此课程(MOOCULUS-2 "Sequences and Series")由Ohio State University于2014年在Coursera平台讲授。
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本系列学习笔记PDF下载（Academia.edu） MOOCULUS-2 Solution
Summary
- Ratio Test
Consider the series $sum_{n=0}^infty a_n$ where each term $a_n$ is positive.
Suppose that $$lim_{n o infty} frac{a_{n+1}}{a_n}=L$$ Then,- if $L < 1$ the series $sum_{n=0}^infty a_n$ converges.
- if $L > 1$ the series diverges.
- if $L=1$ this test is inconclusive.
- Integral Test
Suppose that $f(x)>0$ and is decreasing on the infinite interval $[k,infty)$ (for some $kge1$) and that $a_n=f(n)$. Then the series $$sum_{n=k}^infty a_n$$ converges if and only if the improper integral $$int_{k}^infty f(x)\,dx$$ converges. - $p$-series
Recall that a $p$-series is any series of the form $$sum_{n=1}^infty 1/n^p$$ A $p$-series converges if and only if $p>1$. - Root Test
Consider the series $sum_{n=0}^infty a_n$ where each term $a_n$ is positive. Suppose that $$lim_{n o infty} left(a_n ight)^{1/n}=L$$ Then,- if $L < 1$ the series $sum a_n$ converges,
- if $L > 1$ the series diverges, and
- if $L=1$, then the root test is inconclusive.
- Determine a series convergent or divergent:
- $n^{ ext{th}}$ test, if $$lim_{n oinfty}a_n eq0$$ then it diverges; otherwise, using other tests;
- Comparison test, $p$-series test, ratio test, integral test, root test, and Cauchy test.
Exercises 3.1
1. Compute $$lim_{n oinfty} |a_{n+1}/a_n|$$ for the series $$sum_{n=1}^infty {1over n^2}$$
Solution:
$$lim_{n oinfty} |a_{n+1}/a_n|=lim_{n oinfty}{{1over(n+1)^2}over{1over n^2}}=lim_{n oinfty}{n^2over n^2+2n+1}=1$$
2. Compute $$lim_{n oinfty} |a_{n+1}/a_n|$$ for the series $$sum_{n=1}^infty {1over n}$$
Solution:
$$lim_{n oinfty} |a_{n+1}/a_n|=lim_{n oinfty}{{1over n+1}over{1over n}}lim_{n oinfty}{nover n+1}=1$$
Determine whether each of the following series converges or diverges.
3. $$sum_{n=0}^infty (-1)^{n}{3^nover 5^n}$$
Solution:
By ratio test, we have $$lim_{n oinfty} |a_{n+1}/a_n|=lim_{n oinfty}{3^{n+1}over 5^{n+1}}cdot{5^n over 3^n}={3over5} < 1$$ Thus it converges.
4. $$sum_{n=1}^infty {n!over n^n}$$
Solution:
By ratio test, we have
$$lim_{n oinfty} a_{n+1}/a_n=lim_{n oinfty}{(n+1)!over(n+1)^{n+1}}cdot{n^nover n!}=lim_{n oinfty}{n^nover(n+1)^n}=lim_{n oinfty}({nover n+1})^n$$ On the other hand, $$lim_{n oinfty}(1+{1over n})^n=e$$ Thus $$lim_{n oinfty}({nover n+1})^n={1over e} < 1$$ Therefore it converges.
5. $$sum_{n=1}^infty {n^5over n^n}$$
Solution:
By ratio test, we have
$$lim_{n oinfty} a_{n+1}/a_n=lim_{n oinfty} {(n+1)^5over(n+1)^{n+1}}cdot{n^nover n^5}$$ $$=lim_{n oinfty}({n+1 over n})^5cdot({nover n+1})^ncdot{1over n+1}=1 imes{1over e} imes0=0 < 1$$ Thus it converges.
6. $$sum_{n=1}^infty {(n!)^2over n^n}$$
Solution:
By ratio test, we have $$lim_{n oinfty} a_{n+1}/a_n=lim_{n oinfty}{((n+1)!)^2over (n+1)^{n+1}}cdot{n^nover(n!)^2}$$ $$=lim_{n oinfty}{(n+1)^2over(n+1)^{n+1}}cdot n^n=lim_{n oinfty}(n+1)cdot({nover n+1})^n={1over e}cdotlim_{n oinfty}(n+1) oinfty$$ Thus it diverges.
Exercises 3.2
Determine whether each series converges or diverges.
1. $$sum_{n=1}^infty {1over n^{pi/4}}$$
Solution:
$p$-series, and ${piover4} < 1$ so it diverges.
2. $$sum_{n=1}^infty {nover n^2+1}$$
Solution:
By integral test, we have
$$int_{1}^infty {nover n^2+1}\,dn={1over2}int_{1}^{infty}{d(n^2+1)over n^2+1}={1over2}ln(n^2+1)Big|_{1}^{infty} oinfty$$ Thus it diverges.
3. $$sum_{n=1}^infty {ln nover n^2}$$
Solution:
By integral test, we have $$u=ln n, dv={1over n^2}dnRightarrow du={1over n}dn, v=-{1over n}$$ $$int_{1}^{infty}{ln nover n^2}dn=-{ln nover n}Big|_{1}^{infty}+int_{1}^{infty}{1over n^2}dn=-{ln nover n}Big|_{1}^{infty}-{1over n}Big|_{1}^{infty}=0-(-1)=1$$ Thus it converges.
4. $$sum_{n=1}^infty {1over n^2+1}$$
Solution:
$p$-series test, we have $$sum_{n=1}^infty {1over n^2+1}<sum_{n=1}^infty {1over n^2} o ext{converges}$$ Thus it converges.
5. $$sum_{n=1}^infty {1over e^n}$$
Solution:
This is geometric series and ${1over e}< 1 $, thus it converges.
6. $$sum_{n=1}^infty {nover e^n}$$
Solution:
By ratio test, we have
$$lim_{n oinfty}a_{n+1}/a_n=lim_{n oinfty}{n+1over e^{n+1}}cdot{e^nover n}={1over e}lim_{n oinfty}(1+{1over n})={1over e} < 1$$ Thus it converges.
7. $$sum_{n=2}^infty {1over nln n}$$
Solution:
By integral test, we have
$$int_{2}^{infty}{1over nln n} dn=int_{2}^{infty}{d(ln n)overln n}=ln(ln n)Big|_{2}^{infty} oinfty$$ Thus it diverges.
8. $$sum_{n=2}^infty {1over n(ln n)^2}$$
Solution:
By integral test, we have
$$int_{2}^{infty}{1over n(ln n)^2}dn=int_{2}^{infty}{d(ln n)over(ln n)^2}=-{1overln n}Big|_{2}^{infty}={1overln2}$$ Thus it converges.
9. Find an $N$ so that $$sum_{n=1}^infty {1over n^4}$$ is between $$sum_{n=1}^N {1over n^4}$$ and $$sum_{n=1}^N {1over n^4} + 0.005$$
Solution:
For a series ${a_n}$, we define a continuous function $f(x)$ where $f(n)=a_n$. And we have $$R_n=sum_{i=n+1}^{infty}a_i=a_{n+1}+a_{n+2}+a_{n+3}+cdots$$ Thus the series $$s=sum_{i=1}^{infty}a_1+a_2+cdots+a_n+a_{n+1}+cdots$$ $$=s_n+R_n=sum_{i=1}^{n}a_i+sum_{i=n+1}^{infty}a_i$$ Particularly, $$int_{n+1}^{infty}f(x)dx < R_n < int_{n}^{infty}f(x)dx$$ Hence $$s_n+int_{n+1}^{infty}f(x)dx < s < s_n+int_{n}^{infty}f(x)dx$$ For this question, we have $$int_{n+1}^{infty}{1over x^4}dx < R_n < 0.005$$ $$Rightarrow -{1over3}cdot{1over x^3}Big|_{n+1}^{infty}={1over3}cdot{1over (n+1)^3} < 0.005$$ $$Rightarrow ngeq4$$ R code:
## Q9
f = function(x) 1 / x^4
n = 1e7
limit = sum(f(x = 1:1e7))
for (i in 1:n){
if(sum(f(x = 1:i)) < limit & limit < sum(f(x = 1:i)) + 0.005){
print(i)
break
}
}
# [1] 4
10. Find an $N$ so that $$sum_{n=0}^infty {1over e^n}$$ is between $$sum_{n=0}^N {1over e^n}$$ and $$sum_{n=0}^N {1over e^n} + 10^{-4}$$ Solution: $$int_{n+1}^{infty}{1over e^x}dx < {1over 10^4}Rightarrow -e^{-x}Big|_{n+1}^{infty}=e^{-(n+1)} < 10^{-4}$$ $$Rightarrow e^{n+1} > 10^4Rightarrow n+1 > 4ln10Rightarrow ngeq9$$ R code:
## Q10
## Q10
f = function(x) 1 / exp(x)
n = 1e7
limit = sum(f(x = 0:1e7))
for (i in 0:n){
if(sum(f(x = 0:i)) < limit & limit < sum(f(x = 0:i)) + 1e-4){
print(i)
break
}
}
# [1] 9
11. Find an $N$ so that $$sum_{n=1}^infty {ln nover n^2}$$ is between $$sum_{n=1}^N {ln nover n^2}$$ and $$sum_{n=1}^N {ln nover n^2} + 0.005$$ Solution:
R code:
## Q11
f = function(x) log(x) / x^2
n = 1e7
limit = sum(f(x = 1:1e7))
for (i in 1:n){
if(sum(f(x = 1:i)) < limit & limit < sum(f(x = 1:i)) + 0.005){
print(i)
break
}
}
# [1] 1685
12. Find an $N$ so that $$sum_{n=2}^infty {1over n(ln n)^2}$$ is between $$sum_{n=2}^N {1over n(ln n)^2}$$ and $$sum_{n=2}^N {1over n(ln n)^2} + 0.005$$ Solution: $$int_{n+1}^{infty}{1over xcdot(ln x)^2}dx < 0.005$$ $$Rightarrow -{1overln x}Big|_{n+1}^{infty}={1overln(n+1)} < 0.005$$ $$Rightarrow n+1 > e^{200}Rightarrow ngeq e^{200}$$
Exercises 3.3
Determine whether the series converge or diverge.
1. $$sum_{n=1}^infty {1over 2n^2+3n+5}$$ Solution:
When $ngeq1$ $${1over 2n^2+3n+5} < {1over 2n^2} o ext{converge}$$ By $p$-series test and comparison test, it converges.
2. $$sum_{n=2}^infty {1over 2n^2+3n-5}$$
Solution:
When $ngeq2$ $${1over 2n^2+3n-5} < {1over 2n^2} o ext{converge}$$ By $p$-series test and comparison test, it converges.
3. $$sum_{n=1}^infty {1over 2n^2-3n-5} $$
Solution:
When $ngeq5$ $${1over 2n^2-3n-5}={1over n^2+(n^2-3n-5)} < {1over n^2} o ext{converge}$$ By $p$-series test and comparison test, it converges.
4. $$sum_{n=1}^infty {3n+4over 2n^2+3n+5} $$ Solution: $$int_{1}^{infty}{3x+4over 2x^2+3x+5}dx > int_{1}^{infty}{3x+{9over4}over 2x^2+3x+5}dx={3over4}cdotint_{1}^{infty}{d(2x^2+3x+5)over 2x^2+3x+5}$$ $$={3over4}cdotln(2x^2+3x+5)Big|_{1}^{infty} oinfty$$ By integral test and comparison test, it diverges.
5. $$sum_{n=1}^infty {3n^2+4over 2n^2+3n+5} $$ Solution: $$lim_{n oinfty}{3n^2+4over 2n^2+3n+5}={3over2} eq0$$ By $n^{ ext{th}}$ test, it diverges.
6. $$sum_{n=1}^infty {log nover n}$$
Solution:
When $n > e$ $${log nover n}>{1over n} o ext{diverge}$$ By $p$-series test and comparison test, it diverges.
7. $$sum_{n=1}^infty {log nover n^3}$$
Solution:
Set $u=log x$, $dv={dxover x^3}$, so $du={dxover x}$, $v=-{1over2}cdot{1over x^2}$. $$int_{1}^{infty}{log xover x^3}dx=-{1over2}cdot{1over x^2}cdotlog xBig|_{1}^{infty}+{1over2}int_{1}^{infty}{dxover x^3}$$ $$=-{1over2}cdot{1over x^2}cdotlog xBig|_{1}^{infty}-{1over4}cdot{1over x^2}Big|_{1}^{infty}={1over4} o ext{converge}$$ By integral test, it converges.
8. $$sum_{n=2}^infty {1over log n}$$ Solution: $$log n < nRightarrow {1overlog n} > {1over n} o ext{diverge}$$ By $p$-series test and comparison test, it diverges.
9. $$sum_{n=1}^infty {3^nover 2^n+5^n}$$ Solution: $$lim_{n oinfty}a_{n+1}/a_n=lim_{n oinfty}{3^{n+1}over 2^{n+1}+5^{n+1}}cdot{2^n+5^n over 3^n}$$ $$=3cdotlim_{n oinfty}{2^n+5^n over 2^{n+1}+5^{n+1}}=3cdotlim_{n oinfty}{({2over5})^n+1 over 2cdot({2over5})^n+5}={3over5} < 1$$ By ratio test, it converges.
10. $$sum_{n=1}^infty {3^nover 2^n+3^n}$$ Solution: $$lim_{n oinfty}{3^nover 2^n+3^n}=lim_{n oinfty}{1over ({2over3})^n+1}=1 eq0$$ By $n^{ ext{th}}$ test, it diverges.
Exercises 3.4
1. Prove the root test theorem.
Solution:
First, prove when $L < 1$ it converges. There should be an $r$ such that $L < r < 1$, and there exists some $n$, when $n < N$ we have $${a_n}^{{1over n}}=L < r < 1Rightarrow a_n < r^n o ext{converge}$$ Note that the last expression is a geometric series. By comparison test, ${a_n}$ converges.
Secondly we prove when $L > 1$ it diverges.
There exists some $n > N$ we have $${a_n}^{{1over n}}=L > 1Rightarrow a_n > 1^n=1$$ which means $$lim_{n oinfty}a_n > 1
eq0$$ By $n^{ ext{th}}$ test, it diverges.
Lastly, when $L=1$ it is inconclusive. Such as
- $lim_{n oinfty}({1over n})^{{1over n}}=1$, and it diverges.
- $lim_{n oinfty}({1over n^2})^{{1over n}}=1$, and it converges.
2. Compute $$lim_{n oinfty} |a_n|^{1/n}$$ for the series $sum 1/n^2$.
Solution:
$$lim_{n oinfty} ({1over n^2})^{1/n}=1$$ But it converges by $p$-series test.
3. Compute $$lim_{n oinfty} |a_n|^{1/n}$$ for the series $sum 1/n$.
Solution:
$$lim_{n oinfty} ({1over n})^{1/n}=1$$ By $p$-series test, it converges.
Additional Exercises
Determine whether the series converge or diverge.
1. $$sum_{n=6}^{infty}{15^n over 3^{3n+3}cdot(3n+4)}$$ Solution: $$lim_{n oinfty}a_{n+1}/a_n=lim_{n oinfty}{15^{n+1}over 3^{3n+6}cdot(3n+7)}cdot{3^{3n+3}cdot(3n+4)over 15^n}$$ $$=lim_{n oinfty}{15cdot(3n+4)over 27cdot(3n+7)}={5over9} < 1$$ By ratio test, it converges.
2. $$sum_{k=8}^{infty}{(k+2)cdot k!over 4^k}$$
Solution:
$$lim_{k oinfty}a_{n+1}/a_n=lim_{k oinfty}{(k+3)cdot(k+1)!over4^{k+1}}cdot{4^{k}over(k+2)cdot k!}=lim_{k oinfty}{(k+3)(k+1)over4(k+2)} oinfty$$ By ratio test, it diverges.
3. $$sum_{n=1}^{infty}{10^ncdot n!over(9n)^n}$$ Solution: $$lim_{n oinfty}a_{n+1}/a_n=lim_{n oinfty}{10^{n+1}cdot(n+1)!over9^{n+1}cdot(n+1)^{n+1}}cdot{9^ncdot n^nover10^ncdot n!}$$ $$=lim_{n oinfty}{10(n+1)cdot n^nover9(n+1)^{n+1}}={10over9}cdot({nover n+1})^n={10over9}cdot{1over e} < 1$$ By ratio test, it converges.
4. $$sum_{k=3}^{infty}{3over7k^{1.88}}$$ Solution: $$sum_{k=3}^{infty}{3over7k^{1.88}}=sum_{k=3}^{infty}{3over7}cdot{1over k^{1.88}}$$ By $p$-series test, it converges.
5. $$sum_{i=6}^{infty}{i-2over i^5+3i}$$ Solution: $${i-2over i^5+3i} < {iover i^5+3i} < {iover i^5}={1over i^4}$$ By $p$-series test and comparison test, it converges.
6. $$sum_{m=2}^{infty}({7m+5 over 6m+5})^m$$ Solution: $$lim_{m oinfty}{a_m}^{{1over m}}=lim_{m oinfty}{7m+5over 6m+5}={7over6} > 1$$ By root test, it diverges.
7. $$int_{2}^{infty}{4over (4x+1)^2}dx={1over9}$$
What is the interval of $$sum_{k=2}^{infty}{4over(4k+1)^2}$$
Solution:
Since $$int_{m}^{infty}f(x)dx leq sum_{n=m}^{infty}a_n leq a_m+int_{m}^{infty}f(x)dx$$ That is $$int_{2}^{infty}{4over(4x+1)^2}dxleqsum_{k=2}^{infty}{4over(4k+1)^2}leq a_2+int_{2}^{infty}{4over(4x+1)^2}dx$$ $$Rightarrow {1over9}leq s_kleq{4over81}+{1over9}={13over81}$$ Thus the interval is $[{1over9}, {13over81}]$.