¿What's the value of the following limit?

Calculus Level 3

lim ⁡ n → + ∞ 1 ⋅ 4 ⋅ 7 ⋅ … ⋅ ( 3 n − 2 ) 3 n n ! \lim_{n\to+\infty} \frac{1\cdot 4\cdot 7\cdot\ldots\cdot (3n-2)}{3^nn!}


The answer is 0.

Agustín Valverde Ramos by Agustín Valverde Ramos , 54, Spain

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1 solution

  • Fact 1: the sequence a n = 1 ⋅ 4 ⋅ 7 ⋅ … ⋅ ( 3 n − 2 ) 3 n n ! a_n=\dfrac{1\cdot 4\cdot 7\cdot\ldots\cdot(3n-2)}{3^nn!} decreases and its terms are positive: a n + 1 a n = 3 n + 1 3 n + 3 < 1 \frac{a_{n+1}}{a_n} = \frac{3n+1}{3n+3} <1
  • Fact 2: the sequence b n = 1 ⋅ 4 ⋅ 7 ⋅ … ⋅ ( 3 n − 5 ) 3 n n ! b_n=\dfrac{1\cdot 4\cdot 7\cdot\ldots\cdot(3n-5)}{3^nn!} is sumable, because it is hypergeometric b n + 1 b n = 3 n − 2 3 n + 3 , and 3 − ( − 2 ) > 3 \frac{b_{n+1}}{b_n} = \frac{3n-2}{3n+3}, \quad\text{and} \quad 3-(-2)>3

Therefore, lim ⁡ a n = lim ⁡ b n ( 3 n − 2 ) \lim a_n=\lim b_n(3n-2) must be 0

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