How does it equal -1/12?

Question

So all my friends keep telling me that if you add up all the numbers from 1 to infinity, (1+2+3+4...) then the answer is -1/12. They showed me this proof with infinite sums, but I didn't understand it, so all I'm asking is does it really equal -1/12?

Answer 1

It doesn't. there's a function called the reimann zeta function. It has a complicated definition but $\zeta (s) $ will equal $1/1^s + 1/2^s + 1/3^s+... $ IF that expression has a value. If that expression does not have a value $\zeta (s) $ will have a different value.

As it turns out $\zeta (-1) = -1/12$. Now IF 1+2+3+... converged (it doesn't obviously but if it did) then it would have to be that 1+2+3+... = $\zeta (-1)=-1/12$. But 1+2+3+... DOESN'T converge so this is utterly irrelevant and meaningless.

Here's an analogy. It's a different result but it's a similar idea:

Let $N_x = 1 + x + x^2 + x^3 + ....$. To cut to the chase, if $-1 < x < 1$ then $N_x = \frac 1{1 - x}$. This is because $(1 - x)(1 + x + x^2 + x^3 + ...) = (1 + x + x^2 + x^3 + ...) - (x + x^2 + x^3 + x^4 + ...) = 1$.

So, for example $N_{1/2} = 1 + 1/2 + 1/4 + 1/8 + .... = 2 = \frac 1{1 - 1/2}$. And $N_{-1/2} = 1 - 1/2 + 1/4 - 1/8 + 1/16 - 1/32 + ... = \frac 1{1-(-1/2)} = \frac 2 3$.

Neat, huh?

But does that mean $N_{-1} = 1 -1 + 1 - 1 + 1 - 1 + .... = \frac 1{1-(-1)} = 1/2$? Or that $N_{2} = 1 + 2^2 + 2^3 + 2^4 + .... = \frac 1{1-2} = -1$?

Obviously not. Why not? Well, because when we said $(1 - x)(1 + x + x^2 + x^3 + ...) = (1 + x + x^2 + x^3 + ...) - (x + x^2 + x^3 + x^4 + ...) = 1$, we were assuming $(1 + x + x^2 + x^3 + ...)$ converges to a meaningful value. It does converge to a meaningful value if $-1 < x < 1$ and if so then everything we said was true. But if $|x| \ge 1$ then $(1 + x + x^2 + x^3 + ...)$ doesn't converge to a meaningful value and nothing we said makes any sense.

So it's the same thing with $1 + 2 + 3 + 4 + ....$. IF $1 + 2^{-s} + 3^{-s} + ....$ equals anything than that thing equals $\zeta(s)$. But $1 + 2^1 + 3^1 + ...$ doesn't equal anything. So it doesn't equal $\zeta(-1) = -1/12$.

Answer 2

The "usual" sum of this series is $\infty$, since $$ \sum_{j=1}^\infty j = \lim_{m \to \infty} \sum_{j=1}^m j = \lim_{m \to \infty} \frac{m(m+1)}{2} = \infty $$ Your friend is talking about Ramanujan summation, which gives finite values to some divergent series. Don't get mislead by Numberphile!

The Ramanujan sum of $\sum_{j=1}^\infty j$ corresponds to $\zeta(-1)$, where $\zeta$ is the Riemann $\zeta$ function $$ \zeta(z) = \sum_{j=1}^\infty \frac{1}{z^j} \qquad (\Re (z) > 1) $$ analytically continued to the entire complex plane except for $1$.