Sum of natural numbers equal -1/12 ?

$$\tag{1}1+2+3+4+\cdots =^?-\frac{1}{12}$$

There are many problems with that equality. Can a summation of positive integers lead to a negative quantity ?

The equality implies that

$$\sum_{k=1}^\infty k = -\frac{1}{12}$$

But we already know that

$$S_N = \sum_{k=1}^Nk = \frac{N(N+1)}{2} \to {\infty}$$

Clearly the summation is divergent. By all means of divergence tests. More precisely we have \(S_N = \mathcal{O}(N^2) \). So the summation acts like a parabola that extends to infinity as we increase the number of terms in the summation.

So, what is the origin of (1) ?

Let us look at the zeta function defined as

$$\zeta(s) = \sum_{k=1}^\infty \frac{1}{k^s}$$

We see that

$$\zeta(-1) = \sum_{k=1}^\infty \frac{1}{k^{-1}} = \sum_{k=1}^\infty k$$

Now according to the functional equation

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

Then clearly


It is well know that \( \zeta(2) = \frac{\pi^2}{6}\) and \( \Gamma(2) = 1! = 1 \)

$$\zeta(-1)=-\frac{1}{2\pi^2}\Gamma(2)\zeta(2) = -\frac{1}{12}$$

So we have

$$1+2+3\cdots = -\frac{1}{12}$$

What is wrong with this proof ?!

The problem is that \( \zeta(s) \) is differentiable everywhere except at \( s = 1 \). But the summation formula is not actually convergent unless for \( s >1 \)


$$\zeta(s) = \sum_{k=1}^\infty \frac{1}{n^s}\,\,;\,\, s > 1$$

It doesn’t make sense to say that

$$\zeta(-1) =^? \sum_{k=1}^\infty k$$

This is because the series definition is not defined for all values just for some values but the functional equation holds for \( s\neq 1\).

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