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I was going through $ \lim\limits_{n \to{+}\infty}{\sqrt[n]{n!}}$ is infinite

I don't follow why/how this is true $(n!)^2 = (1 \cdot n) (2 \cdot (n-1)) (3 \cdot (n-2)) \cdots ((n-2) \cdot 3) ((n-1) \cdot 2) (n \cdot 1)\ge n^n$

I read that At least half the terms in the product for n! will be at least n/2 .. so ?

Please advise.

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    $\begingroup$ In each paired product other than the first and last, one of the terms is $\geq n/2$, and the other is $\geq 2$, so the product is $\geq n$. The first and last terms are $n$, so they are also $\geq n$. $\endgroup$ Oct 17, 2014 at 10:13
  • $\begingroup$ Thanks.. I have a thick brain.. but the aha moments are worth it :) $\endgroup$
    – square_one
    Oct 17, 2014 at 10:16
  • $\begingroup$ No, each of the $n$ factors is $\ge n$, so the product is $\ge n^n$. Take as an example $n=6,\;$ then you have $$(1\times 6)\times(2\times 5)\times(3\times 4)\times(4\times 3)\times(5\times 2)\times(6\times 1)$$ and each (factor) is $\ge 6$ $\endgroup$ Oct 17, 2014 at 10:17

2 Answers 2

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$$(n!)^2 = \overbrace{(1 \cdot n)}^{\ge n} \overbrace{(2 \cdot (n-1))}^{\ge 2\cdot n/2} \overbrace{(3 \cdot (n-2))}^{\ge 2\cdot n/2} \cdots \overbrace{((n-2) \cdot 3)}^{\ge 2\cdot n/2} \overbrace{((n-1) \cdot 2)}^{\ge 2\cdot n/2} \overbrace{(n \cdot 1)}^{\ge n}\ge n^n$$

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If $k \le \frac{n-1}2$, then $4k^2 \le n^2 - 2n + 1$, and so

\begin{align} \left(\frac{n + 1}2 + k\right)\left(\frac{n + 1}2 - k\right) & = \left(\frac{n+1}2\right)^2 - k^2 \\ & = \frac{n^2 + 2n + 1 - 4k^2}{4} \\ & \ge \frac{n^2 + 2n + 1 - (n^2 - 2n + 1)}{4} \\ & = n. \end{align}

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