The Sum That Is Not a Sum

Adding infinitely many numbers means something only once we say what the adding is.

A series is a list of numbers, called its terms, written with plus signs between them. The series in the famous claim is the positive whole numbers in order:

The first term is 1, the second term is 2, and the term in position $k$ is the whole number $k$.

You cannot add every term in one step. What you can do is add the first term, then the next, and keep a running total. The running total after $k$ terms is the kth partial sum. The first few are

Every one of them is given by a short formula, which is proved in the fifth section:

The sum of a series, in the ordinary sense, is the number those running totals settle on, if they settle. These do not. Each new term is positive, so each new total is larger than the last, and $k(k+1)/2$ passes any bound once $k$ is large enough. The series $1+2+3+4+\cdots$ has no sum. That is the first claim in the ledger, and nothing below revises it.

The first laboratory view draws exactly these running totals. Choose the series $1+2+3+4+\cdots$ and a number of terms. Each point is the partial sum after that many terms. The points leave the top of the frame because the totals grow.

A famous claim sets the series above equal to $-1/12$. The derivation most people meet, in popular videos and in older manipulations of the same kind going back to Euler, does not start from the partial sums. It starts from two other series and treats each of them as if it were already a number. Here it is, in the order it is usually told. Each line looks like school algebra.

The first series. Write

The terms are $1$ and $-1$, alternating, starting with $1$. Pull off the first term and what remains is $S_1$ again, with the signs flipped:

Then $2S_1=1$, so $S_1=1/2$. People also point at the running totals $1,0,1,0,\dots$, whose average sits at $1/2$. That average is a real calculation, and the third section says what it measures. The step just written is the algebraic one, and it is the step this argument uses.

The second series. Write

The term in position $n$ is $n$ when $n$ is odd and $-n$ when $n$ is even. Add $S_2$ to a copy of itself shifted one place to the right, so that the first term of the copy sits under the second term of the original:

Check the first four columns by hand. Column 1 is $1$. Column 2 is $-2+1=-1$. Column 3 is $3+(-2)=1$. Column 4 is $-4+3=-1$. The sum of the two copies is $S_1$. The previous line called $S_1$ equal to $1/2$, so $2S_2=1/2$ and $S_2=1/4$.

The series we started with. Write

and subtract $S_2$ term by term. Subtracting a negative number is adding, so the pairs are $(1-1)$, $(2-(-2))=(2+2)$, $(3-3)$, $(4-(-4))=(4+4)$, and so on:

The right-hand side is four copies of $1+2+3+\cdots$, because $4\times 1=4$, $4\times 2=8$ and $4\times 3=12$. So it is $4S$. The previous line called $S_2$ equal to $1/4$, and therefore

Move the terms in $S$ to one side: $S-4S=1/4$, so $-3S=1/4$, and

That looks fine. Every step was a rearrangement, a shift by one place, or a subtraction of one series from another. The next section says why those steps were not available, and what $1/2$, $1/4$ and $-1/12$ actually measure.

Every line of that argument treated a series with no sum as a number that obeys ordinary algebra. You may rearrange it, shift it by one place and add it to itself, and subtract term by term. $S_1$, $S_2$ and $S$ are not such numbers. The partial sums of $S$ are $k(k+1)/2$ and they grow without bound. The partial sums of $S_1$ are $1,0,1,0,\dots$ and never settle. The partial sums of $S_2$ are $1,-1,2,-2,3,-3,\dots$ and step outward in both directions. None of the three has a sum, so none of them can be moved from one side of an equation to the other.

The same moves prove something false. Start again from $S=1+2+3+4+\cdots$ and shift it one place by writing a leading zero. If $S$ were a number, a leading zero would not change it, so the shifted copy would still be $S$:

Subtract term by term. The left side is $S-S$, which would be $0$ if $S$ were a number. The right side is $1+(2-1)+(3-2)+(4-3)+\cdots$, that is

Call that series $U$. The manipulation says $U=0$. Shift $U$ in the same way:

Subtract again. The left side is $0$. The right side is $1$. The argument has proved $0=1$.

Nothing about those two shifts was different in kind from "$2S_2=S_1$" or "$S-S_2=4S$". The argument does not show that $S$ equals $-1/12$. If you pretend a series with no sum is a number, the algebra will hand you $-1/12$ and it will also hand you $0=1$.

What is true uses the same three series, and each number measures something else.

$1/2$ is an average of running totals, not a sum. Grandi's series is $S_1$. Its terms are $1,-1,1,-1,\dots$: the term in position $n$ is $1$ when $n$ is odd and $-1$ when $n$ is even. The running totals are $1$, then $1+(-1)=0$, then $0+1=1$, then $1+(-1)=0$. They alternate forever, so this series has no sum either.

The running average of those totals is a different list. After $k$ totals, add the first $k$ running totals and divide by $k$. The plate below does this for four terms, which is few enough to check with a pencil. After an even number of totals the average is exactly $1/2$. After an odd number it is $(k+1)/(2k)$, a little above $1/2$, and that also moves towards $1/2$. Cesàro summation is the name for reporting that limit of averages. It is a statement about averages of running totals. It is not a statement that the terms add up to $1/2$. The equation $S_1=1-S_1$ is the step that is not allowed. The average is the step that is.

On $1+2+3+\cdots$ the same averaging fails. The $k$th partial sum is $k(k+1)/2$, and the average of the first $k$ of them is $(k+1)(k+2)/6$, which grows without bound. Cesàro summation assigns this series nothing. The value $-1/12$ does not come from anywhere near it. Set the second view to this series and the lower row climbs away with the upper one.

$1/4$ is a limit of discounted sums. Take $S_2$, the series $1-2+3-4+\cdots$. Multiply the term in position $n$ by $x^n$, with $x$ a fraction strictly below 1, so that later terms are counted a little less. The discounted series converges, and its sum is an ordinary number:

valid for $|x|<1$. As $x$ rises towards 1, this rises towards $1/4$. That limit is the Abel sum of $S_2$. At $x=99/100$ the value is exactly $9900/39601$, a little under $1/4$. This is a statement about the discounted series and about a limit of those sums. It is not a statement that $1-2+3-4+\cdots$ adds up to $1/4$: at $x=1$ the discount disappears and the partial sums do not settle.

The left panel of the third view prints this sum at the $x$ you choose, and the ladder of values as $x$ rises. It refuses $x=1$. Abel summation is defined as the limit from below, and substituting the endpoint is not one of its steps. The same substitution in the closed form for $1+2+3+\cdots$, which is $x/(1-x{)}^2$, divides by zero. For that series the discounted sums grow without bound as $x$ rises towards 1, so Abel summation assigns it no value either.

$-1/12$ is the constant left after a gentle cutoff, once the part that blows up is removed. Cesàro and Abel both fail for $1+2+3+\cdots$. What reaches $-1/12$ is a third procedure. Multiply the term $n$ by $e^{-n\epsilon }$, with $\epsilon$ a small positive number. The factor is near 1 for small $n$ and fades towards 0 once $n$ is large compared with $1/\epsilon$, so the tail is switched off gently rather than cut after a fixed number of terms. For any $\epsilon >0$ the smoothed series converges, and

Three pieces, and they behave differently as $\epsilon$ shrinks. The piece $1/{\epsilon }^2$ blows up. You are still adding a longer stretch of terms that the cutoff has barely touched, and the sum of that stretch grows like $1/{\epsilon }^2$. The pieces with ${\epsilon }^2$ and higher vanish. In between sits a constant, $-1/12$, which neither blows up nor vanishes.

The constant left after removing the part that blows up means this. Compute the smoothed sum, subtract $1/{\epsilon }^2$, and see what remains as $\epsilon$ gets smaller. The smoothed sum itself becomes huge. The remainder does not: it moves towards $-1/12$ from above. At $\epsilon =1/10$ the remainder is $-0.083291683195$. At $\epsilon =1/100$ it is $-0.083332916668$. And $-1/12=-0.0833\overline{3}$. The right-hand panel of the third view does this subtraction at the $\epsilon$ you choose. The headline on that panel is the remainder, not the smoothed sum and not a partial sum of $1+2+3+\cdots$. The term-count control does not change it. The series on that panel is always $1+2+3+\cdots$, with every term $n$ multiplied by $e^{-n\epsilon }$.

The same constant is the value of the Riemann zeta function at $-1$, after analytic continuation. The series $\sum n^{-s}$ converges only when $s>1$. At $s=-1$ it is $1+2+3+\cdots$, which has no sum. The symbol $\zeta (-1)$ does not denote that series. It denotes the value at $-1$ of the function that agrees with the series where the series converges, and that continues uniquely to the rest of the plane except for a pole at $s=1$. The continued function takes the value $-1/12$ at $-1$. That continuation is not performed on this page.

There is a short arithmetic link between the Abel value $1/4$ and this constant, and the link is the same continuation. The alternating series $\eta (s)=\sum (-1{)}^{n-1}{n}^{-s}$ becomes $1-2+3-4+\cdots$ at $s=-1$, and the Abel sum of that series is $1/4$, computed here. The identity $\eta (s)=(1-2^{1-s})\zeta (s)$ at $s=-1$ has factor $1-2^2=-3$, so dividing $1/4$ by $-3$ gives $-1/12$. The division is arithmetic. The identity holding at $s=-1$ is the continuation, and it is cited, not performed.

The constant is kept because it is stable. In a zeta-regularised calculation, or under a smooth cutoff of this kind, the piece that blows up depends on how the tail is switched off, and the constant $-1/12$ does not. Number theorists meet it as the value of the continued zeta function at $-1$. Physicists meet the same constant when a regularised sum of modes leaves a finite piece after the divergent piece is removed. That agreement is a theorem about those procedures. It is not a theorem about the partial sums of $1+2+3+\cdots$. Those partial sums grow without bound.

$-1/12$ is not the sum of $1+2+3+\cdots$. It is the constant left when that series is smoothed and the part that blows up is removed, and it is the value of a continued function at a point where its series does not converge. Those two are the same number. The partial sums are a third thing, and they do not approach it.

The statements below are the ones the ledger stands behind. The shifting argument in the second section is not among them: it is an argument that fails, and the third section says why.

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, with no qualifier at all, is not any of these, and statement 1 is why.

These are proofs of the statements in the previous section. They do not repair the shifting argument. That argument assumes each series is a number, and the first proof is that $1+2+3+\cdots$ is not.

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 towards $-0.08333$, 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. On the first view each point is a partial sum: the running total after $k$ terms of the series you selected. On the second view the lower row is the average of those partial sums, again after a chosen number of terms. Abel closed forms are evaluated at exact fractions $x$, checked against truncated power series under a rigorous rational tail bound: that panel is a discounted sum, not a partial sum of the undiscounted series. And the smoothed sum $\sum n{e}^{-n\epsilon }$ of $1+2+3+\cdots$, at seven values of $\epsilon$, is computed in BigInt fixed point at 20 decimal places with 25 guard digits, twice by unlike routes. The headline there is the smoothed sum minus $1/{\epsilon }^2$. 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 towards $-0.08333$ 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 people quote with the qualifier removed. 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.