Question

“ 设f（n）=1+1/2+1/3+1/4+……+1/n,是否存在关于自然数n的函数g(n)使f(1)+f(2)+f(3

Answer 1

f(n)-1=1/2+1/3+...+1/n

f(1)+f(2)+...+f(n-1)=1+(1+1/2)+(1+1/2+1/3)+...+[1+1/2+1/3+...1/(n-1)]

=(n-1) + (n-2)/2 + (n-3)/3 +...+ 2/(n-2)+1/(n-1)

=1 + [(n-2)/2+1] + [(n-3)/3+1] + ...+ [2/(n-2)+1] + [1/(n-1)+1]

=1 + n/2 + n/3 + ...+ n/(n-2)+n/(n-1)

=n[1/n + 1/2 + 1/3 + ...+ 1/(n-2)+1/(n-1)]

=n[f(n)-1]

所以：g(n)=n[f(n)-1]/[f(n)-1]=n