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 ∑∞n=1an. The series we get by removing the first N−1 terms, namely ∞∑n=Nan is called a tail of the given series.
- Let N>1 be an integer. The series ∞∑n=1an converges if and only if ∞∑n=Nan 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 an≥0 and bn≥0. Then if lim 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^5\over 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 < 2\cdot 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}=({2\over5})^n That is, the tail series \sum_{n=23}^{\infty}({2\over5})^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^5\over 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={1\over n} We have \lim_{n\to\infty}{a_n\over b_n}=\lim_{n\to\infty}{\frac{n^4 - 3n + 5}{2n^5 + 5n^3 - n^2}\over{1\over n}}=\lim_{n\to\infty}{n^5-3n^2+5n\over2n^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.
作者:赵胤
出处:http://www.cnblogs.com/zhaoyin/
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