上册 3.5 不等式证明 第31题

\section*{解题过程：} （1）先证：$\displaystyle \frac{2 n}{3} \sqrt{n}<\sum_{k=1}^{n} \sqrt{k}$ ． 由 $\sqrt{n}<\int_{n}^{n+1} \sqrt{x} \mathrm{~d} x<\sqrt{n+1}$ 得 $\displaystyle \left.\sqrt{n} \leqslant \frac{2}{3}[(n+1) \sqrt{n+1})-n \sqrt{n}\right] \leqslant \sqrt{n+1}$ ．于是 $$ \frac{2}{3}(n \sqrt{n}-1)=\frac{2}{3}[(2 \sqrt{2}-1)+(3 \sqrt{3}-2 \sqrt{2})+(4 \sqrt{4}-3 \sqrt{3})+\cdots+(n \sqrt{n}-(n-1) \sqrt{n-1})]<\sqrt{2}+\cdots+\sqrt{n} . $$ 故 $$ \frac{2}{3} n \sqrt{n}<\frac{2}{3}+\sqrt{2}+\cdots+\sqrt{n}<1+\sqrt{2}+\cdots+\sqrt{n} . $$ 下证：$\displaystyle \sum_{k=1}^{n} \sqrt{k}<\left(\frac{2 n}{3}+\frac{1}{2}\right) \sqrt{n}$ ． 由 $\displaystyle 1<\frac{2}{3}+\frac{1}{2}$ 知 $k=1$ 时不等式成立． 假设当 $k=n$ 时不等式成立，看 $k=n+1$ 时的情形： 由 $16 n^{3}+24 n^{2}+9 n<16 n^{3}+24 n^{2}+9 n+1$ 得 $$ (4 n+3)^{2} n<(4 n+1)^{2}(n+1) $$ 变形得 $$ \left(\frac{2 n}{3}+\frac{1}{2}\right) \sqrt{n}<\left(\frac{2 n}{3}+\frac{1}{6}\right) \sqrt{n+1} $$ 于是 $$ \left(\frac{2}{3} n+\frac{1}{2}\right) \sqrt{n}+\sqrt{n+1}<\left[\frac{2}{3}(n+1)+\frac{1}{2}\right] \sqrt{n+1} $$ 即 $$ \sum_{k=1}^{n+1} \sqrt{k}=\sum_{k=1}^{n} \sqrt{k}+\sqrt{n+1}<\left(\frac{2}{3} n+\frac{1}{2}\right) \sqrt{n}+\sqrt{n+1} $$ 即 $k=n+1$ 时不等式成立．由数学归纳法得证． （2）由 $\displaystyle \frac{1}{\sqrt{n+1}}<\int_{n}^{n+1} \frac{1}{\sqrt{x}} \mathrm{~d} x<\frac{1}{\sqrt{n}}$ 得 $\displaystyle \frac{1}{\sqrt{n+1}}<2(\sqrt{n+1}-\sqrt{n})<\frac{1}{\sqrt{n}}$ ．于是 $$ \sum_{k=1}^{n} \frac{1}{\sqrt{k}}<1+2[(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+\cdots+(\sqrt{n}-\sqrt{n-1})]=1+2(\sqrt{n}-1)=2 \sqrt{n}-1 $$ 又由 $\displaystyle 2(\sqrt{n+1}-\sqrt{n})<\frac{1}{\sqrt{n}}$ 得 $$ 2[(\sqrt{2}-1)+(\sqrt{3}-\sqrt{2})+\cdots+(\sqrt{n+1}-\sqrt{n})]<1+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{n}} . $$ 于是 $$ 2(\sqrt{n+1}-1)<1+\frac{1}{\sqrt{2}}+\cdots+\frac{1}{\sqrt{n}} . $$ 综上得 $$ 2 \sqrt{n+1}-2 \leqslant \sum_{k=1}^{n} \frac{1}{\sqrt{k}}<2 \sqrt{n}-1 $$ （3）由 $\displaystyle \frac{1}{n+1}<\int_{n}^{n+1} \frac{1}{x} \mathrm{~d} x<\frac{1}{n}$ 得：$\displaystyle \frac{1}{n+1}<\ln (n+1)-\ln n<\frac{1}{n}$ ．于是 $$ \sum_{k=1}^{n} \frac{1}{k}<1+(\ln 2-\ln 1)+(\ln 3-\ln 2)+\cdots+(\ln n-\ln (n-1))=\ln n+1 . $$ 故 $\displaystyle \sum_{k=1}^{n} \frac{1}{k}-\ln n<1$ ． 下证：$\displaystyle \frac{1}{2} \leqslant \sum_{k=1}^{n} \frac{1}{k}-\ln n$ ． 由 $\displaystyle 1+\frac{1}{2}-\ln 2>\frac{1}{2}$ 知，$i=2$ 时不等式成立． 假设当 $i=n$ 时不等式成立，看 $i=n+1$ 时的情形： 由 $\displaystyle \frac{1}{n+1}<\ln \left(1+\frac{1}{n}\right)$ 知 $$ \sum_{k=1}^{n+1} \frac{1}{k}-\ln (n+1)=\sum_{k=1}^{n} \frac{1}{k}-\ln n+\frac{1}{n+1}-\ln (n+1)+\ln n>\frac{1}{2}+\frac{1}{n+1}-\ln \left(1+\frac{1}{n}\right)>\frac{1}{2} . $$ 即当 $i=n+1$ 时不等式成立．由数学归纳法得证． （4）由 $\displaystyle \frac{1}{k+1}<\int_{k}^{k+1} \frac{1}{x} \mathrm{~d} x<\frac{1}{k}$ 得 $\displaystyle \frac{1}{k+1}<\ln (k+1)-\ln k<\frac{1}{k}$ ．让 $k=1,2, \cdots, n$ ，并相加得 $$ \frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n+1}<\ln (n+1)<1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}, $$ 即 $\mathrm{e}^{\lambda}>(n+1)$ ． （5）由 $\displaystyle \frac{1}{n+1}<\int_{n}^{n+1} \frac{1}{x} \mathrm{~d} x<\frac{1}{n}$ 得 $\displaystyle \frac{1}{n+1}<\ln (n+1)-\ln n<\frac{1}{n}$ ．于是 $$ \sum_{k=1}^{n} \frac{1}{k}-\ln (n+1)>\ln (n+1)-\ln (n+1)=0 . $$ 又由 $\displaystyle \frac{1}{(n+1)^{2}}<\int_{n}^{n+1} \frac{1}{x^{2}} \mathrm{~d} x<\frac{1}{n^{2}}$ 得 $\displaystyle \frac{1}{(n+1)^{2}}<\frac{1}{n}-\frac{1}{n+1}<\frac{1}{n^{2}}$ ．于是 $$ \sum_{k=1}^{n} \frac{1}{k^{2}}=1+\sum_{k=2}^{n} \frac{1}{k^{2}}<1+\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\cdots+\left(\frac{1}{n-1}-\frac{1}{n}\right)<2 . $$ 所以 $\displaystyle \frac{1}{2} \sum_{k=1}^{n} \frac{1}{k^{2}}<1$ ． 下证：$\displaystyle \sum_{k=1}^{n} \frac{1}{k}-\ln (n+1) \leqslant \frac{1}{2} \sum_{k=1}^{n} \frac{1}{k^{2}}$ ． 当 $n=1$ 时， $\displaystyle 1-\ln 2 \leqslant \frac{1}{2}$ 成立． 假设当 $i=n$ 时不等式成立，看 $i=n+1$ 时的情形． $$ \sum_{k=1}^{n+1} \frac{1}{k}-\ln (n+2)=\sum_{k=1}^{n} \frac{1}{k}-\ln (n+1)+\frac{1}{n+1}-\ln (n+2)+\ln (n+1)<\frac{1}{2} \sum_{k=1}^{n} \frac{1}{k^{2}}+\frac{1}{n+1}-\ln \left(1+\frac{1}{n+1}\right) . $$ 由于当 $x>0$ 时有 $\displaystyle x-\frac{x^{2}}{2}<\ln (1+x)