調和級数前n項的和 S_n=1/x+1/(x+a)+1/(x+2a)+…++1/(x+(n-1)a)虽然不能表示成任何n的有理函数,但在实际計算过程中,还是可以找到比直接相加更方便的求和办法。由于 d/dx ln x(x+a)(x+2a)…[x+(n-1)a]=d/dx{ln x+ln(x+a)+ln(x+2a)+…++ln[x+(n-1)a]}=1/x+1/(x+a)++1/(x+2a)+…+1/(x+(n-1)a)=S_n,所以 S_n= The sum of the n terms of the harmonic progression before S_n = 1/x+1/(x+a)+1/(x+2a)+...++1/(x+(n-1)a) cannot be expressed as any n Rational function, but in the actual calculation process, you can still find a more convenient summation method than direct addition. Since d/dx ln x(x+a)(x+2a)...[x+(n-1)a]=d/dx{ln x+ln(x+a)+ln(x+2a)+...+ +ln[x+(n-1)a]}=1/x+1/(x+a)++1/(x+2a)+...+1/(x+(n-1)a)=S_n, So S_n=