Linked Questions
10 questions linked to/from How do you prove that $n^n$ is $O(n!^2)$?
61
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12
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$\lim\limits_{n \to{+}\infty}{\sqrt[n]{n!}}$ is infinite
How do I prove that $ \displaystyle\lim_{n \to{+}\infty}{\sqrt[n]{n!}}$ is infinite?
- 1,648
17
votes
5
answers
2k
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Prove that $n! > \sqrt{n^n}, n \geq 3$
Problem Prove that $n! > \sqrt{n^n}, n \geq 3$.
I'm currently have two ideas in mind, one is to use induction on $n$, two is to find $\displaystyle\lim_{n\to\infty}\dfrac{n!}{\sqrt{n^n}}$. ...
- 12.2k
12
votes
3
answers
3k
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Which is greater, $300 !$ or $(300^{300})^\frac {1}{2}$?
Which is greater among $300 !$ and $\sqrt {300^{300}}$ ?
The answer is $300 !$ (my textbook's answer). I do not know how to solve problems involving such large numbers.
- 3,383
17
votes
6
answers
3k
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A question on the Stirling approximation, and $\log(n!)$
In the analysis of an algorithm this statement has come up:$$\sum_{k = 1}^n\log(k) \in \Theta(n\log(n))$$ and I am having trouble justifying it. I wrote $$\sum_{k = 1}^n\log(k) = \log(n!), \ \ n\log(n)...
- 2,555
6
votes
2
answers
27k
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Why is $\log(n!)$ $O(n\log n)$?
I thought that $\log(n!)$ would be $\Omega(n \log n )$, but I read somewhere that $\log(n!) = O(n\log n)$.
Why?
- 3,445
7
votes
1
answer
10k
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solve $\ln(n!) = \Theta(n\ln(n))$ without stirling approximation
My homework was proving this equation which is simple using Stirling approximation. I was wondering if there is any other method to prove it - without Stirling -
I can prove $\ln(n!) = O(n\ln(n))$ ...
- 691
5
votes
7
answers
230
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Some trouble with a proof on $n!/(\sqrt{n})^n \geq 1$
Originally the problem is to prove that $n! \geq n^{n/2}$.
I reduced this to: $n! \geq (\sqrt{n})^n$ so that:
Prove that $\frac{n!}{(\sqrt{n})^n} \geq 1$.
Each term in $n!$ is divided by the $\sqrt{...
- 227
7
votes
4
answers
248
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Show $\left( n!\right)^2 > n^n$. [duplicate]
If $n > 2$, show that $$\left(n!\right)^2 > n^n$$
Although the problem is pretty obvious, I couldn't come up with a rigorous proof. I was thinking some sort of AM-GM, but couldn't build ...
- 1,025
4
votes
4
answers
814
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How to prove this approximation of logarithm of factorial
In other words, how to prove this equation
$$\lg{n!} = \Theta(n\lg{n})$$
- 7,243
6
votes
2
answers
1k
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Prove $\log x!$ is $\Omega (xlogx)$
Find a positive real number $C$ and a nonnegative real number $x_o$ such that
$Cx$$\log x$ $\leq$ $\log x!$ for all real numbers $x > x_o$.
I tried to expand $\log x!$ into $\log 1 + \log2 +\log3 ...
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