# Does the sum of all natural numbers REALLY equal -1/12, as stated by Ramanujan?

## If so, why and how? If not, why and how? Explanations don't need to be too in-depth; a superficial explanation shall suffice.

Jan 13, 2017

No

#### Explanation:

Srinivasa Ramanujan found a method for associating finite values with divergent series - i.e. series which have no finite sum. The method is called "Ramanujan summation".

Expressed very simply, you could write a "proof" like this:

The Maclaurin series expansion of $\frac{1}{1 + x} ^ 2$ is:

$\frac{1}{1 + x} ^ 2 = 1 - 2 x + 3 {x}^{2} - 4 {x}^{3} + \ldots$

which converges for $\left\mid x \right\mid < 1$.

If we put $x = 1$ then we find the non-convergent sum:

$\frac{1}{4} = \frac{1}{1 + 1} ^ 2$

$\textcolor{w h i t e}{\frac{1}{4}} = 1 - 2 + 3 - 4 + 5 - 6 + \ldots$

$\textcolor{w h i t e}{\frac{1}{4}} = \left(1 + 2 + 3 + 4 + 5 + \textcolor{w h i t e}{0} 6 + \ldots\right) -$
$\textcolor{w h i t e}{\frac{1}{4} =} \left(\textcolor{w h i t e}{0 +} 4 \textcolor{w h i t e}{+} \textcolor{w h i t e}{0} + 8 \textcolor{w h i t e}{+} \textcolor{w h i t e}{0} + 12 + \ldots\right)$

$\textcolor{w h i t e}{\frac{1}{4}} = \left(1 + 2 + 3 + 4 + 5 + 6 + \ldots\right) - 4 \left(1 + 2 + 3 + 4 + 5 + 6 + \ldots\right)$

$\textcolor{w h i t e}{\frac{1}{4}} = \left(- 3\right) \left(1 + 2 + 3 + 4 + 5 + 6 + \ldots\right)$

Then divide both ends by $- 3$ to find:

$- \frac{1}{12} = 1 + 2 + 3 + 4 + 5 + 6 + \ldots$

This is not really a proof, but a sort of intuitive motivation and picture of an idea which can be formalised into a method.