Is this considered a disproof of $1+2+3+4\ldots=-\frac{1}{12}$?

269 Views Asked by At

The series "$1+2+3+4\ldots=-\frac{1}{12}$" didn't seem to make sense to me as it breaks clear rules of series

(but I am yet to research if the rules it broke aren't breakable(doesn't work if broken) or is just a selection by convention of mathematicians)

So I tried to disprove it by its own method, manipulation...

I thought "what would happen if I took all those integers on the left($1+2+3+4+5\ldots$) with their inverse sign to the right except $1$(as in $z+x=2y \rightarrow z=2y-x$)" which is what I did, meaning $1= -\frac{1}{12}-2-3-4-5\ldots$, and wanted to check if the solution would actually be $1$. I have taken pictures of my work to not make this long to read:

trying to assign to $2+3+4+5\ldots$ a value

substituting $-\frac{7}{3}$ and subtracting it from $-\frac{1}{12}$

In short, what I did was, like $-\frac{1}{12}$, I tried to assign a value to "$2+3+4+5+6\ldots$".

Now

$S_1=1+2+3+4+5\ldots$

$S_2=2+3+4+5+6\ldots$

so I wanted to subtract $S_1$ from $S_2$ as in

$S_1-S_2=1+2+3+4+5\ldots$

$...........-(2+3+4+5+6\ldots)$(I wanted to indent it but couldn't so I used dots instead)

doing some steps, the assigned value of $2+3+4+5\ldots$ turned out to be $-\frac{7}{3}$(displayed on my image), subtracting $S_2$ from $S_1$ gives us $1$ on one side

(as in $(1+2+3+4+5+6\ldots)-(2+3+4+5+6\ldots) = 1 $ )

and $-\frac{1}{12}-(-\frac{7}{3})$, which makes the equality(or so called): $-\frac{1}{12}+\frac{7}{3}=1$ which gives $2.25=1$. Clearly false!

Is this considered as a disproof for $1+2+3+4+5\ldots=-\frac{1}{12}$ or is there a mistake I did?

Notes:

  1. I followed the steps used in the Numberphile video to disprove it if there are more approaches to this answer I didn't know please inform me.

  2. I did my research and found nothing useful, so I knew an answer to this question would need asking experts about it because it was never raised before

  3. Sorry if my handwriting isn't that clear for some of you and about those cuts on the right of the paper :)

1

There are 1 best solutions below

0
On BEST ANSWER

Any method of summation which is both stable and linear will fail to give a consistent value to the divergent series $1+2+3+\cdots$ (see this wikipedia article, under "Failure of stable linear summation methods"). You've used both properties when you

  • removed $1$ from the beginning of the series and expected the result to change correspondingly (stability)
  • subtracted the two series from eachother by subtracting term by term (linearity)

So it's not unexpected that you get a contradiction, and it does not really disprove anything. You're just assuming a contradiction when you assume both that the series has a consistent value and that you can manipulate series the way you do.