神马作文网微积分正项级数敛散性问题.
来源：学生作业帮 编辑：神马作文网作业帮 分类：数学作业 时间：2024/11/20 05:24:21
微积分正项级数敛散性问题.
请用比值判别法（达朗贝尔判别法）判断敛散性：∑ n^3 ×sin(π/3^n)
请用比值判别法（达朗贝尔判别法）判断敛散性：∑ n^3 ×sin(π/3^n)
既然知道要用比值判别法,那么其实就是一个求极限的问题.
记a[n] = n³sin(π/3^n),则lim{n → ∞} a[n+1]/a[n]
= lim{n → ∞} (n+1)³/n³·sin(π/3^(n+1))/sin(π/3^n)
= (lim{n → ∞} (n+1)/n)³·(lim{n → ∞} sin(π/3^(n+1))/sin(π/3^n))
= lim{n → ∞} sin(π/3^(n+1))/sin(π/3^n)
= (lim{n → ∞} sin(π/3^(n+1))/(π/3^(n+1)))/(3·lim{n → ∞} sin(π/3^n)/(π/3^n))
= 1/3.
最后用到了n → ∞时π/3^n与π/3^(n+1) → 0,以及重要极限lim{x → 0} sin(x)/x = 1.
根据比值判别法,级数∑a[n] = ∑n³sin(π/3^n)收敛.
注:如果等价无穷小代换运用熟练的话,求极限时代换sin(π/3^n) π/3^n,
sin(π/3^(n+1)) π/3^(n+1))写起来会比较简单,当然本质上是一样的.
记a[n] = n³sin(π/3^n),则lim{n → ∞} a[n+1]/a[n]
= lim{n → ∞} (n+1)³/n³·sin(π/3^(n+1))/sin(π/3^n)
= (lim{n → ∞} (n+1)/n)³·(lim{n → ∞} sin(π/3^(n+1))/sin(π/3^n))
= lim{n → ∞} sin(π/3^(n+1))/sin(π/3^n)
= (lim{n → ∞} sin(π/3^(n+1))/(π/3^(n+1)))/(3·lim{n → ∞} sin(π/3^n)/(π/3^n))
= 1/3.
最后用到了n → ∞时π/3^n与π/3^(n+1) → 0,以及重要极限lim{x → 0} sin(x)/x = 1.
根据比值判别法,级数∑a[n] = ∑n³sin(π/3^n)收敛.
注:如果等价无穷小代换运用熟练的话,求极限时代换sin(π/3^n) π/3^n,
sin(π/3^(n+1)) π/3^(n+1))写起来会比较简单,当然本质上是一样的.