If s = 1 / (1 / 1980 + 1 / 1981 +... + 1 / 2001), find the integral part of S

If s = 1 / (1 / 1980 + 1 / 1981 +... + 1 / 2001), find the integral part of S

ninety

S=1/1/1980+1/1981+1/1983…… +1 / 2000, find the integer part of SS
Using inequality to solve the problem of elasticity and compactness

Let s = 1 / (1980 + 1 / 1981 + 1 / 1982 +. 1 / 2000), then the original formula = 1 / S; in the fraction, if the numerator is the same, the greater the denominator is, the smaller the denominator is, and the smaller the denominator is, so there are: ① s = 1 / 1980 + 1 / 1981 + 1 / 1982 +... + 1 / 2000 > 1 / 2000 + 1 / 2000 +... + 1 / 2000 = 21 / 2000; ② s = 1 / 1980 + 1 / 1981

S＝1÷（1÷1980＋1÷1981＋1÷1982＋…… Find the integer part of S

99
Accurate 99.4742
Matlab source program:
h=zeros(20);
a=0;
for i=1980:1999
h=1/i;
a=a+h;
end
s=1/a;
disp(s);
In addition:
1/(20/1980)

If s = 1 / (1 / 1980 + 1 / 1981 + 1 / 1982 +. + 1 / 2007), then the integer part of S is——

s=1/(1/1980+1/1981+1/1982+.+1/2007)
>1/(1/1980+1/1980+1/1980+.+1/2000)
=1/(1/198+1/199+8/2000)
>71
s=1/(1/1980+1/1981+1/1982+.+1/2007)