Math Problem Statement
$\text{用柯西判别法判断正项级数}\sum_{n=1}^\infty\frac{n^n}{3^n\cdot n!}\text{的敛散性}.$
Solution
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Math Problem Analysis
Mathematical Concepts
Series Convergence
Cauchy Root Test
Stirling's Approximation
Formulas
Cauchy Root Test: \(\lim_{n \to \infty} \sqrt[n]{a_n} = L\)
Stirling's Approximation: \(n! \sim \sqrt{2\pi n} \left(\frac{n}{e}\right)^n\)
Formula for the series term: \(a_n = \frac{n^n}{3^n \cdot n!}\)
Theorems
Cauchy Root Test
Stirling's Approximation
Suitable Grade Level
Undergraduate
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