I think I understand intuitively how we can assign a value to the sum of all natural numbers. But of all the proofs that I've seen that show why $\zeta(-1) = -\frac{1}{12}$, none of them use their own tactics to address the divergence of the harmonic sum.

To give an example of what I mean, take a look at one of the most famous proofs for the sum of the natural numbers: $$ c = 1 + 2 + 3 + 4 + ... \\ 4c = 4 + 8 + 12 + 16 + ... \\ -3c = 1 - 2 + 3 - 4 + 5 - ... \\ 1 - 2 + 3 - 4 + ... = \left.\frac{1}{(1 + x)^2}\right|_{x=1} = \frac{1}{4} \\ -3c = \frac{1}{4} \\ c = -\frac{1}{12} $$

If you were considering finding a value for $c$ whilst keeping in mind convergence, you would stop immediately at the first line of that proof. Clearly, $c = \infty$, no doubt about it. Yet if we continue on with the proof, using algebraic manipulations that would seem absurd to anyone considering convergence, we end up with $c = -\frac{1}{12}$.

So obviously we don't mean that the sum of the natural numbers converges to $-\frac{1}{12}$, but rather we mean that if $c$ were to be some number, it would have to satisfy the given algebraic properties ($-3c = \frac{1}{4}$).

This idea of analytic continuation makes sense to me, but what doesn't make sense to me is why it can't be applied to the harmonic sum. If we completed disregarded the convergence of $\sum_{k=1}^\infty{\frac{1}{k}}$ like we did above, are there algebraic manipulations we could apply to the sum that allow us to assign a value to it? Why can't we just define $H = \sum_{k=1}^\infty{\frac{1}{k}}$ to be a number that has certain algebraic properties, like we did with the sum of the natural numbers? Can you prove that $H$ must be $\infty$ using the same process as above?

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    $\begingroup$ For one thing, the zeta function has a pole at $z=1$ (where there harmonic series would live) - so that method actually agrees that the harmonic series diverges. I suspect this means that you either can't derive any algebraic property of $H$ via naive manipulation of the sum (or that, more helpfully, you may be able to derive paradoxical properties) $\endgroup$ – Milo Brandt Jan 12 '15 at 4:13
  • $\begingroup$ @Meelo Could you please elaborate on why the zeta function has a pole at $z = 1$? I get that if you plug it in you get the harmonic series which diverges, but that's exactly the argument I'm confused about in my question. $\endgroup$ – user3002473 Jan 12 '15 at 4:47
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    $\begingroup$ Sort of; there's a series of convergent series converging term-wise to $\sum_{k=1}^{\infty}\frac{1}k$ (and the sums of those series goes to infinity as they approach the harmonic series), whereas there is no such series of sums converging to $\sum_{k=1}^{\infty} k$. So, the analytic continuation is forced to infinity, just from continuity, in the former case, where we can make no such argument to the latter case. $\endgroup$ – Milo Brandt Jan 12 '15 at 5:14
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    $\begingroup$ @whacka You are correct, my mistake. It's also a matter of the odd formatting that makes it less apparent why $c - 4c$ yields the line for $-3c$. For a clearer explanation, see here: en.wikipedia.org/wiki/… $\endgroup$ – user3002473 Jan 12 '15 at 6:16
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    $\begingroup$ I'm flabbergasted that this pseudo-mathematical argument pops up every time again. It is clear that the sum of all naturals diverges and does not exist. Period. $\endgroup$ – Han de Bruijn Apr 18 '15 at 10:08

Letting $$ H=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots $$ and assuming convergence, we have $$ 0=H-2\cdot\frac{1}{2}H=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots\right) - 2\cdot\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\ldots\right)=\\1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\ldots=\ln 2. $$ Since this is a contradiction, the sum must diverge.


To be more clear about what I'm saying here... I'm not saying that there's no way to assign a value to the harmonic series. Clearly there are any number of ways, some sillier and more arbitrary than others. Call a partial function $S$ from infinite sequences to real or complex numbers a summation method if $S(\{a_i\})$ is defined and equal to $\sum_{i=1}^{\infty}a_i$ whenever the sum converges (in the usual sense); and say that $\{a_i\}$ is $S$-summable if $S(\{a_i\})$ is defined. There are a number of properties that you might want your summation method to preserve from ordinary summation. For instance, you might want it to remain linear: $$S(\{ \alpha a_i + \beta b_i \})=\alpha S(\{a_i\}) + \beta S(\{b_i\})$$ whenever $\{a_i\}$ and $\{b_i\}$ are $S$-summable. You might also want it to remain stable under insertion of zeros: if $J:\mathbb{N}\rightarrow\mathbb{N}$ is increasing and $\{a_i\}$ is $S$-summable, then $S(\{a'_i\})=S(\{a_i\})$, where $a'_{J(i)}=a_i$ and $a'_i=0$ for $i\not\in J^{-1}(\mathbb{N})$. I haven't said anything about analytic continuation or whatever other exotic regularization you want to consider. But what I'm saying, above and in the comments, is that no summation method under which either the harmonic series or $1+2+3+\ldots$ is summable can be both linear and stable.

Now, there's a weaker form of stability (under insertion of only finitely many zeros) that you can preserve while making the harmonic series summable, but not while making $1+2+3+\ldots$ summable. So in that specific sense, the latter series (!) is more pathological. In order to sum the natural numbers, you need to give up either linearity or stability, rendering the manipulations in the original post meaningless.

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    $\begingroup$ Is it impossible to derive a contradiction by algebraically manipulating the original divergent sum $1+2+3+\cdots$? (I am not sure what manipulations are being allowed either.) $\endgroup$ – whacka Jan 12 '15 at 6:05
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    $\begingroup$ Assuming convergence, $0=(1+2+3+\ldots)-(0+1+2+3+\ldots)=1+1+1+\ldots$. But then $0=(1+1+1+\ldots)-(0+1+1+1+\ldots)=1$, a contradiction. $\endgroup$ – mjqxxxx Jan 12 '15 at 15:21
  • $\begingroup$ @mjqxxxx I don't understand you're point. In your answer you showed that assuming $H$ converges you get a contradiction, but you also just proved that you get a contradiction when assuming $1 + 2 + 3 + 4...$ converges! Correct me if I'm wrong, but doesn't that mean you didn't really prove that there is any difference between the sums $H$ and $1 + 2 + 3 + 4 + ...$? $\endgroup$ – user3002473 Jan 12 '15 at 16:24
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    $\begingroup$ There's not a difference. In both cases, the summation is achieved by allowing these operations on series: scalar multiplication, insertion of zeroes, and termwise addition. (In summing the natural numbers, you also use analytic continuation.) But these operations are enough to prove a contradiction under the assumption that either series is summable, as shown in this answer and my previous comment. $\endgroup$ – mjqxxxx Jan 13 '15 at 19:19
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    $\begingroup$ @DanielV: No, I'm not reordering any terms, just inserting zeroes and using linearity. Explicitly, $0=(1 + \frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\ldots)-(0+1+0+\frac{1}{2}+0+\frac{1}{3}+ \ldots)=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+ \ldots = \log 2$. $\endgroup$ – mjqxxxx Apr 24 '15 at 2:45

You should note that the Cauchy principlal value of $\zeta(1)$ is $\gamma$ (Euler-Mascheroni constant):


The same value can be obtained using the Ramanujan's summation of harmonic series.

So to directly answer your question, yes, we can assign a value to the sum of harmonic series, and at least the two methods give the same result.

Note that the Ramanujan's sum of $1+2+3+4+...$ is $-\frac1{12}$ so this method can be seen to work in both cases.

  • $\begingroup$ Cool! Why don't we use this as often as we do $1 + 2 + 3 + ... = -\frac{1}{12}$? $\endgroup$ – user3002473 Jan 25 '15 at 14:24
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    $\begingroup$ @user3002473 "As often as we do $1+2+3+\ldots=-\frac1{12}$" would be never for me :) $\endgroup$ – Hagen von Eitzen Apr 19 '15 at 17:01

For anyone who happens upon this question and doesn't understand, like I did, I would like to elaborate on exactly what I specifically was missing in my understanding.

When I posted this question, I didn't know what people meant when they said $\sum_{n=1}^\infty n = -1/12$. Now that I'm older and a bit more experienced, this statement is mathematically wrong. Equating these two quantities is simply a false, pseudo-mathematical inanity. No one told me this when I heard about it, so I always perceived it as mathematical fact.

So what do people actually mean when they write that? Well, we can define the Riemann Zeta Function: $$ \zeta(s) = \sum_{n=1}^\infty\frac1{n^s} $$ for a complex number $s$. We know that this sum converges for $\Re(s)>1$, and it diverges when $s=1$, and $\Re(s)<1$. However, quite frequently we talk about values of the zeta function for which $\Re(s)<1$. This is because we're not really talking about the zeta function, but rather an (and the unique) analytic continuation of the zeta function.

To be specific, if a function $f$ is analytic in a domain $U$, and $V\supset U$, and $F$ is some function analytic on $V$ such that $F(z)=f(z)$ for $z\in U$, then $F$ is an analytic continuation of $f$ to a larger domain. It turns out that this function $F$ is unique. Since the zeta function is analytic in the half-plane $\Re(s)\ge 1$, minus the point $s = 1$, it has an analytic continuation to $\mathbb{C}\backslash\{1\}$, which is uniquely determined by the behaviour of the function $\zeta(s)$ in the domain of convergence of the infinite sum.

One may be curious as to why this analytic continuation cannot be "infinite everywhere" (which would seem to be the only logical choice for such a function given that the sum doesn't converge anywhere outside $\Re(s)>1$). If this were true, then the reciprocal of that analytic continuation would be identically zero on a dense subset of the complex plane, which, by the principle of analytic continuation once again, must imply that the function is identically zero everywhere. But this doesn't make sense since $1/\zeta(s)$ is clearly defined for $\Re(s)>1$.

Hence, there is some meaning to the values $\zeta(s)$ for $\Re(s)<1$, but this does not mean that $\zeta(s) = \sum_{n=1}^\infty\frac1{n^s}$ for all $s$. The sum diverges. The zeta function is extended.

However..., if we abuse our notation a bit, and let $\sum_{n=1}^\infty\frac1{n^s}$ denote the analytic continuation, rather than the actual sum, for $s$ outside the domain of convergence, then we can plug in $s = 1$ to get $$ \zeta(-1) "=" 1 + 2 + 3 + 4 + \cdots $$ and one can prove, properly (i.e. using real math, not pseudo-math), that $\zeta(-1)$ (that is, the analytic continuation of $\zeta$ evaluated at $s=-1$), is indeed equal to $-1/12$.

As for the sum $\sum_{n=1}^\infty\frac1{n}$, this is simply a pole of the Riemann zeta function. The zeta function, on the entire real line, is a meromorphic function (i.e. it is holomorphic/analytic everywhere except on a countable set with no limit points). Functions like these are "allowed" to have singularities on a countable number of disjoint points. They can be thought of as parts where the function "looks like" $1/(s-z)^k$ for some $k$. (There are also "essential singularities" which go to infinity faster than this for any $k$, but the pole of the zeta function is not essential).

Essentially, what this means is that $\sum_{n=1}^\infty n = -1/12$ doesn't contradict some notion of convergence/divergence which implies $\sum_{n=1}^\infty\frac1n = \infty$. Both diverge, we just have a "special way" of evaluating the former sum.


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