1/(1/20+1/21+1/22+1/23+...+1/38+1/39)

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1/(1/20+1/21+1/22+1/23+...+1/38+1/39)
的整数部分是多少?

double c = 20;
double sum = 0;
for (int i = 0; i < 19; i++)
{
sum += 1 / (c + i);
}
Console.WriteLine(1/sum);
1.47023720286133
应该是1吧

1/(1/20+1/21+1/22+1/23+...+1/38+1/39) 把1/20+1/21+1/22+1/23+……+1/38+1/39分之1化简后,整数部分是几. ｜1/21-1/20｜+｜1/22-1/21｜+｜1/23-1/22｜+...｜1/30-1/29｜急关于绝对值｜1/21-1/20｜+｜1/22-1/21｜+｜1/23-1/22｜+...｜1/30-1/29｜ |1/21-1/20|+|1/22-1/21|+|1/23-1/22|+…+|1/30-1/29| 计算:|1/21-1/20|+|1/22-1/21|+|1/23-1/22|...|1/30-1/29| -23-22-21-20-19.-1=? (1-1/19)(1-1/20)(1-1/21)(1-1/22)...(1-1/99)= (19/1-1)(20/1-1)(21/1-1)(22/1)...(99/1-1) 简算:23×21/22+1/22 23×22分之21+22分之1 |1/21-1/20|﹢|1/21-1/22|+|1/23-1/22|+……+|1/30-1/29|解题过程