微积分正项级数敛散性问题.
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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)
![微积分正项级数敛散性问题.](/uploads/image/z/17958630-30-0.jpg?t=%E5%BE%AE%E7%A7%AF%E5%88%86%E6%AD%A3%E9%A1%B9%E7%BA%A7%E6%95%B0%E6%95%9B%E6%95%A3%E6%80%A7%E9%97%AE%E9%A2%98.)
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.