逆天的数学 | 数学科普

前言

所有自然数的和相加为多少呢，即 \(1\)\(+\)\(2\)\(+\)\(3\)\(+\)\(4\)\(+\)\(5\)\(+\)\(6\)\(+\)\(\cdots\)\(=\)\(?\)，估计我们都会说是 \(+\infty\)，但数学家说 \(1\)\(+\)\(2\)\(+\)\(3\)\(+\)\(4\)\(+\)\(5\)\(+\)\(6\)\(+\)\(\cdots\)\(=\)\(-\cfrac{1}{12}\)，那么数学家到底是如何计算的呢？

最简过程

求值：\(S=1-1+1-1+1-1+1-1+\cdots\)

解： \(S=1-1+1-1+1-1+1-1+\cdots\)

即，\(S=\)\(1-(1-1+1-1+1-1+1-1+\cdots)\)

则 \(S=1-S\)，解得 \(S=\cfrac{1}{2}\) .

求值：\(1\)\(+\)\(2\)\(+\)\(3\)\(+\)\(4\)\(+\)\(5\)\(+\)\(6\)\(+\)\(\cdots\)\(=\)\(?\)

解：令 \(S_1=\)\(1\)\(+\)\(2\)\(+\)\(3\)\(+\)\(4\)\(+\)\(5\)\(+\)\(6\)\(+\)\(\cdots\) ①

再令 \(\quad\)\(S_2=\)\(1\)\(-\)\(2\)\(+\)\(3\)\(-\)\(4\)\(+\)\(5\)\(-\)\(6\)\(+\)\(\cdots\) ②

①-② 得到，\(S_1\)\(-\)\(S_2\)\(=\)\(4\)\(+\)\(8\)\(+\)\(12\)\(+\)\(16\)\(+\)\(\cdots\)$=$4(\(1\)\(+\)\(2\)\(+\)\(3\)\(+\)\(4\)\(+\)\(5\)\(+\)\(6\)\(+\)\(\cdots\))\(=\) \(4S_1\)

即 \(-3S_1=S_2\)， 则 \(S_1=-\cfrac{1}{3}S_2\)，以下重点求解 \(S_2\)，

由于 \(S_2=\)\(1\)\(-\)\(2\)\(+\)\(3\)\(-\)\(4\)\(+\)\(5\)\(-\)\(6\)\(+\)\(\cdots\)

则 \(\quad\)\(S_2\)$=$0+\(1\)\(-\)\(2\)\(+\)\(3\)\(-\)\(4\)\(+\)\(5\)\(-\)\(6\)\(+\)\(\cdots\) \(\quad\) 此处用 \(0\) 补位，是为了错位相加，

则 \(2S_2=1-1+1-1+1-1+1-1+\cdots\)，而由上题我们知道，\(1-1+1-1+1-1+1-1+\cdots=\cfrac{1}{2}\)，

故 \(2S_2=\cfrac{1}{2}\)， \(S_2=\cfrac{1}{4}\)，代入 \(S_1=-\cfrac{1}{3}S_2\)，

可得到， \(S_1=-\cfrac{1}{12}\)，即 \(1\)\(+\)\(2\)\(+\)\(3\)\(+\)\(4\)\(+\)\(5\)\(+\)\(6\)\(+\)\(\cdots\)\(=\)\(-\cfrac{1}{12}\)，逆天吧 .