0
votes
1answer
30 views

Show that if $a,k\in \mathbb{Z}$ with $0\leq k \leq a$, then $\binom ak=\frac{a!}{k!(a-k)!}=\binom {a}{a-k}$.

I'm reading Ghorpade's A Course in Calculus and Analysis. Given $a\in \mathbb{R}$ and $k\in \mathbb{Z}$, the binomial coefficient associated with $a$ and $k$ is defined by: $$\binom ak = ...
7
votes
2answers
94 views

Does this really converge to 1/e? (Massaging a sum)

Short version: can we prove that $$\sum_{k=0}^n (-1)^k \binom{n}{k}^2 \frac{k!}{n^{2k}} \to \frac1e$$ as $n \to \infty$? Long version: First, consider $$a_n = \sum_{k=0}^n \frac{(-1)^k}{k!}$$ It is ...
0
votes
2answers
45 views

Prove that the set of all subsets of cardinality k of a set of cardinality n has cardinality n choose k

That title is probably very compound and confusing so here's a picture. I'm onto something here. Intuitively this makes sense. Let $A$ be a set such that $|A| = n$. If we choose the first $k$ ...
2
votes
1answer
90 views

How to evaluate $\sum\limits_{k=0}^{n} \sqrt{\binom{n}{k}} $

Can we find $$ \sum_{k=0}^{n} \sqrt{\binom{n}{k}} \quad$$ This problem asked me my friend about a year ago, but I didn't know how to attack problem. Now, I am interesting in solution. Any suggestion? ...
1
vote
5answers
75 views

Calculate $\lim_{n\to\infty}\binom{2n}{n}2^{-n}$

I would like to show that: $$\lim_{n\to\infty}\binom{2n}{n}2^{-n} = \infty$$ I have gotten as far as: $$ \binom{2n}{n}={(2n)!\over (n!)^2}=({n\over1}+1)({n\over2}+1)(\dots)({n\over n}+1)\ge2^n $$ But ...
3
votes
1answer
71 views

Bertrand's postulate proof

Regarding http://michaelnielsen.org/polymath1/index.php?title=Bertrand%27s_postulate I think the last inequality should be $4^{n/3}\le(2n+1)(2n)^{\sqrt{2n}}$. But even when the RHS is decreased from ...
0
votes
1answer
166 views

Using the general binomial theorem to find a series-like expression for $\sqrt 2$

How do I use the general binomial theorem (i.e. the series expansion of ${(1+x)^\alpha}$ for $ |x|<1$) to show the following? $$\sqrt 2=1+\frac 1{2^2}+\frac{1\cdot3}{2!\cdot{2^4}} ...
8
votes
1answer
362 views

Summation of an Infinite Series: $\sum_{n=1}^\infty \frac{4^{2n}}{n^3 \binom{2n}{n}^2} = 8\pi G-14\zeta(3)$

I am having trouble proving that $$\sum_{n=1}^\infty \frac{4^{2n}}{n^3 \binom{2n}{n}^2} = 8\pi G-14\zeta(3)$$ I know that $$\frac{2x \ \arcsin(x)}{\sqrt{1-x^2}} = \sum_{n=1}^\infty ...
2
votes
0answers
18 views

Majorante functions of class $C^k$ to multinomial coeficientes.

Let's $k_1+\ldots +k_p=1$. What functions of class $c^k$ are upper bounds for multinomial coeficientes $$ \begin{pmatrix} n\\k_1,\ldots,k_p\end{pmatrix}=\frac{n}{k_1!\cdot k_2!\cdot\ldots\cdot k_p!} ...
1
vote
2answers
44 views

Proof that $\lim_{n\rightarrow\infty}\frac{n^\alpha}{(1+p)^n}=0$ from Rudin's Principles of Mathematical Analysis, involving the binomial theorem

We are showing that when $\alpha$ and $p$ are real and $p>0$ then $$\lim_{n\rightarrow\infty}\frac{n^\alpha}{(1+p)^n}=0$$ Proof. Let $k$ be an integer such that $k>0$, $k>\alpha$. Then for ...
10
votes
3answers
293 views

Computing: $\lim\limits_{n\to\infty}\left(\prod\limits_{k=1}^{n} \binom{n}{k}\right)^\frac{1}{n}$

I try to compute the following limit: $$\lim_{n\to\infty}\left(\prod_{k=1}^{n} \binom{n}{k}\right)^\frac{1}{n}$$ I'm interested in finding some reasonable ways of solving the limit. I don't find any ...
4
votes
1answer
118 views

Convergence of a sequence of partial binomial sums

I have a sequence $$a_n = (1-p)^n \sum_{\frac{n}{2}\le k \le n} \binom{n}{k} \left( \frac{p}{1-p} \right)^k.$$ I want to show that $a_n\to 0$ when $n\to\infty$ if $0\le p < \frac{1}{2}$. Here's a ...
6
votes
1answer
1k views

Using the Taylor expansion for ${(1+x)}^{-1/2}$, evaluate $\sum_{n=0}^\infty \binom{2n}{n} a^n$

Using the Taylor expansion for $${(1+x)}^{-1/2}$$ we have $${(1+x)}^{-1/2}= \sum_{n=0}^\infty \binom{-1/2}{n} (x^n)$$ for $|x|<1$. But if $|a| <1$, how can we use the above fact to find ...
7
votes
3answers
261 views

Asymptotic difference between a function and its binomial average

The origin of this question is the identity $$\sum_{k=0}^n \binom{n}{k} H_k = 2^n \left(H_n - \sum_{k=1}^n \frac{1}{k 2^k}\right),$$ where $H_n$ is the $n$th harmonic number. Dividing by $2^n$, we ...