THE 2002 VIETNAMESE MATHEMATICAL OLYMPIAD

The given equation can be written in the form: $$f_n(x)=\frac{-1}{2}+\frac{1}{x-1}+\frac{1}{4x-1}+\cdots +\frac{1}{k^{2}x-1}+\cdots +\frac{1}{n^{2}x-1}=0(1)$$ 1/ It is easily seen that for every $n\in {\mathbb{N}}^{\ast }$, the function $f_n(x)$ is continuous and decreasing on the interval $(1;+\infty )$. Moreover, $f_n(x)\to +\infty$ when $x\to 1^{+}$ and $f_n(x)\to -1/2$ when $x\to +\infty$. Therefore, for every $n\in {\mathbb{N}}^{\ast }$, the equation (1) has a unique root $x_n>1$.

2/ For $n\in {\mathbb{N}}^{\ast }$, we have: $$\begin{array}{rl}{f}_n(4)& =\frac{-1}{2}+\frac{1}{2^2-1}+\frac{1}{4^2-1}+\cdots +\frac{1}{(2k{)}^2-1}+\cdots +\frac{1}{(2n{)}^2-1}\\ & =\frac{1}{2}\left(-1+1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\cdots +\frac{1}{2k-1}-\frac{1}{2k+1}+\cdots +\frac{1}{2n-1}-\frac{1}{2n+1}\right)\\ & =\frac{-1}{2(2n+1)}<0=f_n(x_n).\end{array}$$ But $f_n(x)$ is decreasing on $(1;+\infty )$, so $$x_n<4\forall n\in {\mathbb{N}}^{\ast }.(2)$$ On the other hand, because for every $n\in {\mathbb{N}}^{\ast }$, the function $f_n(x)$ is differentiable on the interval $(1;+\infty )\supset |x_n;4|$, the theorem of Lagrange gives: for every $n\in {\mathbb{N}}^{\ast }$, there exists $t\in (x_n;4)$ such that $$\frac{f_n(4)-f_n(x_n)}{4-x_n}=f_n^{\prime }(t)=\frac{-1}{(t-1{)}^2}+\frac{-4}{(4t-1{)}^2}+\cdots +\frac{-n^2}{(n^{2}t-1{)}^2}<\frac{-1}{9}$$ $$\text{so}\frac{-1}{2(2n+1)(4-x_n)}<\frac{-1}{9}$$ $$\text{i.e.}{x}_n>4-\frac{9}{2(2n+1)}\forall n\in {\mathbb{N}}^{\ast }$$

Thus, $\forall n\in N^{\ast },4-\frac{9}{2(2n+1)}<x_n<4$. The squeeze theorem proves that ${\lim }_{n\to \infty }{x}_n=4$.