1. Introduction
It does make sense and it’s an unambiguous, rigorous mathematical result that the sum, when properly defined, equals [math]-1/12[/math]. To explain how this comes about, we first need to define what we mean by the sum of an infinite series. The definition of such a summation does not follow from the definition of addition of numbers. The definition of addition allows us to compute the sum of two numbers which then allows us to also compute the sum if a finite number of numbers. Clearly, we can never get to the sum of an infinite series by only applying the definition of addition.
We therefore have to provide for a definition of the sum of an infinite series. The standard definition involves the partial series, which is defined by truncating the infinite series at some point. The sum of the partial series is then well defined as it’s a finite summation. If the limit of the partial series where the point where its truncated, is sent to infinity, exists, then one defines the sum of the infinite series to be that limit.
This means that this definition only applies in cases where the limit of the partial series exists. Such series are called “convergent series”. In contrast, if the limit of the partial series does not converge, then the series is called a “divergent series”. The sum of divergent series is then not defined using the limit of partial series. It’s not correct to say that the sum is infinite.
In case of divergent series, one has to resort to different methods to define the sum. Many different methods exist, books have been written on this subject, such as this classic by Hardy. This should dispel any myths about divergent series not being summable or that it’s just a trick that can yield any answer, that it’s all based on shaky mathematics and that the true sum of a divergent series is always either infinite or indeterminate.
However, the treatment of divergent series on the mathematics literature is still very axiomatic in nature, as can clearly be seen from reading Hardy’s classic book on this subject. So, it does come over as somewhat unnatural and ad hoc, even though results such as that the sum of all integers equals [math]-1/12[/math] can be shown to be universal. You’ll see this after studying the topic in detail and reading many pages of terse mathematical treatment. Most people won’t do that and then comments here on Quora that it’s all nonsense because the value of the sum is obviously infinite will be picked up by most people as the correct answer.
The question is then if there’s a simpler more intuitive way to understand that the sum of all integers is [math]-1/12[/math]? I do think there exist an intuitive explanation. We can get there by considering series expansions of functions.
2. Truncated series with a remainder term
We all know about Taylor’s theorem. A function [math]f(x)[/math] that’s [math]n[/math] times differentiable at a point [math]x = a[/math] can be expanded around a point as follows:
[math]\displaystyle f(a+h) = f(a) + f’(a) h + \frac{f’’(a)}{2} h^2 + \cdots +\frac{f^{(n)}}{n!}h^n + R_n(a,h)\tag{1}[/math]
where [math]R_n(a,h)[/math] is the so-called remainder term. In case we only assume [math]n[/math] times differentiability at [math]x = a,[/math] we can write the remainder term as:
[math]\displaystyle R_n(a,h) = u(a,h)h^n[/math]
where the function [math]u(a,h)[/math] has the property that
[math]\displaystyle \lim_{h\to 0}u(a,h) = 0[/math]
We then note that the summation in (1) is a finite summation so it’s well defined using only the definition of addition. If the function is infinitely often differentiable and the remainder term tends to zero, then we may also compute [math]f(a + h)[/math] by summing the infinite series. In this case the standard definition in terms of the limit of partial series applies.
However, the Taylor series (1) is always valid, regardless of whether or not the infinite series in case of functions that are in finitely often differentiable, actually converges. In case the series is divergent, the remainder term won't tend to zero. And besides Taylor series there are other types of power series, like so called asymptotic series. In a rigorous formulation they are also supposed to be truncated at some finite order and the statement is then that some quantity equals the value of the truncated series plus a remainder term and some properties of the remainder term will then be formulated.
The remainder term is, of course, not known. Only some properties are known and that can then be used to estimate the value of the function at the desired point.
3. The sum of a divergent series
We can now formulate the question of finding the sum of a divergent series as follows. We assume that the divergent series is the result of computing a well-defined value of a function attained at some point using a series. The value of the function is then given by truncating the series and adding the (unknown) remainder term. But the order at which one truncates the series is then entirely arbitrary. The definition of the sum of the divergent series is then the value of the function.
When given a divergent series:
[math]\displaystyle S = a_0 +a_1 + a_2 + a_3+\cdots [/math]
we are not told which function was expanded that yielded the given series. We then take the value of [math]S[/math] to be given as:
[math]\displaystyle S = a_0 +a_1 + a_2 + a_3+\cdots + a_n + R_n\tag{2}[/math]
where [math]R_n [/math]is the unknown remainder term. To compute [math]S[/math], we can appeal to analytic continuation. We then assume that we can deform the summation so that the terms of the summation are a function of a parameter [math]t[/math] and we have that [math]S(0) = S[/math], [math]a_k(0) = a_k[/math], [math]R_n(0) = R_n[/math], and:
[math]\displaystyle S(t) = a_0(t) +a_1(t) + a_2(t) + a_3(t)+\cdots + a_n(t) + R_n(t)\tag{3}[/math]
We then assume that there exists a region [math]U[/math] for the parameter [math]t[/math] where [math]S(t)[/math] is a convergent series. We then have that [math]R_n(t)[/math] tends to zero for [math]n[/math] to infinity when[math] t\in U[/math]. And [math]S(t)[/math] for [math]t\in U[/math] is then given by the sum of the convergent series as defined by the limit of the partial series:
[math]\displaystyle S(t) = \sum_{k=0}^{\infty} a_k(t)[/math]
With [math]S(t)[/math] defined, [math]R_n(t)[/math] in (3) is then also defined:
[math]\displaystyle R_n(t) = S(t) - \sum_{k=0}^{n} a_k(t)\tag{4}[/math]
This means that we can analytically continue (3) for finite [math]n[/math] to [math]t = 0[/math], which yields (2) with now [math]S[/math] given by the analytic continuation of the convergent summation for [math]t\in U[/math] to [math]t = 0[/math], and it’s then equal to the r.h.s. of (2) with the remainder term [math]R_n[/math] given by the analytic continuation of [math]R_n(t)[/math] for [math]t\in U[/math] to [math]t = 0[/math].
So, we see that analytic continuation of a divergent summation is a tool for finding the value of the summation as interpreted as the value of a function for which the series is a divergent series. It’s not some ad hoc arbitrary definition, as wrongly claimed by many here on Quora. And it’s then not going to depend on how exactly one performs the analytic continuation. So, it's not true that the sum of the integers is only [math]-1/12[/math] if one uses the zeta function to do the analytic continuation but that one could come up with any other arbitrary value if one performs the analytic continuation in a different way.
In the next section I’m going to derive a formula for the sum of a convergent series in terms of the partial sum. In the section after that I’m going to compute the sum of the integers by invoking that an analytic continuation can be done without specifying this explicitly, which then allows met to use the summation formula for convergent series and get to the answer.
4. Summation formula for convergent infinite series in terms of the partial sum
In case of the sum of natural numbers, we have an explicit formula for the partial sum:
[math]\displaystyle \sum_{k=1}^n k = \frac{1}{2} n (n+1)[/math]
We can then use a relation between the partial sum and the sum of the infinite series that I’ll derive in this section. We start with considering a convergent series whose partial sum [math]P(n)[/math] is known:
[math]\displaystyle P(n) = \sum_{k = 1}^n f(k)[/math]
The partial sum then satisfies the recursion relation:
[math]\displaystyle P(n) - P(n-1) = f(n)\tag{5}[/math]
with initial condition [math]P(1) = f(1)[/math].
We then analytically continue [math]P(n)[/math] from integers to reals such that the limit of [math]P(x)[/math] of [math]x[/math] to infinity on the real axis, exists. This limit is then equal to the sum of the series:
[math]\displaystyle \sum_{k = 1}^{\infty} f(k) = \lim_{R\to\infty}P\left(R\right)\tag{6}[/math]
We then analytically continue the summand [math]f(k)[/math] by analytically continuing the recursion relation (5) for the partial sum:
[math]\displaystyle P(x) - P(x-1) = f(x)\tag{7}[/math]
We then take this to be the definition of [math]f(x)[/math] for real [math]x[/math]. The limit in (6) for [math]R[/math] to infinity remains the same if we replace [math]P\left(R\right)[/math] by [math]P\left(R+x\right)[/math] for any arbitrary [math]x[/math]. One can then also replace [math]P\left(R\right)[/math] by the average of different values of [math]x[/math] of [math]P\left(R+x\right)[/math]. One can also integrate over [math]x[/math] over an interval of length [math]1[/math], doing this yields:
[math]\displaystyle \sum_{k = 1}^{\infty} f(k) = \lim_{R\to\infty}\int_{r-1}^{r} P\left(R+x\right)dx\tag{8}[/math]
where [math]r[/math] is an arbitrary real number.
Using (7) we can write:
[math]\displaystyle \begin{split}P\left(R+x\right) &= f(R+x) + P\left(R -1 +x\right)\\ &=f(R+x) + f(R-1+x)+P\left(R -2 +x\right)\\&=P\left(R -s-1 +x\right)+\sum_{m=0}^s f(R-s+m+x)\end{split}\tag{9}[/math]
where [math]s[/math] is an arbitrary positive integer. Let’s then consider the limit for [math]R[/math] to infinity in (8) for [math]R[/math] restricted to be integers. If we take [math]s = R-1[/math] in (9), then this becomes:
[math]\displaystyle P\left(R+x\right)=P(x)+\sum_{m=0}^{R-1} f(m+1+x)\tag{10}[/math]
If we substitute this in the integral from [math]r-1[/math] to [math]r[/math] in (8), then the [math]P(x)[/math] term gives rise to the integral:
[math]\displaystyle\int_{r-1}^{r} P(x)dx\tag{11} [/math]
And the [math]m[/math]th term from the summation in (10) gives rise to the integral:
[math]\displaystyle\int_{r-1}^{r} f(m+1+x)dx = \int_{r+m}^{r+m+1} f(x)dx\tag{12}[/math]
Summing (12) from [math]m = 0[/math] to [math]R-1[/math], then yields the integral:
[math]\displaystyle\int_{r}^{r+R} f(x)dx\tag{13}[/math]
Adding up (11) and (13) then yields for the integral in (8):
[math]\displaystyle\int_{r-1}^{r} P(x)dx + \int_r^{r+R}f(x) dx[/math]
Taking the limit of [math]R[/math] to infinity in (8) then yields the result:
[math]\displaystyle \sum_{k = 1}^{\infty} f(k) =\int_{r-1}^{r} P(x)dx + \int_r^{\infty}f(x) dx\tag{14}[/math]
5. Summation method for divergent infinite series in terms of the partial sum
Suppose that we’re given a divergent series. Then, as explained in section 3, the sum follows from performing an analytic continuation to a summation that is convergent and then analytically continuing the result back to the case at hand. We should then apply (14) to the convergent summation obtained after analytic continuation. But suppose that we apply (14) directly to the divergent summation. Then we see that the integral in (14) of [math]f(x)[/math] from [math]r[/math] to infinity wil diverge. We can then study the divergence of this integral by replacing the upper limit by [math]R[/math].
We will ten get a formula that will have terms that diverge to infinity as R is sent to infinity. If we then perform the analytic continuation to a region where the summation converges, then all these divergent terms must have changed into terms that tend to zero. In that region the answer is the first integral term in (14) plus the surviving constant term from the second integral. If we then analytically continue this result back to the original summation, then the first integral becomes the same as what it is when we evaluate it directly for the diverging summation.
And the constant term that survives after taking the limit of [math]R[/math] to infinity after analytic continuation to the convergent summation, and then analytically continuing this back the case of the diverging summation will simply yield the constant term from the second integral in (14) truncated at [math]R[/math]. We’re after all analytically continuing back and forth and upon return, we should get back to what we started out with, except that in the convergent region the limit of [math]R[/math] to infinity was taken which erases everything, except the constant term.
We thus need to change (14) and define a function [math]S(R)[/math]:
[math]\displaystyle S(R)=\int_{r-1}^{r} P(x)dx + \int_r^{R}f(x) dx\tag{15}[/math]
We then have that:
[math]\displaystyle \sum_{k = 1}^{\infty} f(k) =\text{Constant term in }S(R)\tag{16}[/math]
In case of the sum over integers, we have: [math]f(x) = x[/math] and [math]P(x) = \frac{1}{2} x (x+1)[/math]. The parameter [math]r[/math] can be chosen in any arbitrary way. If we then choose [math]r = 0[/math], then the integral of [math]f(x)[/math] in (15) becomes [math]\frac{1}{2}R^2[/math], which is a divergent term and doesn't have a constant term. So, we then find that:
[math]\displaystyle \sum_{k = 1}^{\infty}k=\frac{1}{2}\int_{-1}^{0} x(x+1)dx = -\frac{1}{12}[/math]
6. Conclusion
I’ve explained in detail that it does make sense to define the sum of divergent series as whatever the value of the quantity is that upon power series expansion would yield the given series. I’ve pointed out that such series are mathematically well defined despite being divergent, because they will always rigorously appear truncated at some finite order and with a remainder term added.
But given a divergent series and merely assuming that this is the expansion of some unknown function around some unknown point for some unknown value of the expansion parameter, doesn’t help to actually evaluate the value of the summation. As I’ve shown, analytic continuation to a summation that is convergent and then analytically continuing the result back, will evaluate the value of the divergent summation in exactly the correct interpretation, i.e. the value of the function which is represented as the given divergent series.
Then because the answer should not depend on how the analytic continuation is performed, it should be possible to get to the value of the sum of a divergent summation without actually performing an analytic continuation, by merely invoking that one could analytically continue it to a convergent summation. And that’s what I've shown in section 5. Using (15) and (16) one can evaluate the sum of a divergent series directly in terms of the partial sum of the diverging series.