关于调和级数的证明

考虑 (ln (1+x)) 的麦克劳林展开式 (x-dfrac{x^2}{2}+dfrac{x^3}{3}-dfrac{x^4}{4}+dfrac{x^5}{5}-dots)

令 (x=dfrac{1}{v}) 可得 (ln(1+dfrac{1}{v})=dfrac{1}{v}-dfrac{1}{2v^2}+dfrac{1}{3v^3}-dfrac{1}{4v^4}+dots)

故 (dfrac{1}{v}=ln(1+dfrac{1}{v})+dfrac{1}{2v^2}-dfrac{1}{3v^3}+dfrac{1}{4v^4}-dots)

令 (v=1,2,3,dots,n)，并将得到的式子加起来可得：

左边 (=1+dfrac{1}{2}+dfrac{1}{3}+dots+dfrac{1}{n})

右边 (=ln(dfrac{2}{1})+ln(dfrac{3}{2})+dots+ln(dfrac{n+1}{n})+)
(dfrac{1}{2}(1+dfrac{1}{2^2}+dfrac{1}{3^2}+dots+dfrac{1}{n^2})-dfrac{1}{3}(1+dfrac{1}{2^3}+dfrac{1}{3^3}+dots+dfrac{1}{n^3})+dots)

根据对数的性质有 (ln(dfrac{2}{1})+ln(dfrac{3}{2})+dots+ln(dfrac{n+1}{n})=ln(dfrac{2}{1} imesdfrac{3}{2} imesdots imesdfrac{n+1}{n})=ln(n+1))

令 (C=dfrac{1}{2}(1+dfrac{1}{2^2}+dfrac{1}{3^2}+dots+dfrac{1}{n^2})-dfrac{1}{3}(1+dfrac{1}{2^3}+dfrac{1}{3^3}+dots+dfrac{1}{n^3})+dots)

则 (=1+dfrac{1}{2}+dfrac{1}{3}+dots+dfrac{1}{n}=ln(n+1)+C)

现在我们的任务就是求出 (C) 的大致范围。

令 (P(t)=dfrac{1}{t}(1+dfrac{1}{2^t}+dfrac{1}{3^t}+dots+dfrac{1}{n^t}))

则 (C=P(2)-P(3)+P(4)-P(5)+dots)

显然 (P(i)<P(i+1))

一方面 (C=(P(2)-P(3))+(P(4)-P(5))+dots>0)

另一方面 (C=P(2)-(P(3)-P(4))-(P(5)-P(6))+dots<P(2))

而 (P(2)=dfrac{1}{2}(1+dfrac{1}{2^2}+dfrac{1}{3^2}+dots+dfrac{1}{n^2})<dfrac{1}{2}(1+1+dfrac{1}{1 imes 2}+dfrac{1}{2 imes 3}+dots+dfrac{1}{n imes(n+1)}=1-dfrac{1}{2n(n+1)}<1)

故 (0<C<1)

证毕