Question

Let ${\left(a_n\right)}_{n\ge 1}$ be a sequence of non-negative such that $a_n\ge a_{2n}+a_{2n+1},$ for all $n\ge 1.$ Prove that for any positive integer N we can find N consecutive terms of the sequence, all equal to zero.

Answer 1

Let us first show that at least one term of the sequence equals zero. Suppose the contrary, that is, all terms are positive integers.
But then $a_1\ge a_2+a_3\ge a_4+a_5+a_6+a_7\ge \cdot \cdot \cdot \ge a_{2^n}+a_{2^{n+1}}+\cdot \cdot \cdot +a_{2^{n+1}-1}\ge 2^n,$ for all $n\ge 1,$ which is absurd. Thus, at least one term,say $a_k$ equals zero. But then $0=a_k,\ge a_{2k}+a_{2k+1}\ge \cdot \cdot \cdot \ge a_{2^{n}k}+a_{2^{n}k+1}+\cdot \cdot \cdot +a_{2^n+k+2^n-1}$ hence $a_{2^{n}k}=a_{2^{n}k+1}=\cdot \cdot \cdot =a_{2^n+k+2^n-1}=0.$
We found $2^n$ consecutive terms of our sequence,all equal to zero, which clearly proves the claim. Observation An nontrivial example of such a sequence is the following:
$a_n=\{\begin{array}{ccc}1,& ifn=2^k,& forsomeintegerk,\\ 0,& otherwise.& \end{array}$