China Mathematical Competition (Complementary Test)

Proof of necessity: Assume that there exists $\{x_n\}$ satisfying (1)–(3). Notice that the expression in (3) can be written as $$x_n-x_{n-1}=\sum _{k=1}^{2008}{a}_k(x_{n+k}-x_{n+k-1}),n\in \mathbb{N}.$$ As $x_0=0$, we then have $$\begin{array}{rl}{x}_n& =\sum _{l=1}^n(x_l-x_{l-1})=\sum _{l=1}^n\sum _{k=1}^{2008}{a}_k(x_{l+k}-x_{l+k-1})\\ & =\sum _{k=1}^{2008}\sum _{l=1}^n{a}_k(x_{l+k}-x_{l+k-1})\\ & =\sum _{k=1}^{2008}{a}_k(x_{n+k}-x_k).\end{array}$$ From (2) we are able to define $b={\lim }_{n\to \infty }{x}_n$. Let $n\to \infty$ in the above expression. Then we have $$\begin{gathered} b=\sum _{k=1}^{2008}{a}_k(b-x_k)=b\sum _{k=1}^{2008}{a}_k-\sum _{k=1}^{2008}{a}_k{x}_k \\ <b\sum _{k=1}^{2008}{a}_k. \end{gathered}$$ Therefore, ${\sum }_{k=1}^{2008}{a}_k>1$.

Proof of sufficiency: Assume that ${\sum }_{k=1}^{2008}{a}_k>1$. Define a polynomial function by $$f(s)=-1+\sum _{k=1}^{2008}{a}_k{s}^k,s\in [0,1].$$ $f(s)$ is strictly increasing on the interval $[0,1]$. In the meantime, $$f(0)=-1<0,f(1)=-1+\sum _{k=1}^{2008}{a}_k>0,$$ and there then exists a unique $0<s_0<1$ such that $f(s_0)=0$. Now, define $x_n={\sum }_{k=1}^n{s}_0^k$, $n\in \mathbb{N}$. It is easy to see that $\{x_n\}$ satisfies (1) and $$x_n=\sum _{k=1}^n{s}_0^k=\frac{s_0-s_0^{n+1}}{1-s_0}.$$ On the other hand, ${\lim }_{n\to \infty }{s}_0^{n+1}=0$ as $0<s_0<1$. Then we have $$\underset{n\to \infty }{\lim }{x}_n=\underset{n\to \infty }{\lim }\frac{s_0-s_0^{n+1}}{1-s_0}=\frac{s_0}{1-s_0}.$$ This means that $\{x_n\}$ satisfies (2). Finally, we have $$0=f(s_0)=-1+\sum _{k=1}^{2008}{a}_k{s}_0^k.$$ That is to say, ${\sum }_{k=1}^{2008}{a}_k{s}_0^k=1$. Then we have $$\begin{array}{rl}{x}_n-x_{n-1}& =s_0^n\\ & =\left(\sum _{k=1}^{2008}{a}_k{s}_0^k\right)s_0^n\\ & =\sum _{k=1}^{2008}{a}_k{s}_0^{n+k}\\ & =\sum _{k=1}^{2008}{a}_k(x_{n+k}-x_{n+k-1}).\end{array}$$ Therefore, $\{x_n\}$ also satisfies (3). This completes the proof.