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An inequality about floor function
03-04-2010, 04:07 PM
Post: #1
An inequality about floor function
For \(k\geq 8\) show that \(\displaystyle {\sum_{n=1}^{\infty} \left( \left[\frac{2^{k+1}}{3^n}\right]-2\left[\frac{2^k}{3^n} \right] \right) > 0}\quad\) where \([x]\) is the floor function \(\forall x \in \mathbf{R} \left([x] \in \mathbf{Z}\cap (x-1,x]\right) \)

证明 \(k\geq 8\) 时 \(\displaystyle {\sum_{n=1}^{\infty} \left( \left[\frac{2^{k+1}}{3^n}\right]-2\left[\frac{2^k}{3^n} \right] \right) > 0}\quad\) 这里 \([x]\) 是 \(x\) 的整数部分: \(\forall x \in \mathbf{R} \left([x] \in \mathbf{Z}\cap (x-1,x]\right) \)
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