1:20
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    $\begingroup$ @GeraldEdgar, re, surely the same objection can be made to any method of summing divergent series? It is precisely because the familiar mathematics of absolutely convergent series don't guarantee a priori that all these rearrangements will produce the same answer that one is interested in why, despite not ‘having’ to be the same, they nonetheless are the same. $\endgroup$
    – LSpice
    Commented Feb 25, 2023 at 20:04
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    $\begingroup$ You can probably find a sensible way to represent what is going on with this approach: terrytao.wordpress.com/2010/04/10/… The idea is that your changes to the sum are invisible in the formal sum expression, but are captured by the smoothing term. $\endgroup$ Commented Feb 25, 2023 at 20:36

2 Answers 2


Yes there is, in $p$-adics.

You are probably familiar with the relation


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.

enter image description here

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 $p\ne3$.

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    $\begingroup$ Ignorant question: what do you gain from writing $8T(n)+1=(2n+1)^2$ instead of $T(n)=\frac{n(n+1)}2$? $\endgroup$ Commented Feb 25, 2023 at 20:23
  • $\begingroup$ The squares are then converged to zero in a subsequence, by using powers of $p$ (in my $3$-adic example the successive $2n+1$ values are $...0,...00,...000$). Note that introducing the squaring function in this way leads the derived subsequence of triangular to converge to $-1/8$ quadratically. $\endgroup$ Commented Feb 25, 2023 at 21:18
  • $\begingroup$ @OscarLanzi The same argument shows that $T(n)$ admits a subsequence that converges to $0$ in the $p$-adics, no? But why should one prefer the one you're finding? I do not get it, nor how it is related to the question of regrouping. Notice that $T(n)$ does not converge in the $p$-adics (for any $p$) so what you're proposing seems slightly arbitrary at the moment. $\endgroup$
    – Pedro
    Commented Feb 26, 2023 at 10:17
  • $\begingroup$ I put more emphasis on the unique feature resulting from the choice of $-1/8$: quadratic convergence of the subsequences. This leads to a hidden structure in the base-$p$ representation of ordinary triangular numbers, as discussed now for base $3$. $\endgroup$ Commented Feb 26, 2023 at 11:13
  • $\begingroup$ @Pedro If you stop the sum at the end of a group (in the groups of three version, say) then the last term $n$ will be congruent to $-1/2$ mod $3$. If we choose the group so that the sum at the right side of James's identity also stops at the end of a group then $n$ will be congruent to $-1/2$ mod $9$. If we keep iterating the grouping process and stop at an end of the group then $n$ will be congruent to $-1/2$ modulo higher and higher powers of $3$. $\endgroup$
    – Will Sawin
    Commented Feb 27, 2023 at 2:22

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 $0$.

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    $\begingroup$ But see the argument in another answer. We have a natural subsequence convergence to $-1/8$ in most prime bases. $\endgroup$ Commented Feb 25, 2023 at 3:10
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    $\begingroup$ @OscarLanzi since $T_n := 1 + 2 + \cdots + n = n(n+1)/2$, if $n \to -1/2$ in $\mathbf Z_p$ then $T_n \to (-1/2)(-1/2+1)/2 = -1/8$. And more generally if $n \to x$ in $\mathbf Z_p$ then $T_n \to x(x+1)/2$. How is this related to Jim's method of collecting terms together in the partial sums? $\endgroup$
    – KConrad
    Commented Feb 25, 2023 at 4:04

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