Question

1+1/2+1/3+1/4+1/5+1/6+1/7+1/8+.....1/n等于多少？

Answer 1

这是调和级数，没有通项公式，有近似公式
1+1/2+1/3+……+1/n=lnn
ln是自然对数，
当n 趋于无穷时，
1+1/2+1/3+……+1/n=lnn+R
R为欧拉常数，约为0.5772.
推理查看百科上有，不知道你能不能看懂
1665年牛顿在他的著名著作《流数法》中推导出第一个幂级数：

ln(1+x) = x - x2/2 + x3/3 - ...

Euler(欧拉)在1734年，利用Newton的成果，首先获得了调和级数有限多项和的值。结果是：

1+1/2+1/3+1/4+...+1/n= ln(n+1)+r(r为常量)

他的证明是这样的：
根据Newton的幂级数有：

ln(1+1/x) = 1/x - 1/2x^2 + 1/3x^3 - ...
于是：
1/x = ln((x+1)/x) + 1/2x^2 - 1/3x^3 + ...
代入x=1,2,...,n，就给出：
1/1 = ln(2) + 1/2 - 1/3 + 1/4 -1/5 + ...
1/2 = ln(3/2) + 1/2*4 - 1/3*8 + 1/4*16 - ...
......
1/n = ln((n+1)/n) + 1/2n^2 - 1/3n^3 + ...
相加，就得到：
1+1/2+1/3+1/4+...1/n = ln(n+1) + 1/2*(1+1/4+1/9+...+1/n^2) - 1/3*(1+1/8+1/27+...+1/n^3) + ......
后面那一串和都是收敛的，我们可以定义
1+1/2+1/3+1/4+...1/n = ln(n+1) + r
Euler近似地计算了r的值，约为0.577218。这个数字就是后来称作的欧拉常数。

Answer 2

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Answer 3

1