The Sum That Is Not a Sum

Add the whole numbers in order and keep a running total: 1, then 3, then 6, then 10, then 15. That sequence is going in one direction only. Every term you add is positive and larger than the last, so the total after $k$ terms is $k(k+1)/2$, and whatever bound you care to name is passed once $k$ is large enough. The series $1+2+3+4+\cdots$ has no sum. That is the first thing this page shows, it is the first claim in the ledger, and nothing below revises it.

You are probably here because of a sentence that says otherwise. It is worth saying at once what is actually true. There are several standard procedures, each with a careful definition, which take a series that does not converge and assign a number to it anyway. On some series they agree with ordinary addition. On some series they agree with each other and not with ordinary addition, because ordinary addition has no answer to give. And there is a function, the Riemann zeta function, which is not a series at all but agrees with one where that series converges. Smooth away the tail of $1+2+3+\cdots$ and the constant left behind is $-1/12$; the zeta function takes the value $-1/12$ at $-1$. Both of those are true, and both need the words around them.

The laboratory has three panels. The first draws the partial sums, which is the truth about the series and looks like nothing at all: a line leaving the top of the frame. The second averages those partial sums, which rescues $1-1+1-1+\cdots$ and does nothing whatever for $1+2+3+\cdots$. The third does the two things that actually reach $-1/12$: it evaluates a power series just inside where it converges, and it smooths the series with a gentle cutoff and removes the part that blows up.

Here is the shift that makes all of this ordinary rather than mystical.

A series is a list of numbers with plus signs between them. Written down, it is not yet a number: turning it into one requires a rule, and the rule everybody learns is take the limit of the partial sums. That rule is excellent and it is not the only possible one. It is a choice, and it has two properties we would want from any choice: it agrees with finite addition when the list is finite, and it is linear, so that summing $2a_n+b_n$ gives twice the first answer plus the second.

Now ask for other rules with those same two properties. There are many, and the two on this page are the oldest.

Averaging. Instead of asking where the partial sums go, ask where their running averages go. Every convergent series has convergent averages with the same limit, so nothing is lost. And some divergent series gain: the partial sums of $1-1+1-1+\cdots$ flap between 1 and 0 forever, but their averages are $1,\frac{1}{2},\frac{2}{3},\frac{1}{2},\frac{3}{5},\frac{1}{2},\dots$, closing steadily on $\frac{1}{2}$. Averaging damps oscillation. That is exactly what it is for, and it is exactly what it is good for.

Discounting. Instead of adding $a_n$, add $a_n{x}^n$ for some $x$ a little below 1, so that later terms are counted a little less. The result is often a perfectly ordinary convergent series with a closed form, and then you can ask what happens as $x$ rises back to 1. For $1-2+3-4+\cdots$ the closed form is $x/(1+x{)}^2$, which rises to $1/4$. That is Abel summation, and note where the work happened: at $x=0.99$ the series genuinely converges and the closed form is genuinely its sum. Nothing is being smuggled. The only extra step is taking a limit at the end.

Now try both on $1+2+3+\cdots$ and watch them fail. Averaging gives $(k+1)(k+2)/6$, which runs away as fast as the partial sums did. Discounting gives $x/(1-x{)}^2$, which has a pole at $x=1$ and grows without bound. Neither method assigns this series a value. Whatever produces $-1/12$ has to be something else, and the third panel is where that something else lives.

Here is the sentence worth carrying away.

$-1/12$ is not the sum of $1+2+3+\cdots$. It is the constant term that survives when the series is smoothed and its divergent part is removed, and it is the value of a continued function at a point where its defining series does not converge. Those two turn out to be the same number, and that agreement is the interesting fact.

Take the smoothing first, because the laboratory can show it. Switch off the tail gently instead of sharply: sum $n{e}^{-n\epsilon }$ rather than $n$. For any positive $\epsilon$ this converges, and there is an elementary closed form for it,

Expand that in powers of $\epsilon$ and you get

Three pieces, and each behaves differently as $\epsilon$ shrinks. The first blows up, which is the divergence, still there, in its proper place. The last vanishes. In between sits a constant that neither blows up nor vanishes, and that constant is $-1/12$. Subtract the blowing-up part and watch what is left: at $\epsilon =1/10$ it is $-0.08329$, at $\epsilon =1/100$ it is $-0.083333$, falling toward $-1/12=-0.0833\overline{3}$ from above. That is the headline panel, computed in exact fixed-point arithmetic rather than floating point, because the number being looked for sits in the fourth decimal place of a difference between two quantities of size ten thousand.

The other route is the zeta function. The series $\sum n^{-s}$ converges for $s>1$ and defines a function there; that function extends uniquely to an analytic function on the whole plane except at $s=1$; and the extended function's value at $-1$ is $-1/12$. Substituting $s=-1$ into the series gives $1+2+3+\cdots$, which is why the two get confused, but the extension is not the series and does not claim to be. That continuation is not performed on this page. Nothing in a browser continues an analytic function. It is cited, and the ledger says so.

What this page does reach on its own is the step next door. The alternating version $\eta (s)=\sum (-1{)}^{n-1}{n}^{-s}$ becomes $1-2+3-4+\cdots$ at $s=-1$, whose Abel sum is $1/4$, computed here in exact fractions. And $\eta (s)=(1-2^{1-s})\zeta (s)$, so at $s=-1$ the factor is $1-2^2=-3$ and $\zeta (-1)=(1/4)/(-3)=-1/12$. The arithmetic is a division. The identity holding at $s=-1$ is the continuation again, and it is cited too.

Let $(a_n{)}_{n\ge 1}$ be a sequence, $s_k={\sum }_{n=1}^k{a}_n$ its partial sums, and ${\sigma }_k=\frac{1}{k}{\sum }_{j=1}^k{s}_j$ the running averages.

1. For $a_n=n$, $s_k=k(k+1)/2\to \infty$, so the series diverges. Proved, here, in full.
2. For $a_n=(-1{)}^{n-1}$, ${\sigma }_k=\frac{1}{2}$ for even $k$ and ${\sigma }_k=(k+1)/(2k)$ for odd $k$, so ${\sigma }_k\to \frac{1}{2}$. Proved, here, and computed in exact rationals for $k\le 1000$.
3. For $a_n=n$, ${\sigma }_k=(k+1)(k+2)/6\to \infty$. Cesàro summation fails on this series. Proved, here.
4. For $|x|<1$, ${\sum }_{n\ge 1}(-1{)}^{n-1}n{x}^n=x/(1+x{)}^2$, and this tends to $1/4$ as $x\uparrow 1$. So the Abel sum of $1-2+3-4+\cdots$ is $1/4$. Proved, here, with the closed form checked against the truncated series under a rigorous rational tail bound at twenty values of $x$.
5. $\eta (s)=(1-2^{1-s})\zeta (s)$ holds for all $s$ by analytic continuation, and at $s=-1$ gives $\zeta (-1)=\eta (-1)/(-3)=-1/12$. Proved in the literature and cited. The continuation is not performed here; the only step taken on this page is the division by $-3$.
6. ${\sum }_{n\ge 1}n{e}^{-n\epsilon }=\frac{1}{4{\sinh }^2(\epsilon /2)}=\frac{1}{{\epsilon }^2}-\frac{1}{12}+O({\epsilon }^2)$ as $\epsilon \downarrow 0$. Proved in the literature and cited; demonstrated here numerically at 20 decimal places over seven values of $\epsilon$, by two independent computations.
7. $\zeta (-1)=-\frac{1}{12}$ for the analytic continuation of $\sum n^{-s}$. Proved in the literature and cited. Not performed here, and no computation on this page establishes it.

Statement 1 is not superseded by any of the others. Statements 2, 3 and 4 are about summation methods; 6 is about a smoothed series; 5 and 7 are about a function that is not a series. The line people quote, which sets $1+2+3+\cdots$ equal to $-1/12$ with no qualifier at all, is not any of these, and statement 1 is why.

The series diverges. By induction, ${\sum }_{n=1}^{k}n=k(k+1)/2$: true at $k=1$, and adding $k+1$ to $k(k+1)/2$ gives $(k+1)(k+2)/2$. Given any bound $B$, take $k>2B$; then $k(k+1)/2>k/2>B$. So the partial sums are eventually above every bound and the series has no sum. $■$

Grandi's series is Cesàro summable to $\frac{1}{2}$. The partial sums alternate $1,0,1,0,\dots$, so ${\sum }_{j=1}^k{s}_j=\lceil k/2\rceil$. For even $k$ this is $k/2$ and ${\sigma }_k=\frac{1}{2}$ exactly; for odd $k$ it is $(k+1)/2$ and ${\sigma }_k=(k+1)/(2k)=\frac{1}{2}+\frac{1}{2k}$. Both tend to $\frac{1}{2}$, so ${\sigma }_k\to \frac{1}{2}$. $■$

Cesàro summation fails on $1+2+3+\cdots$. Here $s_j=j(j+1)/2$, so

and dividing by $k$ gives ${\sigma }_k=(k+1)(k+2)/6$. This is quadratic in $k$, so it passes every bound and the means diverge. Indeed ${\sigma }_k/s_k\to \frac{1}{3}$, so the averaging did not even change the order of growth. $■$

The Abel sum of $1-2+3-4+\cdots$ is $\frac{1}{4}$. For $|x|<1$ the geometric series gives ${\sum }_{n\ge 0}(-x{)}^n=1/(1+x)$, which may be differentiated term by term inside its disc of convergence. Doing so and multiplying by $x$ gives ${\sum }_{n\ge 1}(-1{)}^{n-1}n{x}^n=x/(1+x{)}^2$. This is a continuous function on $[0,1]$, so the limit as $x\uparrow 1$ is its value at 1, namely $1/4$. $■$

The laboratory checks the closed form rather than trusting it. Truncating at $N$ terms leaves a tail bounded by ${\sum }_{n>N}n{t}^n=t^{N+1}((N+1)-Nt)/(1-t{)}^2$ with $t=|x|$, an exact rational, and the test suite asserts that the closed form and the truncation differ by no more than that bound at twenty different rational $x$.

Why $x=1$ is refused. Abel summation is defined as ${\lim }_{x\uparrow 1}\sum a_n{x}^n$, a limit over points strictly inside the disc. Substituting $x=1$ into $x/(1+x{)}^2$ happens to give $1/4$, the correct value, which is precisely the danger: the same substitution in $x/(1-x{)}^2$ divides by zero, and a step that is only valid when you already know the answer is not a step. The laboratory therefore refuses the endpoint and prints the reason.

What is not proved here. Two things, and they are the two that carry the number to the zeta function. The expansion $\sum n{e}^{-n\epsilon }=1/{\epsilon }^2-1/12+O({\epsilon }^2)$ is a Laurent expansion of $1/(4{\sinh }^2(\epsilon /2))$, cited to Tao; this page evaluates it at seven values of $\epsilon$ in fixed-point arithmetic and shows the leftover falling toward $-0.08333$, which is a demonstration and not a derivation. And the analytic continuation of $\sum n^{-s}$ to $s=-1$, together with the identity $\eta (s)=(1-2^{1-s})\zeta (s)$ holding there, is cited to Riemann and to Titchmarsh. Nothing in this browser continues a function past where its series converges, and no number printed on this page would change if the continuation were different.

What this laboratory computes. Partial sums and Cesàro means of three series, in exact rational arithmetic on BigInt, for up to 120 terms with nothing rounded. Abel closed forms at exact fractions $x$, checked against truncated power series under a rigorous rational tail bound. And the smoothed sum $\sum n{e}^{-n\epsilon }$ at seven values of $\epsilon$, in BigInt fixed point at 20 decimal places with 25 guard digits, computed twice by unlike routes so that the agreement means something. There is no floating-point number anywhere in the mathematics.

What a demonstration at seven values of $\epsilon$ is worth. Not much on its own, and the ledger says so. Seven points falling toward $-0.08333$ is consistent with a limit of $-1/12$ and equally consistent with a limit of $-0.08333$ that happens to be something else, and no number of further points would settle it. The reason to believe the constant is exactly $-1/12$ is the expansion, which is cited. What the seven points genuinely establish is that the code implements the expansion it claims to, since two independent routes to the same number agree to nineteen decimal places, and that is worth having.

The exact path does not give $-1/12$, and that is instructive. Smoothing with $(1-\epsilon {)}^n$ instead of $e^{-n\epsilon }$ stays entirely inside the rationals, and the sum is exactly $1/{\epsilon }^2-1/\epsilon$: no constant term at all. Nothing is broken. That cutoff is $e^{-n{\epsilon }^{\prime }}$ for ${\epsilon }^{\prime }=-\log (1-\epsilon )$, and expanding $1/{\epsilon }^{\prime 2}$ in powers of $\epsilon$ recovers the missing $-1/12$. Which quantity counts as the divergent part is part of the statement, and a reader who assumes $1/{\epsilon }^2$ is obviously the right thing to subtract in any parametrisation gets zero here instead. The constant is universal across smooth cutoffs; the head is not.

Where the mathematics is settled and where the sentence is not. The mathematics is entirely settled and has been for over a century. Every result named on this page is a theorem, and the frontier is elsewhere: the interesting open questions about $\zeta$ are about its zeros, not its value at $-1$. What is not settled is the sentence, which continues to circulate with the qualifier removed, and in that shortened form it contradicts the first claim in the ledger. If you take one thing from this page, take the habit of asking what a stated equality is an equality of.

Read this next to 1, 2, 4, 8, 16, and Then Not 32. There a pattern is overwhelming and then simply wrong at the sixth step. Here a formula is famous and correct, and about something other than what it looks like. Both pages are about the same discipline applied at different moments: pinning down the claim before believing it.