MOOCULUS微积分-2: 数列与级数学习笔记 5. Another comparison test

此课程(MOOCULUS-2 "Sequences and Series")由Ohio State University于2014年在Coursera平台讲授。

PDF格式教材下载 Sequences and Series

本系列学习笔记PDF下载（Academia.edu） MOOCULUS-2 Solution

Summary

• Let $N > 1$ be an integer, and consider a series $sum_{n=1}^infty a_n$. The series we get by removing the first $N-1$ terms, namely $$sum_{n=N}^infty a_n$$ is called a tail of the given series.
• Let $N > 1$ be an integer. The series $$sum_{n=1}^infty a_n$$ converges if and only if $$sum_{n=N}^infty a_n$$ converges. This could be shortened to "The series converges iff a tail of the series converges," or even just to the slogan that convergence depends on the tail.
• Limit Comparison Test Suppose $a_n geq 0$ and $b_n geq 0$. Then if $$lim_{n o infty} frac{a_n}{b_n} = L > 0,$$ the series $sum_{n=1}^infty a_n$ converges if and only if $sum_{n=1}^infty b_n$ converges.

Examples

1. Does the series $$sum_{n=153}^infty frac{1}{n^2}$$ converge?

Solution:

Yes! This series is a tail of the convergent $p$-series $$sum_{n=1}^infty frac{1}{n^2}$$ in this case, $p = 2$.

2. Show that $$sum_{n=0}^infty {n^5over 5^n}$$ converges.

Solution:

First, we can easily prove that $2^n > n^5$ when $n$ is sufficient large (by Mathematical Induction). Suppose that $2^n > n^5$, then when $n > 1$, we have $$(n+1)^5=n^5+5n^4+10n^3+10n^2+5n+1 < n^5+5n^4+10n^3+10n^2+6n$$ $$ < n^5+5n^4+10n^3+16n^2 < n^5+5n^4+26n^3 < n^5+31n^4$$ And thus $(n+1)^5 < 2n^5 < 2cdot 2^n=2^{n+1}$ whenever $n > 31$. Actually we can find the minimal $n$ that $2^n > n^5$, R code:

f1 = function(x) 2^x
f2 = function(x) x^5
for (i in 2:100){
  if (f1(i) > f2(i)){
    print(i)
    break
  }
}
## [1] 23

(Note that when $n=2, 3, cdots$, $2^n < n^5$, thus we search from $n=2$.) Back to this problem, we have $${n^5 over 5^n} < {2^n over 5^n}=({2over5})^n$$ That is, the tail series $$sum_{n=23}^{infty}({2over5})^n$$ converges. By the comparison test, the smalled series $$sum_{n=23}^{infty}{n^5 over 5^n} $$ also converges, so does the original series $$sum_{n=0}^infty {n^5over 5^n}$$

3. Does the series $$sum_{n=52}^infty frac{n^4 - 3n + 5}{2n^5 + 5n^3 - n^2}$$ converges or diverge?

Solution:

By limit comparison test, set $$a_n=frac{n^4 - 3n + 5}{2n^5 + 5n^3 - n^2}, b_n={1over n}$$ We have $$lim_{n oinfty}{a_nover b_n}=lim_{n oinfty}{frac{n^4 - 3n + 5}{2n^5 + 5n^3 - n^2}over{1over n}}=lim_{n oinfty}{n^5-3n^2+5nover2n^5+5n^3-n^2}=frac{1}{2} > 0$$ Thus, $sum a_n$ and $sum b_n$ share the same fate. But $b_n$ is harmonic series which diverges. Hence, $$sum_{n=52}^infty frac{n^4 - 3n + 5}{2n^5 + 5n^3 - n^2}$$ diverges.