I was messing around with some math formulas today and came up with a result that I found pretty neat, and I would appreciate it if anyone could explain it to me.

The formula for an infinite arithmetic sum is $$\sum_{i=1}^{n}a_i=\frac{n(a_1+a_n)}{2},$$ so if you want to find the sum of the natural numbers from $1$ to $n$, this equation becomes $$\frac{n^2+n}{2},$$ and the roots of this quadratic are at $n=-1$ and $0$. What I find really interesting is that $$\int_{-1}^0 \frac{n^2+n}{2}dn=-\frac{1}{12}$$ There are a lot of people who claim that the sum of all natural numbers is $-\frac{1}{12}$, so I was wondering if this result is a complete coincidence or if there's something else to glean from it.

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    See terrytao.wordpress.com/2010/04/10/… for one explanation. – Chappers Apr 26 '17 at 3:57
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    @ThomasRasberry I've looked at that, but I don't think this question is the same. What I'm wondering is if the integral's solution in particular is related to the supposed value for the sum of all natural numbers. – Dibbs Apr 26 '17 at 4:13
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    There are a few issues with this question... (i) The first formula is for a finite arithmetic sum, not an infinite one. (ii) The second formula is not the sum of all natural numbers, it is the sum of the natural numbers from $1$ to $n$. (iii) So at no point do you calculate something that could be interpreted as the sum of all natural numbers, so the "indeed" in the last sentence is unwarranted. – Rahul Jun 26 '17 at 9:55
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    Oh, come on. Coincidences don't happen in mathematics. – Frpzzd Jun 26 '17 at 15:48
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    @Rahul Well, the question still begs if $\int_{-1}^0\frac{x(x+1)}2~\mathrm dx=\zeta(-1)$ is a coincidence or not, neither of which require the bringing of the old conversation on sums of the natural numbers and all that. – Simply Beautiful Art Jun 26 '17 at 17:09
up vote 14 down vote accepted

We have Faulhaber's formula:

$$\sum_{k=1}^n k^p = \frac1{p+1}\sum_{j=0}^p (-1)^j\binom{p+1}jB_jn^{p+1-j},~\mbox{where}~B_1=-\frac12$$

$$\implies f_p(x)=\frac1{p+1}\sum_{j=0}^p(-1)^j\binom{p+1}jB_jx^{p+1-j}$$

We integrate the RHS from $-1$ to $0$ to get

$$I_p=\int_{-1}^0f_p(x)~\mathrm dx=\frac{(-1)^p}{p+1}\sum_{j=0}^p\binom{p+1}j\frac{B_j}{p+2-j}$$

Using the recursive definition of the Bernoulli numbers,


Using the well known relation $B_p=-p\zeta(1-p)$, we get


So no coincidence here!

  • Lol $\ddot \smile$. I was going to answer this one ! ... I was mildly miffed when I saw that @Nilknarf had already done so ... & now you SBA ! – Donald Splutterwit Jun 26 '17 at 17:20
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    :P I can't sit by and let interesting questions go unanswered :P – Simply Beautiful Art Jun 26 '17 at 17:33
  • Hopefully that downvote was a mis-click.... unless someone's found a flaw in my answer. – Simply Beautiful Art Jun 27 '17 at 22:09

Funny that you should use that method to demonstrate that the sum of all natural numbers is $-\frac{1}{12}$. I've seen the same claim demonstrated in a completely different way.

In case you aren't familiar with it, the Riemann Zeta function is defined as $$\zeta(s)=\sum_{n=1}^\infty n^{-s}=\frac{1}{1^s}+\frac{1}{2^s}+\frac{1}{3^s}+...$$ Your infinite sum of positive integers is given by $\zeta(-1)$.

The Riemann zeta function satisfies the reflection formula $$\zeta(1-s)=2(2\pi)^{-s}\cos\big(\frac{\pi s}{2}\big)\Gamma(s)\zeta(s)$$ Which suggests the same thing, if you evaluate it at $s=2$: $$\zeta(1-2)=2(2\pi)^{-2}\cos\big(\frac{2\pi}{2}\big)\Gamma(2)\zeta(2)$$ $$\zeta(-1)=\frac{1}{2\pi^2}\cos(\pi)\Gamma(2)\zeta(2)$$ $$\zeta(-1)=\frac{1}{2\pi^2}\cdot-1\cdot 1! \cdot \frac{\pi^2}{6}$$ $$\zeta(-1)=-\frac{1}{12}$$

However, I am somewhat inclined to think that this is merely a coincidence, as strange as that may sound. This is because the other "strange" values that can be obtained using the $\zeta$ function, such as $$\zeta(0)=-\frac{1}{2}$$ $$\zeta(-2)=0$$ are not "confirmed" or obtained from your integral in any analogous way. I would advise you to look for some ways to produce these values from your integral so that you (and any more potential answerers) will have more to work with.

Just because the integral of your function between its two zeroes is equal to this magical number does not mean that there is any sort of connection. For example, consider the formula for the sum of the first $n$ square numbers: $$\frac{n(n+1)(2n+1)}{6}$$ This has zeroes at $0, -\frac{1}{2},$ and $1$. When I take the integral from $-1$ to $0$ of this, I get $$\int_{-1}^0 \frac{n(n+1)(2n+1)}{6}dn=0$$ Which is equal to our strange value for $\zeta(-2)$, which also is a sum of squares. But wait, there's more!

$\zeta(0)$ is the sum of $1+1+1+...$. The partial sum for this would be the sum of $1$, $n$ times, and so this sum is given by $n$. And guess what? $$\int_{-1}^0 n dn=-\frac{1}{2}$$ Which also agrees with our strange value for $\zeta(0)$.

And this just keeps working! $\zeta(-3)$ is, supposedly, $\frac{1}{120}$. The sum of the first $n$ cubes has the formula $$\frac{n^2(n+1)^2}{4}$$ and guess what? $$\int_{-1}^0 \frac{n^2(n+1)^2}{4} dn=\frac{1}{120}$$

Now I've given you lots of examples to work with, and perhaps to draft a proof that the "weird" values of $\zeta(-s)$ is equal to the integral, from $-1$ to $0$, of the formula for the sum of the first $n$ $s$th powers. And hopefully this answer was helpful enough to you to be deserving of that hefty $50$-point bounty... ahem...

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