无和序列的Erds倒数和的上界

若严格递增的正整数序列A={a_1,a_2…}的每一项a_n都不能表示成它前面的一些不同项之和,则称A为无和序列.通过改进无和序列的性质,定义三类新的序列:k_1序列、K_(21)序列和K_(22)序列.借助于k_1,k_(21)和k_(22)序列,利用Erd(o|¨)s,Levine和O'Sullivan以及Chen的思想和方法,进一步改进无和序列的Erd(o|¨)s倒数和的上界.设A={a_1,a_2,…}是一个无和序列,令ρ(A)=∑_(k=1)~∞1/(a_k),本文将ρ(A)的上界改进到2.86以下,证明了ρ(A)2.8570.     

关于数列{Γ(n+1/2)/√nΓ(n/2)}+∞n=1的极限

用单调有界定理证明了数列{Γ(n+1/2)/√nΓ(n/2)}+∞n=1的奇子列和偶子列极限的存在性,并给出了该数列的极限为1/√2.本文所得结果对帮助学生更好理解概率统计论中t分布密度函数的极限函数的证明有一定指导作用.     

巧用(-1)~n求摆动数列的通项

在数列中,我们经常会碰到求形如:1,-1,1,-1,…或-1,1,-1,1,…等数列的通项,很显然,我们只要利用(-1)n进行符号的调整,就能很快求出数列的通项公式,我们不禁会思考,在其它的摆动数列中,还能不能用(-1)n去求通项?引例(2004年北京卷)定义“等和数列”:在一个数列中,如果每一项与它的后一项的和都为同一个常数,那么这个数列叫做等和数列,这个常数叫做该数列的公和.已知数列{an}是等和数列,且a1=2,公和为5,那么a18的值为,且这个数列的前21项和S21的值为.分析由等和数列的定义,易知a2n-1=2,a2n=3(n=1,2,…),故a18=3.当n为偶数时,Sn=52n;当n为奇数…     

推广的Wallis数列的单调性

讨论了推广的Wallis数列{(n+c)(1/2)∫π/20sin~nxdx}(n≥1,c为非负常数)的单调性.黄永忠等(2016)证明了当0≤c≤1/2该数列严格递增;当1/2c≤1该数列对于充分大的n严格递减.本文给出了此结论的一个新的简洁证明,并对相关问题做了讨论.进一步,证明了当且仅当c2π~2-16/16-π~2=0.609945…,推广的Wallis数列为严格递减数列.     

数表与数列问题的求解

数表问题历来是竞赛的热点.这类问题题型新颖,信息性强,往往与数列整合在一起,可以较好地考查学生的观察问题、处理数据、阅读理解、获取信息、归纳推理等能力.本文谈谈如何求解这类题的一些方法.1利用特殊数列沟通纵横联系数表中的横与列都可以看作数列.如果题目已知某项或某列是等差或等比数列,应沟通数表中各数的纵横联系,利用特殊数列的通项公式、求和公式等来求解.例1(南充市2004年高三模拟题)n2个正数排成n行n列(如表1),其中每行数都成等差数列,每列数都成等比数列,且所有公比相同.已知a24=1,a42=18,a43=136,求∑k=n1akk.a11a12a13a14…     

关于一些新的Smarandache数列

主要目的是利用初等方法研究LCM序列和SLOS数列的性质,并给出一个包含这两个数列的恒等式及渐近公式.结论:证明了L(2n)L(n)=SLO2S.S L2OnS(+22n 1--1)1和ln(L(n))=n+O n exp-(l nc(lnln n n))5153.     

新题征展 (77)

A题组新编1.已知集合A={1,2,3,4},B={0,1,-1},现建立从A到B的映射f:x→f(x).(1)若A、B分别为函数的定义域和值域,则这样的不同函数有个;(2)若f(1)相似文献   

从一道训练题谈数列积式不等式的证明策略

题目 (武汉市2011届高中毕业生五月供题训练(三)理科第21题)已知数列{an}的前n项和为Sn,且Sn=1/2na(n+1)(n∈ N+),其中a1=1.(1)求数列{an}的通项公式;(2)设bk＝(a1a3…a2k-1)/(a2a4…a2k)(n∈N+).证明:bn＜1/√2an+1这是一道融数列、不等式与函数为一体的综合问题,主要考查学生的思维能力.第(2)问的证明具有一定的难度,从证法上看,它注重通性通法,也不回避特殊技巧,既可用大众化的常规证法,也可用证明不等式的一些特殊技巧,很好地区分了考生思维的层次性.由第(1)问可知an=n,从而原不等式即为:1/2·3/4·5/6·…·.     

Levine-O'Sullivan序列的几个结论

Levine-O'Sullivan序列Q={q_1,q_2,..}定义为:q_1=1,q_n=max{(k+1)(n-q_k)}(1≤k≤n-1,n≥2).序列Q在研究无和序列的性质时扮演了关键角色,发挥了重要的作用.本文研究了Levine-O'Sullivan序列Q中的项q_n的大小估计,对Chen提出的关于序列Q的两个猜想进行了讨论,并证明了序列Q的倒数和满足3.028464ρ(Q)3.029891.     

一类递推数列的周期性

设k,m是适合k>2的正整数,p=2cos(2π)/k.本文证明了:如果数列A={an}n=0∞满足递推关系an+2m=pan+m-an(n≥0),则A是周期数列,它的最小正周期是km的约数.另外,给出了最小正周期小于km的非零数列的例子.     

$A$-序列倒数和的一个注记

An infinite integer sequence {1 ≤ a1 〈 a2 〈 ... } is called A-sequence, if no ai is sum of distinct members of the sequence other than ai. We give an example for the A-sequence, and the reciprocal sum of elements is∑1/ai〉 2.065436491, which improves slightly the related upper bounds for the reciprocal sums of sum-free sequences.     

On the reciprocal sum of a sum-free sequence

Let A = {1≤a1相似文献   

On a question about sum-free sequences

An increasing sequence of positive integers {n₁, n₂, …} is called a sum-free sequence if every term is never a sum of distinct smaller terms. We prove that there exist sum-free sequences {n_k} with polynomial growth and such that lim_k→∞ n_k+1/n_k = 1.     

On the structure of large sum-free sets of integers

A set of integers is called sum-free if it contains no triple (x, y, z) of not necessarily distinct elements with x + y = z. In this paper, we provide a structural characterisation of sum-free subsets of {1, 2,..., n} of density at least 2/5 ? c, where c is an absolute positive constant. As an application, we derive a stability version of Hu’s Theorem [Proc. Amer. Math. Soc. 80 (1980), 711–712] about the maximum size of a union of two sum-free sets in {1, 2,..., n}. We then use this result to show that the number of subsets of {1, 2,..., n} which can be partitioned into two sum-free sets is Θ(2^4n/5), confirming a conjecture of Hancock, Staden and Treglown [arXiv:1701.04754].     

Topology and topological sequence entropy

Let X be a compact metric space and T:X-→X be continuous.Let h*(T)be the supremum of topological sequence entropies of T over all the subsequences of Z+and S(X)be the set of the values h*(T)for all the continuous maps T on X.It is known that{0}■S(X)■{0,log 2,log 3,...}∪{∞}.Only three possibilities for S(X)have been observed so far,namely S(X)={0},S(X)={0,log 2,∞}and S(X)={0,log 2,log 3,...}∪{∞}.In this paper we completely solve the problem of finding all possibilities for S(X)by showing that in fact for every set{0}?A?{0,log 2,log 3,...}∪{∞}there exists a one-dimensional continuum XAwith S(XA)=A.In the construction of XAwe use Cook continua.This is apparently the first application of these very rigid continua in dynamics.We further show that the same result is true if one considers only homeomorphisms rather than continuous maps.The problem for group actions is also addressed.For some class of group actions(by homeomorphisms)we provide an analogous result,but in full generality this problem remains open.     

LIMIT THEOREM FOR A CLASS OF SEQUENCE OF WEAKLY DEPENDENT RANDOM VARIABLES

In this paper, we consider a central limit theorem for the sequence of stationary m-dependent random variables, the variance of which is possibly infinite. Theorem. Let {Xn, n=l, 2,...} be a sequence of stationary m-dependent random variables with means zero. The following conditions are satisfied. (i) \[{M^2}\int_{{\text{|}}{X_1}| > M} {dP} /\int_{{X_1}| < M} {X_1^2} dP \to 0{\kern 1pt} {\kern 1pt} {\kern 1pt} (M \to \infty )\] (ii) \[\int_{\{ {X_1}| < M,|{X_i}| < M} {X_1^{}} {X_i}dP/\int_{|{X_1}| < M} {X_1^2} dP \to 0{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} (M \to \infty )\] then there are constants Bsubsub>0, such that \[\frac{1}{{{B_n}}}\sum\limits_{i = 1}^n {{X_1}} \] converges in distribution N(0,1).     

反常积分敛散性的根值判别法

由于积分与级数在理论上是统一的,因此有关正项级数的根式判别法可被推广以判别无穷限积分和瑕积分的敛散性.设f(x)是[a,+∞)上的非负函数,li mx→+∞xf(x)=ρ,则当ρ1时,反常积分∫a+∞f(x)dx收敛,而当ρ1时,反常积分∫a+∞f(x)dx发散;设f(x)是(a,b]上的非负函数,a为瑕点,xli→ma+(f(x))x-a=ρ,则当ρ1时,反常积分∫abf(x)dx收敛,而当ρ1时,反常积分∫baf(x)dx发散.     

折叠立方体图的邻点可区别全色数

简单图G的全染色是指对G的点和边都进行染色．称全染色为正常的如果没有相邻或关联元素染同一种颜色．简单图G=（VE）的正常全染色^称为它的邻点可区别全染色如果对任意两个相邻顶点u、v,有H（u）≠H（v）,其中H（u）={（u））U{^（uw）｜uw∈E（G））而H（v）={h（u）}U{h（vx）｜vx∈E（G））．G...     

图Gn的优美表示

将给出三个结果：（i）如果图G是SZ（｜S｜=n≥2）上的整数和图,那么0∈S当且仅当图G至少有一个（n-1）度顶点;（ii）图G（G≠K2）是至少有两个零点的整数和图当且仅当G■K2·Gn;（iii）设图G（G≠K2）是SZ上的整数和图,｜S｜=n＋2,n∈N＋.若图G至少有两个零点,则S={mx｜m=-1,0,1,2,…,n;x∈Z且x≠0}.     

一类幂级数的和函数计算公式

设{an}是r阶等差数列,{bn(x)}是等比数列,根据幂级数和函数的定义,同时使用数学归纳法,可导出幂级数∞∑n=1 anbn(x)的和函数的一个计算公式.