Question

What's the intuition behind the equation 1+2+3+⋯=−1121+2+3+⋯=−112 ?

Answer 1

The sum clearly diverges. This is indisputable. The point of the claim above, that


`1+2+3+\cdots=-\frac1{12}`

is to demonstrate that a sum of infinitely many terms can be manipulated in a variety of ways to end up with a contradictory result. It's an artifact of trying to do computations with an infinite number of terms.

The mathematician Srinivasa Ramanujan famously demonstrated the above as follows: Suppose the series converges to some constant, call it
`C`. Then


`\begin{matrix}C&=&1&+2&+3&+4&+5&+6&+\cdots\\4C&=&&+4&&+8&&+12&+\cdots\\-3C&=&1&-2&+3&-4&+5&-6&+\cdots\end{matrix}`

Now, recall the geometric power series


`\displaystyle\sum_(n\ge0)x^n=1+x+x^2+x^3+\cdots=\frac1{1-x}`

which holds for any
`|x|<1`. It has derivative


`\displaystyle\sum_(n\ge1)nx^(n-1)=1+2x+3x^2+4x^3+\cdots=\frac1{(1-x)^2}`

Taking
`x=-1`, we end up with


`1+2(-1)+3(-1)^2+4(-1)^3+\cdots=1-2+3-4+\cdots=\frac14`

and so


`-3C=\frac14\implies C=-\frac1{12}`

But as mentioned above, neither power series converges unless
`|x|<1`. What Ramanujan did was to consider the sum
`1-2+3-4+\cdots` as a limit of the power series evaluated at
`x=-1`:


`\displaystyle-3C=\lim_(x\to-1^+)\sum_(n\ge1)nx^(n-1)=\lim_(x\to-1^+)\frac1{(1-x)^2}=\frac14`

then arrived at the conclusion that
`C=-\frac1{12}`.

But again, let's emphasize that this result is patently wrong, and only serves to demonstrate that one can't manipulate a sum of infinitely many terms like one would a sum of a finite number of terms.