Where is the mistake in proving 1+2+3+4+… = -1/12? https://www.youtube.com/watch?v=w-I6XTVZXww#t=30
As I watched the video on YouTube of proving sum of $$1+2+3+4+\cdots= \frac{-1}{12}$$
Even we know that the series does not converge.
First I still can't prove what wrong in infinite sum of $$1-1+1-1+1+\cdots= \frac{1}{2}$$
and I want to know more about of what mistake did they make in the video.
I did research on our forum about this topic. But still not clearly understand about of that proving.
Thank you.
(Sorry about my bad English )
 A: It's not a mistake, just an unfamiliar usage of notation.
The series $1+2+3+4+\cdots$ does not converge, you're right. One doesn't usually write a number to the right of the equals sign in "$1+2+3+4+\cdots=$", because that would usually denote the classical sum of the series, which does not exist.
The video essentially computes the zeta regularized sum, which is a different beast, and which can exist when the classical sum does not. When they write $1+2+3+4+\cdots=-1/12$, they are stating that the regularized sum of the series is $-1/12$, which is correct. This notation is intentionally... shall we say, provocative!
A: The thing is that you need to first decide on your definitions.
The thing is that the operation $+$ is defined on a pair of real numbers. Given a number $x$ and a number $y$, then the operation $+$ tells you what the number $x+y$ is.
Using this, you can also calculate what the sum of any finite amount of real numbers is, by adding parentheses:
$$a_1+a_2+\cdots + a_n = a_1+(a_2+(\cdots + (\cdots + a_n))\cdots )$$
Now, because the finite sum is still a real number, you can manipulate it like other real numbers. For example, if $x=a_1 +\cdots + a_n$, and if I know that $a_2 + \cdots + a_n = 1$, then I also know that $x = a_1 + 1$ and I can calculate $x$. That is, algebraic manipulation works on finite sums.
However, there is no clear and natural way to define an infinite sum. What this means is that, for a general sequence $a_1,a_2,\dots$, the sum $$\sum_{i=1}^\infty a_i$$
is NOT defined.
Usually, we ovecome this problem by saying:

For a sequence $a_n$, if the limit $$\lim_{n\to\infty}\sum_{i=1}^na_i$$ exists, then the infinite sum $$\sum_{i=1}^\infty a_i$$ is equal to that limit.

However, with this definition, $1-1+1-1+\dots$ does NOT exist!
A: The series they are adding are not convergent in the Cauchy sense. There are operating in some naïve way.
The thing with the $-1/12$ is, in a sense, not incorrect. The Riemann zeta function is defined as $\zeta(s) = \displaystyle\sum_{n=1}^\infty \dfrac{1}{n^s}$ for $s\in \mathbb{C}$ such that $\Re(s) > 1$. For $s = -1$ you get $1+2+3+\ldots $ but the definition I explained does not apply. On the other hand, $\zeta$ can be extended to a meromorphic function defined in (almost) all the complex plane, including $s = -1$. In this case $\zeta(-1) = -1/12$.
A: As you spotted, a first weakness in the "proof" is when stating that Grandi's series converges to $\dfrac12$, by taking "the average". This is a completely arbitrary justification. We all know that this sum just does not converge and you can't assign a value to it.
The other steps are equally invalid as the other series do not converge absolutely so that the term-wise manipulations are not justified.
A: In proving that 1-1+1-1+... = 1/2, you add two divergent series, which can sometimes produce a convergent series. However, here, (I'll mark our sum I = 1-1+1-1+...) I+I = 2*I = 2-2+2-2+... and that still diverges, although we shifted one of those to get that 2*I = 1. That's a contradiction.
That is the main problem with this proof. 
Hope this helped!
