Does the equality $1+2+3+.... = -\frac{1}{12}$ lead to a contradiction?

Question

Is $1+2+3+4+5.... = -\frac{1}{12}$ self-contradictory ? I've heared much that $1+2+3+.... = -\frac{1}{12}$, although the fact that this series is diverging. I saw a proof of it by a physicist. In fact, I thought about it a little bit and found that I can derive a contradiction from this equation.

Suppose $x=1+2+3+.... $

So, $x=1+2+3+4+.... = 1 + ( 2+ 3 + 4 + 5 + ... ) $

$= 1+ ( [1 + 1] + [1+2] + [1+3] + [ 1+4] +... ) = 1+ (1 + [1+1] + [2+1] + [3+1] + ...)$

(I have moved the square paraenthese only here one step to the right )

$=1+ (1+2+3+4+5+...)= 1+x$

So, $x=1+x$

So, $0=1$, a contradiction.

This is not the only way to derive the contradiction, I can get a contradiction using another mathod.

My question is, If this equality holds in fact, How can we deal with my contradiction?

I heared that the equality is proved using zeta function which I know nothing about. Is that proof valid? and if yes, How come? I mean, it seems to be self-contradictory then How can we prove something false? this means that mathematics is not sound!( usning mathematical logic terms)

For me, the fault is that, We pre-supposed that this series has a value. In fact, it doesn't and here is the gap. Is that true?

Answer 1

The series does not have that as its value, so any attempts to equate the two sides will not result in a contradiction for the reason that they do not have any actual value. Instead what you have done is show that it cannot converge.

What's going on here is a process called analytic continuation, meaning that a different function is being evaluated here, one that agrees with $\sum_{n=1}^\infty{\frac{1}{n^s}}$ for all the values of $s$ for which the series converges, but is continuous on all of $\mathbb{C}$ (and, in particular, exists for every $s\in \mathbb{C}\setminus\{1\}$). This function, when evaluated at $s=-1$ has the value $-1/12$. This analytic continuation is also unique, meaning that if the series did converge, it would need to converge to this value.

It's very unfortunate that this is always regarded as some kind of unintuitive equality, and some youtube channels like Numberphile make it seem like it is an equality. However, it is important to stress that it is not an equality, and instead just the only possible value that it could be if it did converge (which it doesn't).

Answer 2

Short answer: the 'series' on the left hand side is only a symbol, for which the ordinary rules of arithmetic do not necessarily hold.

Example: the series

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

converges (in the ordinary sense) to $\log 2$. But according to a celebrated theorem by Riemann, by rearranging the terms (which you can always do with a finite sum without changing the value) we can get any real value we want (or even a divergent series)!

Similarly, divergent series such as yours (for which a number of sophisticated extended summation techniques are available) do not necessarily obey things like commutativity or associativity.