设an=∫(0-π/4)(tanx)^ndx.求级数∑(an+a(n+2))/n的和.证明当λ>0时,∑an/n^λ收敛

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/10 19:19:22

a[n]+a[n+2] = ∫{0,π/4} (tan(x))^n dx+∫{0,π/4} (tan(x))^(n+2) dx
= ∫{0,π/4} (tan(x))^n·(1+tan²(x)) dx
= ∫{0,π/4} (tan(x))^n·(1/cos²(x)) dx
= ∫{0,π/4} (tan(x))^n·(tan(x))' dx
= (tan(π/4))^(n+1)/(n+1)-(tan(0))^(n+1)/(n+1)
= 1/(n+1).
因此(a[n]+a[n+2])/n = 1/(n(n+1)) = 1/n-1/(n+1).
∑{1 ≤ n} (a[n]+a[n+2])/n = ∑{1 ≤ n} (1/n-1/(n+1)) = 1.
由a[n] = ∫{0,π/4} (tan(x))^n dx > 0,∑a[n]/n^λ是正项级数.
又a[n] = 1/(n+1)-a[n+2] < 1/(n+1) < 1/n,故a[n]/n^λ < 1/n^(1+λ).
而当λ > 0,级数∑1/n^(1+λ)收敛,根据比较判别法,∑a[n]/n^λ也收敛.

设an=∫(0-π/4)(tanx)^ndx.求级数∑(an+a(n+2))/n的和.证明当λ>0时,∑an/n^λ收敛级数收敛性的证明求：设∑an^2收敛,证明：∑an/n绝对收敛? 设级数∑An收敛,且lim(nAn)=a,证明∑n(An-A(n+1))收敛级数an^2收敛,证明级数an除以n收敛（an>0) 设数列{nan}收敛,且级数∑an收敛,证明级数∑n(an-an-1)也收敛设数列{nan}收敛,级数∑n(an-an-1)也收敛,证明级数∑an收敛证明级数绝对收敛若级数∑an绝对收敛,且an≠-1(n=1,2,…),证明：级数∑an/(1+an)收敛. 设级数∑(an)^2收敛则级数∑an/n是收敛还是发散设{nAn}收敛,且级数An收敛,证明:级数n(An-An-1)也收敛级数∑Bn,∑An-A（n-1）收敛,证明∑An*Bn收敛若正项级数∑(1到n)an收敛,则∑(1到n)根号an/n收敛,求证明. 设无穷级数∞∑n=1(an)2和∞∑n=1(bn)2均收敛,证明无穷级数∞∑n=1(an*bn)是绝对收敛.