Linked Questions

2
votes
2answers
552 views

Is there a “simple” way of proving Stirlings formula? [duplicate]

Is there any way to derive Stirlings formula that only requires some undergraduate knowledge of calculus, real analysis and perhaps some identitets involving the gamma function, maybe Wallis product, ...
1
vote
2answers
62 views

Estimate how much large is $\lfloor \log{n!\rfloor}$ [duplicate]

Is there a simple method to find or estimate how large $$\lfloor \lg{n!\rfloor}$$ is ? I'd like to find (or estimate) how much digits $2017!^{2017}$ has, or how much is big that number . I tried ...
1
vote
0answers
40 views

Limit of Stirling's approximation as n goes to infinity. [duplicate]

I would like to see some detailed solution for $$\frac{n!}{\sqrt{2\pi n} \left(\frac{n}{e}\right)^n}$$ as $n\to\infty$. I know that the answer is 1 but i am not sure why? Here is what is tried: I ...
673
votes
43answers
96k views

Different methods to compute $\sum\limits_{k=1}^\infty \frac{1}{k^2}$ (Basel problem)

As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem) $$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}.$$ However, Euler was Euler ...
206
votes
28answers
94k views

Evaluating the integral $\int_0^\infty \frac{\sin x} x \ dx = \frac \pi 2$?

A famous exercise which one encounters while doing Complex Analysis (Residue theory) is to prove that the given integral: $$\int_0^\infty \frac{\sin x} x \, dx = \frac \pi 2$$ Well, can anyone prove ...
22
votes
4answers
6k views

Tough integrals that can be easily beaten by using simple techniques

This question is just idle curiosity. Today I find that an integral problem can be easily evaluated by using simple techniques like my answer to evaluate \begin{equation} \int_0^{\pi/2}\frac{\cos{x}}{...
7
votes
4answers
9k views

Summation of logs

Are there any useful identities for quickly calculating the sum of consecutive logs? For example $\sum_{k=1}^{N} log(k)$ or something to this effect. I should add that I am writing code to do this (as ...
2
votes
4answers
453 views

Prove that $1.49<\sum_{k=1}^{99}\frac{1}{k^2}<1.99$

It can be proven by induction that $$\sum_{k=1}^{n}\frac{1}{k^2}\leq2-\frac{1}{n}$$ From here, we can easily acquire the upper bound of the sum $$\sum_{k=1}^{99}\frac{1}{k^2}$$ letting $n=100$. ...
5
votes
3answers
275 views

Evaluate the limit $\lim_{n\to\infty}\frac{n}{\ln n}\left(\frac{\sqrt[n]{n!}}{n}-\frac{1}{e}\right)$

Evaluate $$\lim_{n\to\infty}\frac{n}{\ln n}\left(\frac{\sqrt[n]{n!}}{n}-\frac{1}{e}\right).$$ This sequence looks extremely horrible and it makes me crazy. How can we evaluate this?
0
votes
5answers
174 views

prove that $\lim\limits_{n\to\infty}\frac{3e^n}{n!}=0$ [duplicate]

Why is $\lim\limits_{n\to\infty}\frac{3e^n}{n!}=0$? $e^n=e\cdot e\cdots e$, $n$-times. I tried to find a constant $C>0$ and a $N\in \mathbb{N}$ such that $\frac{3e^n}{n!}\le C\frac{e}{n}$ for $n\...
6
votes
3answers
169 views

Bounds for $\binom{n}{cn}$ with $0 < c < 1$.

Are there really good upper and lower bounds for $\binom{n}{cn}$ when $c$ is a constant $0 < c < 1$? I know that $\left(\frac{1}{c^{cn}}\right) \leq \binom{n}{cn} \leq \left(\frac{e}{c}\right)^...
8
votes
3answers
289 views

Stirling approximation note

During my study to Stirling approximation I find this formula $n! \approx \sqrt{2\pi n} n^{n}e^{-n} $ but we know that $ 0! =1 $ And in this formula if we replace every $ n $ with $ 0$ we will ...
2
votes
4answers
952 views

Limit $c^n n!/n^n$ as $n$ goes to infinity

Let $c>0$ be a real number. I would like to study the convergence of $a_n := c^n n!/n^n$, when $n \to \infty$, in function of $c$. I know (from this question) that $n!>(n/e)^n$, so that $c^n n!/...
5
votes
5answers
128 views

Proof that $n!\leq {(\frac{n+1}{2})}^{n}$ [duplicate]

I don't know what to do with this. Nothing works. I hope somebody can help me to find a decision $$n!\leq {\left(\frac{n+1}{2}\right)}^{n}$$
0
votes
6answers
439 views

If $n$ is a positive integer, Prove that $\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt\frac{2329}{3600}.$

If $n$ is a positive integer, Prove that $$\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt\frac{2329}{3600}.$$ please don't refer to the famous $1+\frac1{2^2}+\frac1{3^2}+\dotsb=\frac{\pi^2}6$. I am ...

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