所有正整数之和为-1/12的证明

构造辅助公式 $T_1 = 1 - 1 + 1 - 1 + \dots + 1 - 1$ 和 $T_2 = 1 - 2 + 3 - 4 + \dots$.

$$T_1 = 1 - 1 + 1 - 1 + \dots \\ T_2 = 1 - 2 + 3 - 4 + \dots \\ T = \sum_i^n i = 1 + 2 + 3 + \dots$$

显然, $T_1$的值求得为0.

$$\begin{aligned} 1 - T_1 &= 1 - (1 - 1 + 1 - 1 + \dots) \\ &= 1 - 1 + 1 - 1 + \dots \\ &= T_1\\ \\ 2T_1 &= 1 \\ T_1 &= \frac{1}{2} \end{aligned}$$

接下来, 计算$T_2 + T_2$, 将第二个$T_2$向后移一位.

$$\begin{aligned} T_2 + T_2 &= 1 - 2 + 3 - 4 + \dots + (n-1) - n + \\ &\ \ \ \ \ \ \ \ \ \ \ \ 1 - 2 + 3 - 4 + \dots + (n-1) - n \\ &= 1 - (2 - 1) + (3 - 2) - (4 - 3) + \dots \\ &= 1 - 1 + 1 - 1 + \dots \\ &= T_1 \\ \\ 2T_2 &= \frac{1}{2} \\ T_2 &= \frac{1}{4} \end{aligned}$$

最后计算$T - T_2$:

$$\begin{aligned} T - T_2 &= 1 + 2 + 3 + \dots \\ & - (1 - 2 + 3 - 4 + \dots) \\ &= 4 + 8 + 12 + 16 + \dots \\ &= 4(1 + 2 + 3 + \dots) \\ &= 4T \\ \\ 3T &= -T_2 \\ T &= -\frac{1}{12} \end{aligned}$$

由此证得, $1 + 2 + 3 + \dots = -\frac{1}{12} $.