高数.an=3/2*∫上为(n/n+1)下为0 x^(n-1)*(1+x^n)^1/2 dx 则 lim(n趋于无...an=3/2*∫上为(n/n+1)下为0 x^(n-1)*(1+x^n)^1/2 dx则 lim(n趋于无穷大) n*an=?
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/28 12:02:54
![高数.an=3/2*∫上为(n/n+1)下为0 x^(n-1)*(1+x^n)^1/2 dx 则 lim(n趋于无...an=3/2*∫上为(n/n+1)下为0 x^(n-1)*(1+x^n)^1/2 dx则 lim(n趋于无穷大) n*an=?](/uploads/image/z/3937231-55-1.jpg?t=%E9%AB%98%E6%95%B0.an%3D3%2F2%2A%E2%88%AB%E4%B8%8A%E4%B8%BA%28n%2Fn%2B1%29%E4%B8%8B%E4%B8%BA0+x%5E%28n-1%29%2A%281%2Bx%5En%29%5E1%2F2+dx+%E5%88%99+lim%28n%E8%B6%8B%E4%BA%8E%E6%97%A0...an%3D3%2F2%2A%E2%88%AB%E4%B8%8A%E4%B8%BA%28n%2Fn%2B1%29%E4%B8%8B%E4%B8%BA0+x%5E%28n-1%29%2A%281%2Bx%5En%29%5E1%2F2+dx%E5%88%99+lim%28n%E8%B6%8B%E4%BA%8E%E6%97%A0%E7%A9%B7%E5%A4%A7%29+n%2Aan%3D%3F)
高数.an=3/2*∫上为(n/n+1)下为0 x^(n-1)*(1+x^n)^1/2 dx 则 lim(n趋于无...an=3/2*∫上为(n/n+1)下为0 x^(n-1)*(1+x^n)^1/2 dx则 lim(n趋于无穷大) n*an=?
高数.an=3/2*∫上为(n/n+1)下为0 x^(n-1)*(1+x^n)^1/2 dx 则 lim(n趋于无...
an=3/2*∫上为(n/n+1)下为0 x^(n-1)*(1+x^n)^1/2 dx
则 lim(n趋于无穷大) n*an=?
高数.an=3/2*∫上为(n/n+1)下为0 x^(n-1)*(1+x^n)^1/2 dx 则 lim(n趋于无...an=3/2*∫上为(n/n+1)下为0 x^(n-1)*(1+x^n)^1/2 dx则 lim(n趋于无穷大) n*an=?
an=3/2*∫上为(n/n+1)下为0 x^(n-1)*(1+x^n)^1/2 dx
=3/2*∫上为(n/n+1)下为0 1/n*(1+x^n)^1/2 d(x^n+1) (因为d(x^n+1)=n*x^(n-1)dx)
=3/2*{(1/n)*(2/3)*(1+x^n)^(3/2)|[0,n/(n+1)]}
=1/n*(1+(n/n+1)^n)^(3/2)-1/n
所以,lim(n趋于无穷大)n*an=
lim(n趋于无穷大){(1+(1-1/(1+n))^n)^3/2-1}
现在主要是(1-1/(1+n))^n这部分,变化一下
(1-1/(1+n))^n=[(1+1/-(1+n))^(-(1+n))]^(-1)*[(1+1/-(1+n))^(-1)],
当n趋于无穷大时,(1+1/-(1+n))^(-(1+n))=e,(1+1/-(1+n))^(-1)=1;
所以,(1-1/(1+n))^n=[(1+1/-(1+n))^(-(1+n))]^(-1)*[(1+1/-(1+n))^(-1)]=e^(-1)=1/e;
所以,lim(n趋于无穷大){(1+(1-1/(1+n))^n)^3/2-1}=(1+1/e)^(3/2)-1
即lim(n趋于无穷大)n*an=(1+1/e)^(3/2)-1