Theorem 1. The sum of all natural numbers are $-\frac{1}{12}$, or $$\sum_{n=1}^{\infty}n=-\frac{1}{12}.$$
Proof. Let S1, S2 and S3 be following sequences. $$\begin{align*} S_1 &= 1 - 1 + 1 - 1 + 1 - 1 + \cdots \\ S_2 &= 1 - 2 + 3 - 4 + 5 - 6 + \cdots \\ S_3 &= 1 + 2 + 3 + 4 + 5 + 6 + \cdots \end{align*}$$ Despite the fact that S1 is a divergent sequence, the following property may apply. $$\begin{align*} S_1 &= 1 - 1 + 1 - 1 + 1 - 1 + \cdots \\ S_1 &= \,\,\,\,\,\text{ }\text{ }\text{ } 1 - 1 + 1 - 1 + 1 - \cdots \\ 2S_1 &= 1 \\ \therefore S_1 &= \frac{1}{2} \end{align*}$$ Similarly, S2 may be manipulated. $$\begin{align*} S_2 &= 1 - 2 + 3 - 4 + 5 - 6 + \cdots \\ S_2 &= \,\,\,\text{ }\text{ }\text{ }\text{ } 1 - 2 + 3 - 4 + 5 - \cdots \\ 2S_2 &= 1 - 1 + 1 - 1 + 1 - 1 + \cdots \\ 2S_1 &= S_1 \\ 2S_1 &= \frac{1}{2} \\ \therefore S_2 &= \frac{1}{4} \end{align*}$$ Lastly, the relationship between S2 and S3 may be investigated. $$\begin{align*} S_3 &= 1 + 2 + 3 + 4 + 5 + 6 + \cdots \\ S_2 &= 1 - 2 + 3 - 4 + 5 - 6 + \cdots \\ S_3 - S_2 &= 4 + 8 + 1 2 + 16 + \cdots \\ &= 4(1 + 2 + 3 + 4 + \cdots) \\ &= 4S_3 \\ 3S_3 &= -S_2 \\ 3S_3 &= -\frac{1}{4} \\ \therefore S_3& = -\frac{1}{12} \end{align*}$$ ◻