1
vote
1answer
23 views

Using Stirling's approximiation to show that $(\log(\log n))!$ is $O(n^k)$

I am trying to show the following: Prove, using Stirling's approximiation, that $(\log(\log n))!$ is $O(n^k)$ for some positive constant $k$. Stirling's approximation is $$n!=\sqrt{2\pi ...
1
vote
1answer
53 views

Prove or disprove: $(\frac{1}{n})^n(1 - \frac{1}{n})^{n^2-n} \simeq \frac{1}{n!}$ as $n \rightarrow \infty$.

Prove or disprove: $(\frac{1}{n})^n(1 - \frac{1}{n})^{n^2-n} \simeq \frac{1}{n!}$ as $n \rightarrow \infty$. I'm trying to prove the statement by building on my observation that $(1-\frac{1}{n})^n$ ...
3
votes
0answers
58 views

Which series of numbers effectively translates the factorial to the exponential function?

We have the relation of the Bernoulli numbers $$B_{2n} = (-1)^{n+1}\frac {2(2n)!} {(2\pi)^{2n}} \left(1+\frac{1}{2^{2n}}+\frac{1}{3^{2n}}+\cdots\;\right).$$ For $n>1$, the right hand sum ...
2
votes
1answer
37 views

Problem finding limit - which function is asymptotically larger

I have a homework question, so please don't answer fully but I would appreciate a push in the right direction. Basically we need to figure out if $n^{n+\frac{1}{2}}e^{-n}$ is larger,smaller, or equal ...
3
votes
1answer
109 views

Euler numbers grow $2\left(\frac{2}{ \pi }\right)^{2 n+1}$-times slower than the factorial?

Stirling's approximation of the factorial for even numbers is given by $$ (2n)! \sim \left(\frac{2n}{e}\right)^{2n}\sqrt{4 \pi n}. \tag{1} $$ Further, the Euler numbers grow quite rapidly for large ...
3
votes
1answer
123 views

Asymptotics for sums involving factorials

This question is rather general, but I have recently encountered the following situation in a variety of different settings. Let us suppose that we are given a complicated sum involving factorials ...
1
vote
0answers
138 views

Using the gamma function as an upper and lower bound to the logarithm of a factorial function.

I am trying to find an upper and lower bound for the following function: $$f(x) = \ln(\lfloor\frac{x}{b_1}\rfloor!) - \ln(\lfloor\frac{x}{b_2}\rfloor!) - \ln(\lfloor\frac{x}{b_3}\rfloor!)$$ where ...
0
votes
3answers
251 views

Maple Error on Asymptotic Analysis of $\ln(n)!$

In Maple, the command asmypt($f$,$x$) computes the asymptotic expansion of the function $f$ with respect to the variable $x$ (as $x \rightarrow \infty$). The command asympt(ln(n)!,n); gives the ...
3
votes
6answers
524 views

Proof by contradiction that $n!$ is not $O(2^n)$

I am having issues with this proof: Prove by contradiction that $n! \ne O(2^n)$. From what I understand, we are supposed to use a previous proof (which successfully proved that $2^n = O(n!)$) to find ...
0
votes
1answer
129 views

How to compare big numbers that are outcome of different functions.

How is the best way to compare big numbers? They are result of two functions with different asymptotic growth. For example: Googleplex which is $10^{{10}^{100}}$ to $1000!$
2
votes
2answers
2k views

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?
14
votes
2answers
353 views

On the Limit of Stirling's Approximation

I have recently proven the following curious identity: For real $x \geqslant 1$, \begin{align} \lfloor x \rfloor! = x^{\lfloor x \rfloor} e^{1-x} e^{\int_{1}^{x} \text{frac}(t)/t \ dt} \end{align} ...
13
votes
6answers
749 views

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!), \ \ ...
21
votes
9answers
3k views

What is the purpose of Stirling's approximation to a factorial?

Stirling approximation to a factorial is $$ n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n. $$ I wonder what benefit can be got from it? From computational perspective (I admit I don't ...
14
votes
6answers
3k views

Stirling's formula: proof?

Suppose we want to show that $$ n! \sim \sqrt{2 \pi} n^{n+(1/2)}e^{-n}$$ Instead we could show that $$\lim_{n \to \infty} \frac{n!}{n^{n+(1/2)}e^{-n}} = C$$ where $C$ is a constant. Maybe $C = ...
5
votes
3answers
630 views

Asymptotics of terms and errors in Stirling's Approximation

I have two related questions. Both are related to the asymptotics of Stirling's approximation, which is why I have included them in the same question. I will separate the questions if it is deemed ...