共查询到10条相似文献,搜索用时 15 毫秒
1.
Values of new series sum(((2n-1)!ζ(2n))/(2n + 2k)!)α2n from n=1 to ∞,sum(((2n-1)!ζ(2n))/(2n+2k +1)!)β2n from n=1 to ∞ are given concerning ζ(2k + 1),where k is a positive integer,α can be taken as 1,1/2,1/3,2/3,1/4,3/4,1/6,5/6 and β can be taken as 1,1/2.Some previous results are included as special cases in the present paper and new series converges more rapidly than those exsiting results for α = 1/3,or α = 1/4,or α = 1/6. 相似文献
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Values of new series sum(((2n-1)!ζ(2n))/(2n + 2k)!)α2n from n=1 to ∞,sum(((2n-1)!ζ(2n))/(2n+2k +1)!)β2n from n=1 to ∞ are given concerning ζ(2k + 1),where k is a positive integer,α can be taken as 1,1/2,1/3,2/3,1/4,3/4,1/6,5/6 and β can be taken as 1,1/2.Some previous results are included as special cases in the present paper and new series converges more rapidly than those exsiting results for α = 1/3,or α = 1/4,or α = 1/6. 相似文献
3.
林甲富 《数学的实践与认识》1999,(3)
本文用初等的方法研究sum from n=1 to(1/n~(2m))(m∈N)的求和问题。这个问题最先由Euler[8]解决。文献[1][6]给出了另两种求解方法。特别地,对于m=1的情形,即sum from n=1 to ∞(1/n~2)=((π~2)/6),已有许多不同的证明方法,可见文献[2][3][4][5]以及那里的参考文献。本文的想法,主要受文献[5][6]的启发而来的。 相似文献
4.
The object of this note is to prove the followingTheorem Let{a_n}and{b_n}be sequences of real numbers such that0<∑∑a_n~2<+∞and0<∑b_n~2<+∞.Then we have the inequalitysum from m=1 to∞sum from n=1 to∞a_mb_n/m+n<{sum from n=1 to∞(π-θ/n~(1/2)a_n~2}~1/2{sum from n=1 to∞(π-θ/n~(1/2)b_n~2}~1/2 (1)whereθ=3/2~(1/2)-1=1.121320343. 相似文献
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《中国科学 数学(英文版)》2015,(10)
Let λf(n) be the n-th normalized Fourier coefficient of a holomorphic Hecke eigenform f(z) ∈Sk(Γ).We establish that, for any ε 0,1/Xintegral from n=1 to x|sum λ~2f~((n~2)) from n≤x to - c_2x|2dx ?ε X154/101+ε,which improves previous results. 相似文献
6.
吴永锋 《数学的实践与认识》2008,38(19)
利用概率方法给出了形如sum from k=1 to n(1/k)>π/4(sum from k=1 to n((-1)k-1Cnk)1/(k~1/2))与sum from k=1 to n(1/k)<2~(1/2)(sum from k=1 to n((-1)k-1Cnk)1/k2)1/2的组合不等式. 相似文献
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定理1 对于x_k>0,y_k>0,(k=1,2,…,n),则: sum from k=1 to n (x_k~2/ y_k)≥(sum from k=1 to n x_k)~2/sum from k=1 to n y_k (*) 证明由柯西不等式得; sum from k=1 to n y_k·sum from k=1 to n ((x_k~2)/ y_k)≥(sum from k=1 to n x_k)~2 ∴sum from k=1 to n (x_k~2/y_k)≥(sum from k=1 to n x_k)~2/sum from k=1 to n y_k(等号当且仅当x_1/y_1=x_2/y_2=…=x_n/y_n时成立。) 运用上题的结论我们可以解答近几年来国内外有较大难度的一串竞赛题,灵活地运用不等式(*)能收到“一点带一面,一题牵一串”的效果。下面略举几例。以供读者参考。 相似文献
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为什么要证明不等式k~(1/2)+1/(k+1)~(1/2)>(k+1)~(1/2)下面通过实例来说明,高中数学第三册P.147.3(4)题:求证1/1~(1/2)+1/2~(1/2)+…+1/n~(1/2)>n~(1/2)(n>1)。我们用数学归纳法来证明。 (1)当n=2时不等式左边=1/1~(1/2)+1/2~(1/2)=(2+2~(1/2))/2右边=2~(1/2)=(2~(1/2)+2~(1/2))/2,显然不等式成立。 (2)假设当n=k(k>1)时不等式成立, 相似文献
10.
文[1]中讨论了利用差分多项式求sum from k=1 to n f(k)的一个方法。本文将给出直接求sum from k=0 to n f(k)的一个计算公式,作为特例,并给出求自然数方幂和的一个计算公式。设f(k)是K的m(m∈N)次多项式。定义P_m(x)=1/m! x(x-1)…(x-m+1),称为m阶差分多项式,P_0(x)=1称为零阶差分多项式。 相似文献