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CN-LEAGUE 2008 Problem 3

    For all k=1,2,…,2008k=1,2,\ldots,2008 , ak>0a_k>0 .Prove that iff ∑k=12008ak>1\sum_{k=1}^{2008}a_k>1 ,there exists a function f:N→Rf:N\rightarrow R satisfying (1) 0=f(0)<f(1)<f(2)<…0=f(0)<f(1)<f(2)<\ldots ; (2) f(n)f(n) has a finite limit when nn approaches infinity; (3) f(n)−f(n−1)=∑k=12008akf(n+k)−∑k=02007ak+1f(n+k)f(n)-f({n-1})=\sum_{k=1}^{2008}a_kf({n+k})-\sum_{k=0}^{2007}a_{k+1}f({n+k}) ,for all n=1,2,3,…n=1,2,3,\ldots .

    Dataset text AI-MO AoPS. Quoted with source.

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