fimo_2010_algebra_p4.json

{
    "problem_name": "fimo_2010_algebra_p4",
    "informal_statement": "A sequence $x_{1}, x_{2}, \\ldots$ is defined by $x_{1}=1$ and $x_{2 k}=-x_{k}, x_{2 k-1}=(-1)^{k+1} x_{k}$ for all $k \\geq 1$. Prove that $x_{1}+x_{2}+\\cdots+x_{n} \\geq 0$ for all $n \\geq 1$.",
    "informal_proof": "We start with some observations. First, from the definition of $x_{i}$ it follows that for each positive integer $k$ we have\n\n$$\nx_{4 k-3}=x_{2 k-1}=-x_{4 k-2} \\quad \\text { and } \\quad x_{4 k-1}=x_{4 k}=-x_{2 k}=x_{k} \\text {. }\n$$\n\nHence, denoting $S_{n}=\\sum_{i=1}^{n} x_{i}$, we have\n\n$$\n\\begin{gathered}\nS_{4 k}=\\sum_{i=1}^{k}\\left(\\left(x_{4 k-3}+x_{4 k-2}\\right)+\\left(x_{4 k-1}+x_{4 k}\\right)\\right)=\\sum_{i=1}^{k}\\left(0+2 x_{k}\\right)=2 S_{k}, \\\\\nS_{4 k+2}=S_{4 k}+\\left(x_{4 k+1}+x_{4 k+2}\\right)=S_{4 k} .\n\\end{gathered}\n$$\n\nObserve also that $S_{n}=\\sum_{i=1}^{n} x_{i} \\equiv \\sum_{i=1}^{n} 1=n(\\bmod 2)$\n\nNow we prove by induction on $k$ that $S_{i} \\geq 0$ for all $i \\leq 4 k$. The base case is valid since $x_{1}=x_{3}=x_{4}=1, x_{2}=-1$. For the induction step, assume that $S_{i} \\geq 0$ for all $i \\leq 4 k$. Using the relations (1)-(3), we obtain\n\n$$\nS_{4 k+4}=2 S_{k+1} \\geq 0, \\quad S_{4 k+2}=S_{4 k} \\geq 0, \\quad S_{4 k+3}=S_{4 k+2}+x_{4 k+3}=\\frac{S_{4 k+2}+S_{4 k+4}}{2} \\geq 0 .\n$$\n\nSo, we are left to prove that $S_{4 k+1} \\geq 0$. If $k$ is odd, then $S_{4 k}=2 S_{k} \\geq 0$; since $k$ is odd, $S_{k}$ is odd as well, so we have $S_{4 k} \\geq 2$ and hence $S_{4 k+1}=S_{4 k}+x_{4 k+1} \\geq 1$.\n\nConversely, if $k$ is even, then we have $x_{4 k+1}=x_{2 k+1}=x_{k+1}$, hence $S_{4 k+1}=S_{4 k}+x_{4 k+1}=$ $2 S_{k}+x_{k+1}=S_{k}+S_{k+1} \\geq 0$. The step is proved."
}