2
votes
2answers
53 views

The sum of binomial coefficients up to $k\le n/3$ does not exceed the $k$th coefficient

How would you prove the following (for when $k\leq\frac{n}{3}$)? $$\sum_{i=0}^{k-1} \binom ni \le \binom nk$$
0
votes
1answer
26 views

binominal inequality - checking

I have to show that $\displaystyle \binom{n}{k}<\binom{n}{l}$ for $\displaystyle 0 \le k <l \le\frac{n}{2}$ where $n,k,l$ are an integers. I think I solved it but I'm not sure if my approach is ...
8
votes
4answers
154 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.
2
votes
2answers
81 views

How to prove $n! > n^a$ for all $a\in \mathbb{R}$ (for sufficiently large $n$)?

I've encountered a proof which claims $n! > n^2$ for sufficiently large $n$. I tried using induction to prove it for an arbitrary $a$: $n! > n^a$. Lets assume the claim is true for $n$: $n! ...
6
votes
4answers
147 views

How prove this inequality $(1+\frac{1}{16})^{16}<\frac{8}{3}$

show that $$(1+\dfrac{1}{16})^{16}<\dfrac{8}{3}$$ it's well know that $$(1+\dfrac{1}{n})^n<e$$ so $$(1+\dfrac{1}{16})^{16}<e$$ But I found this $e=2.718>\dfrac{8}{3}=2.6666\cdots$ ...
0
votes
0answers
45 views

$\sum$ of binomial coefficients inequality

Let $m,n$ be positive integers with $m>n$. When is it true that $$m\cdot 5^{m-1}\cdot 3+\binom{m}{3}\cdot 5^{m-3}\cdot 3^3\cdot 2+\cdots +\binom{m}{2k+1}\cdot m^{m-2k-1}\cdot 3^{2k+1}\cdot ...
4
votes
2answers
99 views

Binomial expansion inequality

In a paper I am reading, there is a step that seems to come from the following inequality: $$(1+x)^\alpha \le 1+2^\alpha x,$$ where $0<x<1$. (Also, $3\le \alpha \le 9/2$ in the context of the ...
0
votes
2answers
70 views

Show that $\binom{n}{k}< \binom{n}{k+1}$ if and only if $k < (n-1)/2$ [closed]

Show that $\binom{n}{k} < \binom{n}{k+1}$ if and only if $k < \frac{n-1}{2}$ and then use this to deduce that the maximum of $\binom{n}{k}$ for $k=0,1,\dots,n$ is $\binom{n}{\lfloor ...
1
vote
3answers
39 views

Greatest value of the binomial coefficient. [duplicate]

How should I prove the greatest value of the binomial coefficient $C(n,r)$ occurs for $r=\left[\cfrac{(n+1)}{2}\right]$ ?
2
votes
0answers
32 views

Inequality in matroid theory

Working on a proof in matroid theory I found there is a smooth map from an open set of $(\mathbb{C}^{\ast})^{(d−1)(n−d−1)}$ to a disjoint union of tori $(S^{1})^{\binom{n}{d}-n}.$ As a direct ...
0
votes
2answers
60 views

Largest term in binomial expansion of $(1+n)^k$

The binomial expansion of $(1+n)^k$ is $$(1+n)^k=1+\binom{k}{1}n+\binom{k}{2}n^2+\cdots+\binom{k}{k}n^k.$$ If $n=1$, then the term in the middle is the largest, i.e. when $i=\lfloor k/2\rfloor$ and ...
1
vote
2answers
45 views

Let $C=\dfrac{1}{k}\left [\binom{2k-1}{k}-1 \right ]$ where $k \ge 3$. Show that $C\ge 3$.

I have a problem: Let $C=\dfrac{1}{k}\left [\binom{2k-1}{k}-1 \right ]$ where $k \ge 3$. Show that $C\ge 3$. Any help will be appreciated! Thanks!
2
votes
1answer
92 views

How prove this inequality $\frac{1}{n}\sum_{k=0}^{n-1}\binom{k}{a}\binom{k}{b}\le\frac{1}{a+b+1}\binom{n}{a}\binom{n}{b}$

let $a,b,n$ be positive integer numbers,and such $a,b\le n$, show that $$\dfrac{1}{n}\sum_{k=0}^{n-1}\binom{k}{a}\binom{k}{b}\le\dfrac{1}{a+b+1}\binom{n}{a}\binom{n}{b}$$ this inequality maybe ...
3
votes
1answer
45 views

How prove $\binom{n}{m}\le\left(\frac{en}{m}\right)^m$ [duplicate]

Show that $$\binom{n}{m}\le\left(\dfrac{en}{m}\right)^m$$ where $0<m\le n,m,n\in N^{+}$ My idea: since ...
2
votes
3answers
81 views

Set of solutions for a binomial inequality

I bumped into the following inequality: $${a-b\choose c}{a\choose c}^{-1} \le \exp\left(-\frac{bc}{a}\right)$$ Playing with it a little bit, trying to bound it asymptotically for large $a$'s, using ...
3
votes
1answer
51 views

Very loose bound on sum of first binomials

Let $n\geq k\geq 2$. Is it always true that $$\binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{k}\leq n^k?$$ The left-hand side is dominated by the term $\dfrac{n^k}{k!}$, so the statement should be true. ...
1
vote
1answer
46 views

Upper bound for ${n \choose cn}$

Is it true that for any $0<c<1/2$ and sufficiently large $n'$, there exists a $d <2$ such that ${n \choose cn} < d^n$ for all $n>n'$? Clearly we have to assume $cn$ is an integer. I ...
0
votes
1answer
49 views

Binomial coefficient - first two terms, proof of inequality

I've seen the following and I'm not sure whether it is true or not, and if yes, why it holds. $(1-p)^x \geq 1-p x$ for $p\in (0,1)$ and $x>0$. Do I need some additional Information to prove ...
4
votes
3answers
152 views

How do you prove ${n \choose k}$ is maximum when k is $ \lceil \frac n2 \rceil$ or $ \lfloor \frac n2\rfloor $?

How do you prove ${n \choose k}$ is maximum when k is $ \lceil \frac{n}{2} \rceil $ or $ \lfloor \frac{n}{2} \rfloor$ ? This link provides a proof of sorts but it is not satisfying. From what I ...
1
vote
1answer
44 views

Lower bound functional binomial r.v.

I am trying to find a bound of the type $\mathbb{E}(|B-\frac{N}{2}|) \geq C \sqrt{N}$ Where $B$ is a binomial variable with parameters $(N,\frac{1}{2})$. The bound doesn't need to be very tight in ...
1
vote
1answer
92 views

Convexity of Binomial Term

I am reading a book on the probabilistic method, and the following claim was made: $\dbinom{y}{n}$ is convex. Why is this the case?
12
votes
3answers
328 views

Prove that $\,\,\displaystyle\inf_{n\in\mathbb N}\sum_{k=0}^{p}\lvert\sin{(n+k)^p}\rvert>0$

For any positive integer number $p$, show that $$\inf\left\{ {\left\vert\sin{(n^p)}\right\vert+\left\vert\sin{(n+1)^p}\right\vert+\cdots+ \left\vert\,\sin{(n+p)^p}\right\vert\, :\,n\in ...
1
vote
0answers
55 views

Increasing fraction of sum of binomial coeffients

Let $n$ be a positive integer. Show that the quantity $$ \displaystyle \frac{ \displaystyle \sum_{i=1}^n { n+k \choose i-1 } }{ \displaystyle \sum_{i=1}^n { n+k+1 \choose i } } $$ is ...
2
votes
1answer
61 views

Proving Bernoulli's Inequality for $h<0$

I'm answering question 19 of chapter two of Spivak's Calculus and I can't seem to think of a way of doing it. I don't want to look up the answer so I thought I'd ask for a hint as to the general ...
2
votes
1answer
84 views

Is anyway to prove this: $\prod_{k=1}^{n}(a_{k})< (1/n^n)*(\sum_{k=1}^{n}(\sqrt{1+a_{k}*a_{k+1}}))^n$

$$ \prod\limits_{k=1}^{n}a_{k} < {1 \over n^{n}}\left(\,\sum_{k = 1}^{n}\,\sqrt{1+a_{k}\,a_{k+1}\,}\,\right)^n $$ ak and n are positive real number greater than 0. EDIT: a_{k+1} becomes a_{1} ...
1
vote
2answers
46 views

prove that for $n \ge 4, {{2n}\choose{n}} \ge n\cdot2^n$

Prove that for $n \ge 4$ $${{2n}\choose{n}} \ge n\times2^n$$ I tried like that: $T_4$: ${{8}\choose{4}} = 70 \ge 4\times2^4$ = 64 so it's ok $T_{n+1}$: $$\frac{(2n+2)!}{(n+1)!)(n+1)!} \ge ...
2
votes
0answers
55 views

Upper bound for tail of binomial expansion

Let $P,R,T$ be integer constants with $PR$ much greater than $T$. Suppose I flip a coin $PR$ times, each time (independent of other times) getting heads with probability $1/P$. The probability that I ...
0
votes
2answers
64 views

Prove an upper bound for the binomials

This is (supposed to be) an upper bound on the binomial coefficient: $$ \binom{n}{k} \le \frac{n^n}{k^k(n-k)^{n-k}}$$ If we prove it by induction for all integers $0 \le k \le n/2$, we can easily ...
1
vote
1answer
38 views

Direct proof that $\sum_{k=0}^m \binom{n}{k} \leq n^m$

Is there a short direct proof that $\sum_{k=0}^m \binom{n}{k} \leq n^m$ ? I can prove it by showing it is true for $m=2$ and then proving by induction. Is there a direct non-inductive proof?
6
votes
3answers
206 views

Short and intuitive proof that $\left(\frac{n}{k}\right)^k \leq \binom{n}{k}$

The simple inequality that $\left(\frac{n}{k}\right)^k \leq \binom{n}{k}$ has a number of different proofs. But is there a particularly intuitive, short and elegant proof that uses the natural ...
3
votes
1answer
99 views

Simplest proof that $\binom{n}{k} \leq \left(\frac{en}{k}\right)^k$

The inequality $\binom{n}{k} \leq \left(\frac{en}{k}\right)^k$ is very useful in the analysis of algorithms. There are a number of proofs online but is there a particularly elegant and/or simple proof ...
0
votes
1answer
68 views

Inequality with a sum and factorial

For a homework assignment we have the following question that I'm stuck on. Let $ 0 \leq y \leq 1 $ be given. $\forall m \in \mathbb{N}$, define $ \displaystyle S_m(y)=\sum_{k=0}^m \binom{m}{k}y^k$. ...
4
votes
1answer
233 views

Proving $\binom{2n}{n}\ge\frac{2^{2n-1}}{\sqrt{n}}$

Prove that $$\binom{2n}{n}\ge\dfrac{2^{2n-1}}{\sqrt{n}}$$ By the way: I have see $$\binom{2n}{n}\ge\dfrac{4^n}{2n}=\dfrac{2^{2n-1}}{n}$$ proof: Applying the binomial theorem ...
12
votes
7answers
509 views

Prove that $2^n < \binom{2n}{n} < 2^{2n}$

Prove that $2^n < \binom{2n}{n} < 2^{2n}$. This is proven easily enough by splitting it up into two parts and then proving each part by induction. First part: $2^n < \binom{2n}{n}$. The ...
3
votes
1answer
270 views

Average absolute value of sum with Rademacher random variables

Let $a_1, \ldots, a_n $ be independent Rademacher random variables with distribution $P(a_i=1) = P(a_i=-1) = \frac 12$. Estimate from below $$E \left|\sum_{i=1}^n a_i\right|.$$ I've reduced this ...
5
votes
2answers
262 views

Proof of inequality $\sum\limits_{k=0}^{n}\binom n k\frac{5^k}{5^k+1}\ge\frac{2^n\cdot 5^n}{3^n+5^n}$

Show that $$\sum_{k=0}^{n}\binom n k\frac{5^k}{5^k+1}\ge\frac{2^n\cdot 5^n}{3^n+5^n}$$ where $$\binom n k=\frac{n!}{k!(n-k)!}$$
6
votes
1answer
238 views

Lower bound on binomial coefficient

I encountered the following claim $$\frac{1}{n+1}2^{nH_2(k/n)} \le \binom{n}{k} \le 2^{nH_2(k/n)}$$ where $H_2$ is the binary entropy function. The upper bound is rather well known but how does one ...
1
vote
0answers
231 views

proving inequality for combinatorial sum

If somone can prove the following for every $d\leq r$ (for $d=0,1$ its easy, see below, the case d=r may be also simple, I didn't find something helpful) $$\frac{(d!)^2}{2^{n-2d}}\sum_{k=0}^{n}{n ...
5
votes
4answers
204 views

Proving $\binom{2n}{n}\le 4^n$ for all $n$ by smallest counterexample

Prove $$\binom{2n}{n}\le 4^n$$ for all natural numbers $n$ by smallest (minimal) counterexample. My attempt: First, $$\binom{2n}n = \frac{(2n)!}{(n!)^2} \le 4^n\;.$$ We know that $x\ne 0$ because ...
16
votes
4answers
549 views

Proving that $\sum_{i=0}^{n}\binom{n}{i}i^{n-i}(n-i)^{i}\le\frac{1}{2}n^n$

How can we prove that $$\displaystyle\sum_{i=0}^{n}\binom{n}{i}i^{n-i}(n-i)^{i}\le\dfrac{1}{2}n^n$$ where $\displaystyle\binom{n}{i}=\dfrac{n!}{i!(n-i)!}$. This inequality is very interesting. I ...
1
vote
0answers
176 views

how to solve the following expectation? closed-form expression or approximation

Suppose there is a binomial random variable $X\sim B(n-1,p)$,how to solve the following expectation $$E[(1- b^{X})^{m}]$$ where $b\in (0,1]$ and $m\in \mathbb{N} $ are all constants.I have tried my ...
2
votes
0answers
114 views

Inequality with binomial coefficient

Let $n$ be a natural number, $m\in [-n, n]$. Let $p=0,\ldots, \frac{n+m}{2}$. Show, that for all $p$, $$ {n \choose \left[{\frac{n+m}{2}}\right]}\geq \frac{2^{n+1/2}}{\sqrt{n-p/2}}. $$ Thank you for ...
1
vote
1answer
149 views

Inequality for binomial coefficients

Let $m \leq n, n \leq N$ and $0\leq k \leq m$. I am wondering what is the dependence of $n$ and $N$ that for all $m, k$ $$ \frac{{N-m \choose n-k}}{{N \choose n}}\leq 1. $$ Thank you for your help.
0
votes
2answers
81 views

Inequality with binomial coefficients

Let $n, l \in \mathbb{N}, l\leq n$ be fixed. Let $k\in \mathbb{N}$ with $0 \leq k \leq l$. How to show the following? $$ {2n-l\choose n-k}\leq {2n-l \choose \frac{2n-l}{2}} $$
2
votes
1answer
84 views

Wellner Inequality

Working on an exercise from Shorack's Probability for Statisticians, Ex 4.6 (Wellner): Suppose $T \simeq$ Binomial$(n,p)$. Then use the inequality $$\mu(|X| \ge \lambda) \le ...
1
vote
0answers
108 views

Binomial coefficient intervals (inequality)

For given $N$, $x$ and $k$ such that $0\leq x<N$ and $2\leq k\leq \left\lfloor \frac{N+1-2x}{2}\right\rfloor $, does it exist $p,$ $2\leq p\leq \left\lfloor \frac{N+1}{2}\right\rfloor $ such that ...
4
votes
1answer
234 views

Inequality involving sums of fractions of products of binomial coefficients

Let $n\in\mathbb{N}$. For $0\le l\le n$ consider \begin{equation} b_l:=4^{-l} \sum_{j=0}^l \frac{\binom{2 l}{2 j} \binom{n}{j}^2}{\binom{2 n}{2 j}}\text{.} \end{equation} Do you know a technique how ...
2
votes
3answers
458 views

Two inequalities with binomial coefficients

I have two inequalities that I can't prove: $\displaystyle{n\choose i+k}\le {n\choose i}{n-i\choose k}$ $\displaystyle{n\choose k} \le \frac{n^n}{k^k(n-k)^{n-k}}$ What is the best way to prove ...
3
votes
2answers
277 views

Lower Bound of Central Binomial Coefficients

I would like to prove by induction the following inequality: $\frac{4^n}{n+1} < \binom{2n}{n}$, for all natural numbers n > 1. Any hints?
1
vote
3answers
160 views

Combinatorial inequality $\binom{n}{j}\leqslant 2^n$

I was trying to prove (or to find a counterexample) of the following inequality: $$\binom{n}{j}\leqslant 2^n$$ As I coudn't find a proof/counterexample, I tested some numbers and could see it ...