∑(n 6) (n∧3 n∧2-4n 5)的敛散性-搜答案-拍题作业网

1/(n+1)(n+2)+1/(n+2)(n+3)+1/(n+3)(n+4)=1/(n+1)-1/(n+2)+1/(n+2)-1/(n+3)+1/(n+3)-1/(n+4)=1/(n+1)-1/(n+

由于lim((1+n)/(1+n²))/(1/n)=lim(n²+n)/(1+n²)=1所以此级数和1/n有相同敛散性1/n发散,所以此级数发散

n-1=再问：29/20n-4n=

1/n(n+1)+1/(n+1)(n+2)＋1/(n+2)(n+3)+1/(n+3)(n+4)+.+1/(n+99)(n+100)=1/n-1/(n+1)+1/(n+1)-1/(n+2)+...+1/

利用（1+1/n）^n在n趋于无穷极限为e.构造[1+(-6)/(3n^2+4)]^[(3n^2+4)/(-6)]形式.结果为e^(-2)

先看着图片先,可能不清晰.

=lim[1+1/(n+3)]^2n=lim[1+1/(n+3)]^2(n+3)·lim[1+1/(n+3)]^(-6)=e^2·1=e^2

再问：错的，答案是三分之一再答：

这个就是二项式定理的逆用1+2C(n,1)+4C(n,2)+...+2^nC(n,n)=1*C（n,0)+2C(n,1)+4C(n,2)+...+2^nC(n,n)=（1+2）^n=3^n明教为您解答

证明：（1）当n=1时，左边=1×2×3=6，右边=1×2×3×44＝6=左边，∴等式成立．（2）设当n=k（k∈N*）时，等式成立，即1×2×3+2×3×4+…+k×(k+1)×(k+2)＝k(k+

［n/2]+[n/3]+[n/4]+[n/5]+[n/6]=(30n+20n+15n+12n+10n)/60=87n/60=29n/60题目是不是打错了..等于29吧?这样n=60再问：是69~~~└

lim(n->∞)[(n^2+3n-8)/(4n^2+2n+3)]=lim(n->∞)[(1+3(1/n)-8(1/n^2))/(4+2(1/n)+3(1/n^2))]=1/4

原式=(x^3n)^2×y^(2n)^2=4^2×3^2=144原式=(xy)^10n第二个写错了吧

后项比前项=[2^(n+1)×(n+1)!/(n+1)^(n+1)]/2^(n)×(n)!/(n)^(n)]=2/(1+1/n)^n趋于2/e

lim【n→∞】(2n²-3n+1)/(n+1)×sin(1/n)=lim【n→∞】(2n²-3n+1)/(n+1)×(1/n)=lim【n→∞】(2n²-3n+1)/(

先证明对于任意x≠0,1+xf(0)=1>0,即1+x

lim（n→∞）(3∧n-2∧n)/((3∧n+1)-(2∧n+1))分子分母同除以3^n,得lim（n→∞）(1-(2/3)∧n)/((3-((2/3)∧n×2)=(1-0)/(3-0)=1/3

x∧3n=43n=logx4...以x为底,4的对数6n=2logx4=logx4^2=logx16y∧2n=32n=logy34n=logy9(1)x∧6n×y∧4n=x^(logx16)*y^(l

(n+1)(n+2)/1+(n+2)(n+3)/1+(n+3)(n+4)/1=(n+1)(n+2)+(n+2)(n+3)+(n+3)(n+4)=(n+2)(n+1+n+3)+n^2+7n+12=(n+