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