若∑(∞,n=1)(An)²收敛,则∑(∞,n=1)(An n)是否收敛

1.an-an-1=2(n-1)-1=2(n-1)2n-2=-12n=2-12n=1n=1/22.3+(n-1)(-2)=-2n-53-2n+2=-2n-55=-5题目有错,无解.3.2+(n-1)x

1）1/3,1/52)倒数变换一下即可证明从该步骤得到an=1/(2n-1)3)T=(1/1*1/3+1/3*1/5+1/5*1/7+……+[1/(2n-3)][1/(2n-1)]=1/2(1-1/3

lim[(2n-1)an]=lim{[(2n-1)/n]*n*an}因为llim(2n-1)/n=2所以lim[(2n-1)an]=2lim(n*an)=1推知：lim(n*an)=1/2

a(n+1)-an=[(n+2)-(n+1)]/(n+1)(n+2)=(n+2)/(n+1)(n+2)-(n+1)/(n+1)(n+2)=1/(n+1)-1/(n+2)所以an-a(n+1)=1/n-

a(n+1)=an+3得到为等差数列an=3n-1联立所以n=670

An=[2n/(3n+1)]BnAn-1=[2n/(3n+1)]Bn-1lim(n→∞)an/bn=lim(n→∞)[An-An-1]/[Bn-Bn-1]=lim(n→∞)[2n/(3n+1)][Bn

liman=liman+1an+1=根号下an+6即liman+1=根号下liman+1+6liman+1=3或-2-2舍去（显然an>0)所以lim=3

不知道你的题目是不是这样

a(n+1)-an=2na100-a99=2*100a99-a98=2*99a98-a97=2*98.a2-a1=2*2上式进行相加得到a100-a1=2*2+2*3+.+2*100＝2*（2+3+4

a_(n+1)=(1+1/(n+1))^(n+1)=(1/n+1/n+...+1/n+1/(n+1))^(n+1)>[(n+1)(1/((n^n*(n+1)))开(n+1)次方根]^(n+1)(均值不

n充分大时有|an|1/2从而|1/1+an|

证明：∑an绝对收敛,∴an->0,那么存在N>0,使得n>N时,有|an|1+an>1/2=>1/(1+an)|an|/(1+an)∑|an/(1+an)|∑an/(1+an)收敛

an+1=(1-1/n+1)an则an+1=（n/n+1）an则an+1=（n/n+1）an=（n/n+1）*（n-1/n）an-1=...=n/n+1*（n-1/n）*..1/2*a1=1/n+1所

2a(n+1)-an=n-2/n(n+1)(n+2)2a(n+1)-2/(n+1)(n+2)=an-1/n(n+1)[a(n+1)-1/(n+1)(n+2)]/[an-1/n(n+1)]=1/2bn=

利用stolz定理,是最简单的做法结论是明显的~如果不用stolz定理,做法其实也不难~lim(n→∞)a(n+1)/a(n)=a根据定义：对任意ε>0,存在N>0,当N>N,就有|a(n+1)/a(

http://www.math.org.cn/forum.php?mod=viewthread&tid=28241&extra=

要证明一个命题是真命题,就是证明凡符合题设的所有情况,都能得出结论.要证明一个命题是假命题,只需举出一个反例说明命题不能成立.

An=n（n-1）/2+1

n+1=an+1+2n+2b1=a1+2=1an+an+1+4n+2=0bn+bn+1=0bn+1=-bn{bn}为等比数列公比为-1bn=（-1）^(n-1)an+2n=bn=（-1）^(n-1)a

设lima(n)=Aa(n+1)=根号(a(n)+6)两边平方,得(a(n+1))^2=a(n)+6令n趋向无穷大,两边求极限,得A^2=A+6解得A=3或-2由题设易证a(n)恒≥0,故A≥0所以l