China Mathematical Competition (Complementary Test)

We first prove that $$\frac{x}{1+x}<\ln (1+x)<x,x>0.\stackrel{◯}{1}$$ Let $$\begin{gathered} h(x)=x-\ln (1+x), \\ g(x)=\ln (1+x)-\frac{x}{1+x}. \end{gathered}$$ Then, for $x>0$, $$\begin{gathered} h^{\prime }(x)=1-\frac{1}{1+x}>0, \\ {g}^{\prime }(x)=\frac{1}{1+x}-\frac{1}{(1+x{)}^2}=\frac{x}{(1+x{)}^2}>0. \end{gathered}$$ Therefore, $$h(x)>h(0)=0,g(x)>g(0)=0.$$ This completes the proof of the inequalities ①.

Now let $x=\frac{1}{n}$ in ①. We have $$\frac{1}{n+1}<\ln \left(1+\frac{1}{n}\right)<\frac{1}{n}.\stackrel{◯}{2}$$ Let $$x_n=\sum _{k=1}^n\frac{k}{k^2+1}-\ln n.$$ Then $$\begin{array}{rl}{x}_n-x_{n-1}& =\frac{n}{n^2+1}-\ln \left(1+\frac{1}{n+1}\right)\\ & <\frac{n}{n^2+1}-\frac{1}{n}\\ & =-\frac{1}{n(n^2+1)}<0.\end{array}$$ Therefore, $x_n<x_{n-1}<\cdots <x_1=\frac{1}{2}$.

Furthermore, $$\begin{array}{rl}\ln n& =(\ln n-\ln (n-1))+(\ln (n-1)-\ln (n-2))\\ & +\cdots +(\ln 2-\ln 1)+\ln 1\\ & =\sum _{k=1}^{n-1}\ln \left(1+\frac{1}{k}\right).\end{array}$$ Consequently, $$\begin{array}{rl}{x}_n& =\sum _{k=1}^n\frac{k}{k^2+1}-\sum _{k=1}^{n-1}\ln \left(1+\frac{1}{k}\right)\\ & =\sum _{k=1}^{n-1}\left(\frac{k}{k^2+1}-\ln \left(1+\frac{1}{k}\right)\right)+\frac{n}{n^2+1}\\ & >\sum _{k=1}^{n-1}\left(\frac{k}{k^2+1}-\frac{1}{k}\right)\\ & =-\sum _{k=1}^{n-1}\frac{1}{(k^2+1)k}\\ & >-\sum _{k=1}^{n-1}\frac{1}{(k+1)k}\\ & =-1+\frac{1}{n}>-1.\end{array}$$ This completes the proof.