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a condition for the existence of a sequence

Source: China second round 2008 p3

March 3, 2012
functionalgebra unsolvedalgebra

Problem Statement

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:NRf: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(n1)=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.