Linked Questions

2
votes
3answers
439 views

Lower bounds for ${{2n} \choose {n}}$ [duplicate]

What are common lower bounds for ${{2n} \choose {n}}$? Edit: I made a mistake in my original question. It doesn't change my question but there is no reason for me to include the mistake.
1
vote
1answer
221 views

What is the non-inductive proof of this inequality? [duplicate]

$$\dfrac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2 \cdot 4 \cdot 6 \cdots (2n)} < \dfrac{1}{\sqrt{3n+1}}.$$ However I've non-inductive proof of $\dfrac{1 \cdot 3 \cdot 5 \cdots(2n-1)}{2 \cdot 4 \cdot 6 \...
3
votes
1answer
233 views

Theta asymptotic for $\binom{2m}{m}$ [duplicate]

Show that $\binom{2m}{m} = \Theta\left(\frac{2^{2m}}{\sqrt{m}}\right)$ without using Stirling's approximation.
1
vote
0answers
80 views

Approximation of factorial - Stirling formula [duplicate]

Possible Duplicate: Elementary central binomial coefficient estimates How can I prove that $$ \binom{n}{n/2} = \Theta\left(\frac{2^n}{\sqrt n}\right) $$ I tried with Stirlings approximation ...
2
votes
0answers
70 views

How to prove the inequality $\frac{4^m}{4\sqrt{m}}\le\binom{2m}{m}$ using chebyshev inequality [duplicate]

What is a good approach to proving this inequality: $$\dfrac{4^m}{4\sqrt{m}}\le\binom{2m}{m}$$ using the Chebyshev inequality: https://en.wikipedia.org/wiki/Chebyshev's_inequality I thought about ...
1
vote
2answers
40 views

Prove using induction the inequality. [duplicate]

$\forall$:n∈${N}$ $\binom{2n}{n}$ $\ge \frac{4^n}{2n+1}$ I tried with no any success...
0
votes
0answers
24 views

Random symmetric walk. [duplicate]

So there's this assignment I'm doing. Let p=1/2. I already proved that for a random walk P(X_n=k) = (n over (n+k)/2) * 2^(-n) Now I need to prove that lim n->inf (n^(1/2))P(X_2n=2k)) = 1/pi. Given ...
60
votes
10answers
12k 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 = \sqrt{...
6
votes
5answers
9k views

Calculating the limit of $[(2n)!/(n!)^2]^{1/n}$ as $n$ tends to $\infty$

Analysis textbook by Shanti Narayan, is asking to prove the limit $\lim {\left({\dfrac{(2n)!}{(n!)^2}}\right)}^{1/n} \to \frac{1}{4}$ as $n \to \infty$. I tried but was unable to find the solution. ...
12
votes
4answers
679 views

Inequality $\binom{2n}{n}\leq 4^n$

I would like to prove the following inequality, for $n=0,1,2,...$, $$ \binom{2n}{n}\leq 4^n.$$ I already proved it by induction, and I'm looking for another proof.
4
votes
6answers
384 views

Approximate solution: factorial and exponentials

If z= $\dbinom{200}{100}/(4^{100})$, what is the value of z? The options are: a. $z<1/3$ b. $1/3<z<1/2$ c. $1/2<z<2/3$ d. $2/3<z<1$ How should I go about solving these type ...
15
votes
3answers
352 views

Computing $\lim\limits_{n\to\infty} \frac{\sqrt{n}}{4^{n}}\sum\limits_{k=1}^{n} \binom{2n-1}{n-k}\frac{ 1}{(2k-1)^2+\pi^2}$

What tools would you recommend me for computing the limit below? $$\lim_{n\to\infty} \frac{\sqrt{n}}{4^{n}}\sum_{k=1}^{n}\frac{\displaystyle \binom{2n-1}{n-k}}{(2k-1)^2+\pi^2}$$ As soon as any ...
4
votes
6answers
904 views

Limit $\lim_{n\to \infty} \frac{1}{2n} \log{2n\choose n}$ [duplicate]

$\lim_{n\to \infty} \frac{1}{2n} \log{2n\choose n}$ I could not approach it beyond these simple steps, $\lim_{n\to \infty} \frac{1}{2n} \log(\frac{2n!}{(n!)^2})$ $=\lim_{n\to \infty} \frac{1}{2n}...
4
votes
5answers
512 views

Proof without induction of the inequalities related to product $\prod\limits_{k=1}^n \frac{2k-1}{2k}$

How do you prove the following without induction: 1)$\prod\limits_{k=1}^n\left(\frac{2k-1}{2k}\right)^{\frac{1}{n}}>\frac{1}{2}$ 2)$\prod\limits_{k=1}^n \frac{2k-1}{2k}<\frac{1}{\sqrt{2n+1}}$ ...
4
votes
5answers
286 views

Convergence of $\sum \frac{(2n)!}{n!n!}\frac{1}{4^n}$

Does the series $$\sum \frac{(2n)!}{n!n!}\frac{1}{4^n}$$ converges? My attempt: Since the ratio test is inconclusive, my idea is to use the Stirling Approximation for n! $$\frac{(2n)!}{n!n!4^n} \...

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