Heuristics

Passage from Ramanujan's first notebook describing the "constant" of the series

Srinivasa Ramanujan presented two derivations of "1 + 2 + 3 + 4 + ⋯ = −+1/12" in chapter 8 of his first notebook.^[7][8][9] The simpler, less rigorous derivation proceeds in two steps, as follows.

The first key insight is that the series of positive numbers 1 + 2 + 3 + 4 + ⋯ closely resembles the alternating series 1 − 2 + 3 − 4 + ⋯. The latter series is also divergent, but it is much easier to work with; there are several classical methods that assign it a value, which have been explored since the 18th century.^[10]

In order to transform the series 1 + 2 + 3 + 4 + ⋯ into 1 − 2 + 3 − 4 + ⋯, one can subtract 4 from the second term, 8 from the fourth term, 12 from the sixth term, and so on. The total amount to be subtracted is 4 + 8 + 12 + 16 + ⋯, which is 4 times the original series. These relationships can be expressed using algebra. Whatever the "sum" of the series might be, call it c = 1 + 2 + 3 + 4 + ⋯. Then multiply this equation by 4 and subtract the second equation from the first:

${\displaystyle {\begin{alignedat}{7}c={}&&1+2&&{}+3+4&&{}+5+6+\cdots \\4c={}&&4&&{}+8&&{}+12+\cdots \\c-4c={}&&1-2&&{}+3-4&&{}+5-6+\cdots \end{alignedat}}}$

The second key insight is that the alternating series 1 − 2 + 3 − 4 + ⋯ is the formal power series expansion of the function 1/(1 + x)² but with x defined as 1. (This can be seen by equating 1/1 + x to the alternating sum of the nonnegative powers of x, and then differentiating and negating both sides of the equation.) Accordingly, Ramanujan writes

${\displaystyle -3c=1-2+3-4+\cdots ={\frac {1}{(1+1)^{2}}}={\frac {1}{4}}.}$

Dividing both sides by −3, one gets c = −+1/12.

Generally speaking, it is incorrect to manipulate infinite series as if they were finite sums. For example, if zeroes are inserted into arbitrary positions of a divergent series, it is possible to arrive at results that are not self-consistent, let alone consistent with other methods. In particular, the step 4c = 0 + 4 + 0 + 8 + ⋯ is not justified by the additive identity law alone. For an extreme example, appending a single zero to the front of the series can lead to a different result.^[1]

One way to remedy this situation, and to constrain the places where zeroes may be inserted, is to keep track of each term in the series by attaching a dependence on some function.^[11] In the series 1 + 2 + 3 + 4 + ⋯, each term n is just a number. If the term n is promoted to a function n^−s, where s is a complex variable, then one can ensure that only like terms are added. The resulting series may be manipulated in a more rigorous fashion, and the variable s can be set to −1 later. The implementation of this strategy is called zeta function regularization.

Zeta function regularization

Plot of ζ(s). For s > 1, the series converges and ζ(s) > 1. Analytic continuation around the pole at s = 1 leads to a region of negative values, including ζ(−1) = −+1/12.

In zeta function regularization, the series ${\textstyle \sum _{n=1}^{\infty }n}$ is replaced by the series ${\textstyle \sum _{n=1}^{\infty }n^{-s}}$. The latter series is an example of a Dirichlet series. When the real part of s is greater than 1, the Dirichlet series converges, and its sum is the Riemann zeta function ζ(s). On the other hand, the Dirichlet series diverges when the real part of s is less than or equal to 1, so, in particular, the series 1 + 2 + 3 + 4 + ⋯ that results from setting s = –1 does not converge. The benefit of introducing the Riemann zeta function is that it can be defined for other values of s by analytic continuation. One can then define the zeta-regularized sum of 1 + 2 + 3 + 4 + ⋯ to be ζ(−1).

From this point, there are a few ways to prove that ζ(−1) = −+1/12. One method, along the lines of Euler's reasoning,^[12] uses the relationship between the Riemann zeta function and the Dirichlet eta function η(s). The eta function is defined by an alternating Dirichlet series, so this method parallels the earlier heuristics. Where both Dirichlet series converge, one has the identities:

${\displaystyle {\begin{alignedat}{7}\zeta (s)&{}={}&1^{-s}+2^{-s}&&{}+3^{-s}+4^{-s}&&{}+5^{-s}+6^{-s}+\cdots &\\2\times 2^{-s}\zeta (s)&{}={}&2\times 2^{-s}&&{}+2\times 4^{-s}&&{}+2\times 6^{-s}+\cdots &\\\left(1-2^{1-s}\right)\zeta (s)&{}={}&1^{-s}-2^{-s}&&{}+3^{-s}-4^{-s}&&{}+5^{-s}-6^{-s}+\cdots &=\eta (s).\end{alignedat}}}$

The identity ${\displaystyle (1-2^{1-s})\zeta (s)=\eta (s)}$ continues to hold when both functions are extended by analytic continuation to include values of s for which the above series diverge. Substituting s = −1, one gets −3ζ(−1) = η(−1). Now, computing η(−1) is an easier task, as the eta function is equal to the Abel sum of its defining series,^[13] which is a one-sided limit:

${\displaystyle -3\zeta (-1)=\eta (-1)=\lim _{x\to 1^{-}}\left(1-2x+3x^{2}-4x^{3}+\cdots \right)=\lim _{x\to 1^{-}}{\frac {1}{(1+x)^{2}}}={\frac {1}{4}}.}$

Dividing both sides by −3, one gets ζ(−1) = −+1/12.

Cutoff regularization

The series 1 + 2 + 3 + 4 + ⋯

After smoothing

Asymptotic behavior of the smoothing. The y-intercept of the parabola is −+1/12.^[1]

The method of regularization using a cutoff function can "smooth" the series to arrive at −+1/12. Smoothing is a conceptual bridge between zeta function regularization, with its reliance on complex analysis, and Ramanujan summation, with its shortcut to the Euler–Maclaurin formula. Instead, the method operates directly on conservative transformations of the series, using methods from real analysis.

The idea is to replace the ill-behaved discrete series ${\displaystyle \textstyle \sum _{n=0}^{N}n}$ with a smoothed version

${\displaystyle \sum _{n=0}^{\infty }nf\left({\frac {n}{N}}\right),}$

where f is a cutoff function with appropriate properties. The cutoff function must be normalized to f(0) = 1; this is a different normalization from the one used in differential equations. The cutoff function should have enough bounded derivatives to smooth out the wrinkles in the series, and it should decay to 0 faster than the series grows. For convenience, one may require that f is smooth, bounded, and compactly supported. One can then prove that this smoothed sum is asymptotic to −+1/12 + CN², where C is a constant that depends on f. The constant term of the asymptotic expansion does not depend on f: it is necessarily the same value given by analytic continuation, −+1/12.^[1]

Ramanujan summation

The Ramanujan sum of 1 + 2 + 3 + 4 + ⋯ is also −+1/12. Ramanujan wrote in his second letter to G. H. Hardy, dated 27 February 1913:

"Dear Sir, I am very much gratified on perusing your letter of the 8th February 1913. I was expecting a reply from you similar to the one which a Mathematics Professor at London wrote asking me to study carefully Bromwich's Infinite Series and not fall into the pitfalls of divergent series. ... I told him that the sum of an infinite number of terms of the series: 1 + 2 + 3 + 4 + ⋯ = −+1/12 under my theory. If I tell you this you will at once point out to me the lunatic asylum as my goal. I dilate on this simply to convince you that you will not be able to follow my methods of proof if I indicate the lines on which I proceed in a single letter. ..."^[14]

Ramanujan summation is a method to isolate the constant term in the Euler–Maclaurin formula for the partial sums of a series. For a function f, the classical Ramanujan sum of the series ${\displaystyle \textstyle \sum _{k=1}^{\infty }f(k)}$ is defined as

${\displaystyle c=-{\frac {1}{2}}f(0)-\sum _{k=1}^{\infty }{\frac {B_{2k}}{(2k)!}}f^{(2k-1)}(0),}$

where f^(2k−1) is the (2k − 1)-th derivative of f and B_2k is the 2k-th Bernoulli number: B₂ = 1/6, B₄ = −+1/30, and so on. Setting f(x) = x, the first derivative of f is 1, and every other term vanishes, so^[15]

${\displaystyle c=-{\frac {1}{6}}\times {\frac {1}{2!}}=-{\frac {1}{12}}.}$

To avoid inconsistencies, the modern theory of Ramanujan summation requires that f is "regular" in the sense that the higher-order derivatives of f decay quickly enough for the remainder terms in the Euler–Maclaurin formula to tend to 0. Ramanujan tacitly assumed this property.^[15] The regularity requirement prevents the use of Ramanujan summation upon spaced-out series like 0 + 2 + 0 + 4 + ⋯, because no regular function takes those values. Instead, such a series must be interpreted by zeta function regularization. For this reason, Hardy recommends "great caution" when applying the Ramanujan sums of known series to find the sums of related series.^[16]

Failure of stable linear summation methods

A summation method that is linear and stable cannot sum the series 1 + 2 + 3 + ⋯ to any finite value. (Stable means that adding a term at the beginning of the series increases the sum by the value of the added term.) This can be seen as follows. If

$${\displaystyle 1+2+3+\cdots =x,}$$

then adding 0 to both sides gives

$${\displaystyle 0+1+2+3+\cdots =0+x=x}$$

by stability. By linearity, one may subtract the second equation from the first (subtracting each component of the second line from the first line in columns) to give

$${\displaystyle 1+1+1+\cdots =x-x=0.}$$

Adding 0 to both sides again gives

$${\displaystyle 0+1+1+1+\cdots =0,}$$

and subtracting the last two series gives

$${\displaystyle 1+0+0+0+\cdots =0,}$$

contradicting stability.

Therefore, every method that gives a finite value to the sum 1 + 2 + 3 + ⋯ is not stable or not linear.^[17]