Let f(n) be the largest integer k such that n! is divisible by \(n^k\). Prove that \[ \lim_{n\to \infty} \frac{(\log n)\cdot \max_{2\le i\le n} f(i)}{n \log\log n}=1.\]
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2011-7 Factorial,
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Let f(n) be the largest integer k such that n! is divisible by \(n^k\). Prove that \[ \lim_{n\to \infty} \frac{(\log n)\cdot \max_{2\le i\le n} f(i)}{n \log\log n}=1.\]
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