A sum read from inside

Two of the most quoted sums in mathematics are $1 - \tfrac12 + \tfrac13 - \tfrac14 + \dots = \log 2$ and $1 - \tfrac13 + \tfrac15 - \tfrac17 + \dots = \pi/4$. Both are usually derived the same way, and the derivation has a gap in exactly the place nobody looks.

The first comes from the power series $x - x^2/2 + x^3/3 - \dots$, which equals $\log(1 + x)$ for every $x$ strictly between $-1$ and $1$. Put $x = 1$ and the series becomes the sum on the left and the function becomes $\log 2$. But $x = 1$ is not strictly between $-1$ and $1$. It is the edge of the interval where the power series is known to equal the function, and at the edge the usual theorems about power series say nothing at all.

The picture suggests the value at the edge is the limit from inside, and the question is when that is true. Abel answered it in 1826.

A formula that holds inside the interval #

The series for $\log(1 + x)$ comes from the geometric series. For $|t| < 1$, $1/(1 + t) = 1 - t + t^2 - t^3 + \dots$, and integrating from 0 to $x$ term by term gives $\log(1 + x) = x - x^2/2 + x^3/3 - \dots$, because an integral undoes a slope one power at a time. The same move applied to $1/(1 + t^2)$ gives $\arctan x = x - x^3/3 + x^5/5 - \dots$.

Both steps are justified inside the interval. A power series converges uniformly on every closed interval strictly inside its radius of convergence, here 1, and uniform convergence lets integrals pass through the sum. At $x = 1$ the geometric series $1 - 1 + 1 - \dots$ does not converge at all, so the derivation breaks down exactly where the famous sums are taken.

The series for $\log 2$ does converge at $x = 1$, by the alternating series test: the terms shrink and alternate in sign. So there is a sum. What is not yet known is that the sum is $\log 2$. The function $\log(1 + x)$ is continuous at $x = 1$ and the partial sums are continuous there, but a limit of continuous functions need not be continuous, and a limit that failed to be continuous at the edge would put the sum somewhere other than $\log 2$.

Uniform all the way to the edge #

For this series the gap closes with a single estimate. On $[0, 1]$ the terms $x^n/n$ decrease as $n$ grows, for every $x$ at once, and an alternating series with decreasing terms stops within its first omitted term of its sum. So after $N$ terms the partial sum is within $x^{N+1}/(N+1)$ of $\log(1 + x)$, and that is at most $1/(N+1)$ — one bound for every $x$ in the closed interval, the edge included.

A bound that serves every point at once is uniform convergence, and a uniform limit of continuous functions is continuous. So the sum of the series is a continuous function on the closed interval, it agrees with $\log(1 + x)$ everywhere inside, and two continuous functions agreeing inside agree at the edge too. The value at 1 is $\log 2$.

The arctangent series goes the same way and gives $1 - \tfrac13 + \tfrac15 - \dots = \pi/4$, the series Leibniz found in 1674 and that Madhava’s school in Kerala had reached more than two centuries earlier. It converges painfully slowly: twenty terms are still a hundredth out, which is the $1/(2N + 1)$ of the alternating estimate made visible.

Why the interval stops at 1 #

Both series stop converging beyond 1, and the reason is not always where it looks. For the logarithm it is visible on the real line: $\log(1 + x)$ runs off to minus infinity as $x$ falls to $-1$, and a power series centred at 0 cannot reach past a point where its function does that. The arctangent has no such point. It is smooth along the whole real line and stays between $-\pi/2$ and $\pi/2$, and its series stops at 1 all the same.

The obstruction is off the line. The fraction $1/(1 + t^2)$ that the series was integrated from becomes infinite at $t = i$ and $t = -i$, both at distance 1 from the centre, and a power series converges on the largest disc round its centre that avoids every such point — in every direction at once, including along the real line where nothing is visibly wrong. So the edge at which Leibniz’s series sits is placed by two complex numbers that the interval never meets, and it takes Abel’s theorem to make the value there the one the smooth real function takes.

The other end of the interval behaves differently. At $x = -1$ the logarithm series becomes $-(1 + \tfrac12 + \tfrac13 + \dots)$, the harmonic series with its sign changed, which grows without bound — and $\log(1 + x)$ also runs off to minus infinity there. A series of terms that are all of one sign always tends to its sum from inside, even when the sum is infinite, so divergence at one edge and convergence at the other are both read correctly; the theorem is needed only where the signs are mixed.

Half the first omitted term, and what averaging buys #

The alternating estimate promised that twenty terms of the Leibniz series would be within $1/41$ of $\pi/4$. The figure finds them within 0.0125, which is $1/80$ — about half of what was promised, and not by coincidence. When the terms of an alternating series shrink smoothly, consecutive partial sums overshoot and undershoot the limit by nearly equal amounts, so the limit sits almost exactly halfway between them, and the error after $N$ terms is close to half the first term left out. For the Leibniz series that is very nearly $1/(4N)$: 0.012492 at twenty terms, and 0.0025 at a hundred.

That makes the Leibniz series a famously poor way to compute $\pi$. Six correct decimals of $\pi$ need the error in $\pi/4$ below a quarter of a millionth, and the partial sums first get there after a million terms.

The same observation suggests the repair. If the limit sits halfway between consecutive partial sums, average them. The average of the twentieth and twenty-first partial sums is within 0.0003 of $\pi/4$, forty times closer than either sum, and the error of such averages falls like $1/(8N^2)$ rather than $1/(4N)$ — so about seven hundred terms, averaged in pairs, do what a million terms do alone.

Averaging partial sums is not only a way of giving values to series that have none; it is also the cheapest way of hurrying along a series that converges slowly. Abel’s theorem is built on an average of exactly this kind, with weights that change smoothly instead of two neighbours taken equally.

Abel’s theorem, without the alternating signs #

The estimate used the alternating signs and the decreasing terms. Abel’s theorem needs neither.

Abel’s theorem. If $a_0 + a_1 + a_2 + \dots$ converges to $s$, then $a_0 + a_1 x + a_2 x^2 + \dots$ converges uniformly on $[0, 1]$, and its value tends to $s$ as $x$ rises to 1.

The proof is a rearrangement of the sum called summation by parts. Write $s_n = a_0 + \dots + a_n$ for the partial sums. Then, for $x$ strictly between 0 and 1,

$$\sum_{n \ge 0} a_n x^n = (1 - x)\sum_{n \ge 0} s_n x^n,$$

which is checked by multiplying out the right-hand side and watching consecutive terms cancel. The weights $(1 - x)x^n$ are positive and add to exactly 1. So the value of the power series at $x$ is an average of the partial sums, with weights that sit mostly on the first few when $x$ is small and spread out over later and later ones as $x$ approaches 1.

If the partial sums settle on $s$, then an average that puts almost all its weight on late partial sums is close to $s$, and the value at $x$ tends to $s$. Abel’s own version of the argument bounds the tails of the series in the same way for every $x$ at once, which gives the uniform convergence; he proved it in his 1826 paper on the binomial series, the same paper in which he pointed out that Cauchy’s theorem on continuous limits had exceptions.

A series with no sum, read the same way #

The averaging does not need the partial sums to settle. It needs only that their average settles, and that is a weaker thing.

The series $1 - 1 + 1 - 1 + \dots$ has partial sums $1, 0, 1, 0, \dots$, so it has no sum. Its power series is the geometric series $1 - x + x^2 - \dots = 1/(1 + x)$, which tends to $1/2$ as $x$ rises to 1. Abel’s theorem does not apply, because its hypothesis — a convergent series at the edge — fails, and the partial sums never come uniformly close to the function near 1: the largest gap on the closed interval is $1/2$ for every number of terms.

So the theorem runs in one direction only. A convergent series at the edge forces the limit from inside to equal its sum. A limit from inside does not force a sum to exist. The series was argued over by Grandi in 1703, who took its value to be $1/2$, and by Leibniz, who defended the half on the grounds that stopping at a random point gives 1 or 0 equally often. Abel’s theorem makes the half precise without endorsing it.

The weights that produce the half #

The summation-by-parts identity turns the half into an honest average. The partial sums alternate between 1 and 0, and the value of the power series at $x$ is the total weight sitting on the ones. The weights on the even positions add to $(1 - x)(1 + x^2 + x^4 + \dots) = (1 - x)/(1 - x^2)$, which is $1/(1 + x)$. At $x = 0.8$ that is 0.5556, a little more than half, because the first partial sum carries the largest single weight and it is a 1. At $x = 0.97$ the weight has spread across dozens of partial sums, the ones and the zeros almost balance, and the share is 0.5076.

Taking that limit as the value of the series is Abel summation, and it is a consistent way of giving values to some series that have none. It is regular: by Abel’s theorem, whenever a series already has a sum, Abel summation returns that sum. It is linear, since limits of power series add. And it reaches further than ordinary summation, because averaging can settle what the partial sums themselves never do.

A growing series given a value #

Abel summation reaches series whose terms do not even shrink.

The series $1 - 2 + 3 - 4 + \dots$ has terms that grow and partial sums $1, -1, 2, -2, 3, \dots$ that swing further out at every step. Its power series is $1 - 2x + 3x^2 - \dots = 1/(1 + x)^2$, the derivative of the geometric series with a sign changed, and that tends to $1/4$. Euler gave the series the value $1/4$ in 1749, by exactly this route.

The value is not arbitrary, and there is a second way to see it. Multiplying $1 - x + x^2 - \dots$ by itself, collecting equal powers, gives $1 - 2x + 3x^2 - \dots$, so the series of growing terms is the square of Grandi’s series in the sense of power series. If Grandi’s series is worth a half, its square ought to be worth a quarter, and Abel summation agrees with itself: both values come from the same functions, $1/(1 + x)$ and its square, evaluated at the edge.

When reading from inside is enough #

Abel’s theorem goes from a sum to a limit. Theorems that go the other way — from a limit inside the interval back to an honest sum — need an extra hypothesis to rule out series like Grandi’s, and they are called Tauberian theorems after the first of them.

Tauber’s theorem (1897): if the power series tends to $s$ as $x$ rises to 1, and the terms satisfy $n a_n \to 0$, then the series converges to $s$. Grandi’s series fails the condition, since $n a_n$ is $\pm n$, which is exactly what lets its limit from inside exist without a sum. Hardy and Littlewood weakened the condition to a one-sided bound on $n a_n$, and Wiener built a general theory of such converses in the 1930s.

That theory has a use nobody expected when Tauber wrote. The prime number theorem — that the number of primes below $N$ is about $N/\log N$, a count with no exact formula — was given one of its shortest proofs through a Tauberian theorem of Wiener and Ikehara: a function built from the primes is shown to behave well as its variable approaches an edge, and the Tauberian step converts that behaviour into a statement about the primes themselves. The question of when a sum can be read from inside turns out to be a question about how primes are spaced.

Where the reading goes wrong #

Abel summation gives nothing to series that grow in one direction. The series $1 + 1 + 1 + \dots$ becomes $1/(1 - x)$, which runs off to infinity as $x$ rises to 1, and so does $1 + 2 + 3 + \dots$. The values sometimes quoted for such series come from other, stronger methods, and Abel’s method does not reach them.

The order of the terms is part of the question. The power series attaches $x^n$ to the $n$-th term, so rearranging the terms changes the function. A convergent series whose terms do not converge absolutely can be rearranged to add to anything, and each rearrangement has its own power series and its own limit at the edge.

The approach to the edge is along the interval. In the complex plane a power series with radius 1 can be approached at $z = 1$ from many directions, and Abel’s theorem survives for approaches that stay within a fixed angle of the radius. Along paths that creep in tangent to the circle it can fail, and none of the figures, which live on the real interval, shows that.

What the figures cannot show #

Every curve is a finite partial sum. The claims are about infinitely many terms, and the uniform bounds — $1/(N + 1)$, $1/(2N + 1)$, the fixed gap of a half — are what the drawn sums are measured against, on a grid of two thousand points and at $x = 1$ itself.

The functions are identified, not discovered. That $x - x^2/2 + \dots$ sums to $\log(1 + x)$ is checked at $x = 1/2$ by adding four hundred terms, which confirms the pairing of series and function; it is the derivation, not the check, that says the pairing holds across the interval.

And the weights are shown for one series. Grandi’s partial sums are ones and zeros, which is what lets the average be read as a share of coloured bars. For a general series the same average is over partial sums of any size, and no bar chart shows that as plainly.

The question it leaves: which divergent series deserve a value #

Abel summation is one method among several, and each gives values to a different collection of series. Averaging the partial sums with equal weights, as Cesàro did, handles Grandi’s series and gives the same half; Frobenius showed in 1880 that whenever Cesàro’s averages settle, Abel’s limit exists and agrees. The same equal-weight averaging repairs the partial sums of a Fourier series at a jump, which is Fejér’s theorem. Other methods reach further still, down to series that converge nowhere and still determine a function.

What no method settles is which values are the right ones. Any method that is linear, and unchanged when a leading term is deleted, and gives Grandi’s series a value $S$ at all, must have $S = 1 - S$ — the series is 1 followed by its own negative — so that value is forced to be a half. For series such as $1 + 2 + 3 + \dots$, methods that assign values do not all agree, and the question of what such a value means is answered differently in number theory and in physics.

Continuity at the edge is the whole question #

A power series knows its function everywhere inside its interval, and the famous sums are all taken at the edge, where it knows nothing. Reading the value there by continuity is legitimate exactly when the convergence is uniform up to the edge, and Abel’s theorem says that happens whenever the series at the edge converges at all.

The converse fails, and the failure is productive rather than embarrassing: a limit from inside is an average of partial sums, averages settle more easily than the sums themselves, and taking the limit anyway gives values to series that have none. When a formula is proved inside a region and used on its boundary, the boundary is where the proof has to be done again — and what is found there is often a new definition, not merely a missing step.