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 \(\ln(\dfrac{2}{1})+\ln(\dfrac{3}{2})+\dots+\ln(\dfrac{n+1}{n})=\ln(\dfrac{2}{1}\times\dfrac{3}{2}\times\dots\times\dfrac{n+1}{n})=\ln(n+1)\)

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

故 \(0<C<1\)

证毕