Response to bnelo12’s question on Reddit (or the Internet storm on $1 + 2 + \ldots = -1/12$)

bnelo12 writes (slightly paraphrased)

Can you explain exactly how $latex {1 + 2 + 3 + 4 + \ldots = – \frac{1}{12}}$ in the context of the Riemann $latex {\zeta}$ function?

We are going to approach this problem through a related problem that is easier to understand at first. Many are familiar with summing geometric series

$latex \displaystyle g(r) = 1 + r + r^2 + r^3 + \ldots = \frac{1}{1-r}, $

which makes sense as long as $latex {|r| < 1}$. But if you’re not, let’s see how we do that. Let $latex {S(n)}$ denote the sum of the terms up to $latex {r^n}$, so that

$latex \displaystyle S(n) = 1 + r + r^2 + \ldots + r^n. $

Then for a finite $latex {n}$, $latex {S(n)}$ makes complete sense. It’s just a sum of a few numbers. What if we multiply $latex {S(n)}$ by $latex {r}$? Then we get

$latex \displaystyle rS(n) = r + r^2 + \ldots + r^n + r^{n+1}. $

Notice how similar this is to $latex {S(n)}$. It’s very similar, but missing the first term and containing an extra last term. If we subtract them, we get

$latex \displaystyle S(n) – rS(n) = 1 – r^{n+1}, $

which is a very simple expression. But we can factor out the $latex {S(n)}$ on the left and solve for it. In total, we get

$latex \displaystyle S(n) = \frac{1 – r^{n+1}}{1 – r}. \ \ \ \ \ (1)$

This works for any natural number $latex {n}$. What if we let $latex {n}$ get arbitrarily large? Then if $latex {|r|<1}$, then $latex {|r|^{n+1} \rightarrow 0}$, and so we get that the sum of the geometric series is

$latex \displaystyle g(r) = 1 + r + r^2 + r^3 + \ldots = \frac{1}{1-r}. $

But this looks like it makes sense for almost any $latex {r}$, in that we can plug in any value for $latex {r}$ that we want on the right and get a number, unless $latex {r = 1}$. In this sense, we might say that $latex {\frac{1}{1-r}}$ extends the geometric series $latex {g(r)}$, in that whenever $latex {|r|<1}$, the geometric series $latex {g(r)}$ agrees with this function. But this function makes sense in a larger domain then $latex {g(r)}$.

People find it convenient to abuse notation slightly and call the new function $latex {\frac{1}{1-r} = g(r)}$, (i.e. use the same notation for the extension) because any time you might want to plug in $latex {r}$ when $latex {|r|<1}$, you still get the same value. But really, it’s not true that $latex {\frac{1}{1-r} = g(r)}$, since the domain on the left is bigger than the domain on the right. This can be confusing. It’s things like this that cause people to say that

$latex \displaystyle 1 + 2 + 4 + 8 + 16 + \ldots = \frac{1}{1-2} = -1, $

simply because $latex {g(2) = -1}$. This is conflating two different ideas together. What this means is that the function that extends the geometric series takes the value $latex {-1}$ when $latex {r = 2}$. But this has nothing to do with actually summing up the $latex {2}$ powers at all.

So it is with the $latex {\zeta}$ function. Even though the $latex {\zeta}$ function only makes sense at first when $latex {\text{Re}(s) > 1}$, people have extended it for almost all $latex {s}$ in the complex plane. It just so happens that the great functional equation for the Riemann $latex {\zeta}$ function that relates the right and left half planes (across the line $latex {\text{Re}(s) = \frac{1}{2}}$) is

$latex \displaystyle \pi^{\frac{-s}{2}}\Gamma\left( \frac{s}{2} \right) \zeta(s) = \pi^{\frac{s-1}{2}}\Gamma\left( \frac{1-s}{2} \right) \zeta(1-s), \ \ \ \ \ (2)$

where $latex {\Gamma}$ is the gamma function, a sort of generalization of the factorial function. If we solve for $latex {\zeta(1-s)}$, then we get

$latex \displaystyle \zeta(1-s) = \frac{\pi^{\frac{-s}{2}}\Gamma\left( \frac{s}{2} \right) \zeta(s)}{\pi^{\frac{s-1}{2}}\Gamma\left( \frac{1-s}{2} \right)}. $

If we stick in $latex {s = 2}$, we get

$latex \displaystyle \zeta(-1) = \frac{\pi^{-1}\Gamma(1) \zeta(2)}{\pi^{\frac{-1}{2}}\Gamma\left( \frac{-1}{2} \right)}. $

We happen to know that $latex {\zeta(2) = \frac{\pi^2}{6}}$ (this is called Basel’s problem) and that $latex {\Gamma(\frac{1}{2}) = \sqrt \pi}$. We also happen to know that in general, $latex {\Gamma(t+1) = t\Gamma(t)}$ (it is partially in this sense that the $latex {\Gamma}$ function generalizes the factorial function), so that $latex {\Gamma(\frac{1}{2}) = \frac{-1}{2} \Gamma(\frac{-1}{2})}$, or rather that $latex {\Gamma(\frac{-1}{2}) = -2 \sqrt \pi.}$ Finally, $latex {\Gamma(1) = 1}$ (on integers, it agrees with the one-lower factorial).

Putting these together, we get that

$latex \displaystyle \zeta(-1) = \frac{\pi^2/6}{-2\pi^2} = \frac{-1}{12}, $

which is what we wanted to show. $latex {\diamondsuit}$

The information I quoted about the Gamma function and the zeta function’s functional equation can be found on Wikipedia or any introductory book on analytic number theory. Evaluating $latex {\zeta(2)}$ is a classic problem that has been in many ways, but is most often taught in a first course on complex analysis or as a clever iterated integral problem (you can prove it with Fubini’s theorem). Evaluating $latex {\Gamma(\frac{1}{2})}$ is rarely done and is sort of a trick, usually done with Fourier analysis.

As usual, I have also created a paper version. You can find that here.

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4 Responses to Response to bnelo12’s question on Reddit (or the Internet storm on $1 + 2 + \ldots = -1/12$)

  1. Cal Harding says:

    Great article! I think you may have a typo, though. The line after you mention the Basel problem, it should be gamma(1/2) = (-1/2) gamma(-1/2), as t in this case is minus half, not a half. Unless I am mistaken, of course.

    Again, great job!

    • David Lowry-Duda says:

      Oh, you’re right! I make this mistake all the time. Thank you for the correction, and I’ve updated the article.

  2. David Lowry-Duda says:

    I should have guessed that a post derived from reddit would garner the most spam.

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