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
- Suppose $(a_n)$ is a sequence with associated series $$sum_{k=1}^infty a_k$$ The sequence of partial sums associated to these objects is the sequence $$s_n = sum_{k=1}^n a_k$$
- Consider the series $$sum_{k=1}^infty a_k$$ This series converges if the sequence of partial sums $$s_n = sum_{k=1}^n a_k$$ converges. More precisely, if $$lim_{n o infty} s_n = L$$ we then write $$sum_{k=1}^infty a_k = L$$ and say, "the series $sum_{k=1}^infty a_k$ converges to $L$."
If the sequence of partial sums diverges, we say that the series diverges. - A series of the form $$sum_{k=0}^infty a_0 \, r^k$$ is called a geometric series.
- Suppose $a_0
eq 0$. Then for a real number $r$ such that $|r| < 1$, the geometric series $$sum_{k=0}^infty a_0\, r^k$$ converges to $frac{a_0}{1-r}$.
For a real number $r$ where $|r| geq 1$, the aforementioned geometric series diverges. - Consider the series $$sum_{k=0}^infty a_k$$ and suppose $c$ is a nonzero constant. Then $$sum_{k=0}^infty a_k$$ and $$sum_{k=0}^infty c\,a_k$$ share a common fate: either both series converge, or both series diverge.
Moreover, when $$sum_{k=0}^infty a_k$$ converges, $$sum_{k=0}^infty c \, a_k = c cdot sum_{k=0}^infty a_k$$ - Suppose $$sum_{k=0}^infty a_k$$ and $$sum_{k=0}^infty b_k$$ are convergent series. Then $$sum_{k=0}^infty (a_k+b_k)$$ is convergent, and $$sum_{k=0}^infty (a_k+b_k)=left( sum_{k=0}^infty a_k ight)+left(sum_{k=0}^infty b_k ight)$$
- If $$sum_{k=0}^infty a_k$$ converges then $$lim_{n oinfty}a_n=0$$
- Consider the series $$sum_{k=0}^infty a_k$$ If the limit $$lim_{n oinfty}a_n$$ does not exist or has a value other than zero, then the series diverges.
We'll usually call this theorem the "$n^{ ext{th}}$ term test." - The series $$sum_{n=1}^infty {1over n} = frac{1}{1} +frac{1}{2} + frac{1}{3} + frac{1}{4} + cdots$$ is called the harmonic series.
- Consider the series $$sum_{k=0}^infty a_k$$ Assume the terms $a_k$ are non-negative. If the sequence of partial sums $s_n = a_0 + cdots + a_n$ is bounded, then the series converges.
- Comparison Test
Suppose that $a_n$ and $b_n$ are non-negative for all $n$ and that, for some $N$, whenever $n geq N$, we have $a_n leq b_n$.
If $$sum_{n=0}^infty b_n$$ converges, so does $$sum_{n=0}^infty a_n$$ If $$sum_{n=0}^infty a_n$$ diverges, so does $$sum_{n=0}^infty b_n$$ - Cauchy Condensation Test
Suppose $(a_n)$ is a non-increasing sequence of positive numbers. The series $$sum_{n=1}^infty a_n$$ converges if and only if the series $$sum_{n=0}^infty left( 2^n a_{2^n} ight)$$ converges. - $p$-series Test $$sum_{n=1}^{infty} {1over n^p} egin{cases}mbox{converges} & mbox{when $p>1$}\ mbox{diverges} & mbox{when $pleq1$} end{cases}$$
Exercises 2.7
1. Explain why $$sum_{n=1}^infty {n^2over 2n^2+1}$$ diverges.
Solution:
By $n^{ ext{th}}$ test, $$lim_{n oinfty} {n^2over 2n^2+1}={1over2} eq0$$ Therefore it diverges.
2. Explain why $$sum_{n=1}^infty {5over 2^{1/n}+14}$$ diverges.
Solution:
By $n^{ ext{th}}$ test, $$lim_{n oinfty} {5over 2^{1over n}+14}={5over1+14}={1over3} eq0$$ Thus it diverges.
3. Explain why $$sum_{n=1}^infty {3over n}$$ diverges.
Solution:
$$sum_{n=1}^infty {3over n}=3cdotsum_{n=1}^{infty}{1over n}$$ which is a harmonic series.
4. Compute $$sum_{n=0}^infty {4over (-3)^n}- {3over 3^n}$$
Solution:
Geometric series: $$sum_{n=0}^infty {4over (-3)^n}- {3over 3^n}$$ $$=sum_{n=0}^{infty} 4cdot(-{1over3})^n-3cdot({1over3})^n$$ $$=4cdot{1over 1-(-{1over3})}-3cdot{1over 1-{1over3}}$$ $$=4 imes{3over4}-3 imes{3over2}=-{3over2}$$
5. Compute $$sum_{n=0}^infty {3over 2^n}+ {4over 5^n}$$
Solution:
Geometric series: $$sum_{n=0}^infty {3over 2^n}+ {4over 5^n}$$ $$= 3cdot{1over 1-{1over2}}+4cdot{1over 1-{1over5}}=6+5=11$$
6. Compute $$sum_{n=0}^infty {4^{n+1}over 5^n}$$
Solution: Geometric series: $$sum_{n=0}^infty {4^{n+1}over 5^n}$$ $$=sum_{n=0}^{infty}4cdot({4over5})^n=4 imes{1over 1-{4over5}}=20$$
7. Compute $$sum_{n=0}^infty {3^{n+1}over 7^{n+1}}$$
Solution:
Geometric series: $$sum_{n=0}^infty {3^{n+1}over 7^{n+1}}$$ $$=sum_{n=0}^infty {3over7}cdot{3^{n}over 7^{n}}={3over7} imes{1over 1-{3over7}}={3over4}$$
8. Compute $$sum_{n=1}^infty left({3over 5} ight)^n$$
Solution:
Geometric series: $$sum_{n=1}^infty left({3over 5} ight)^n$$ $$=sum_{n=0}^infty left({3over 5} ight)^n-1$$ $$={1over 1-{3over5}}-1={3over2}$$ Alternatively, $$sum_{n=1}^infty left({3over 5} ight)^n={{3over5}over 1-{3over5}}={3over2}$$
9. Compute $$sum_{n=1}^infty {3^nover 5^{n+1}}$$
Solution:
Geometric series: $$sum_{n=1}^infty {3^nover 5^{n+1}}$$ $$=sum_{n=0}^infty {1over5}cdot{3^nover 5^{n}}-{1over5}$$ $$={1over5} imes{1over 1-{3over5}}-{1over5}={3over10}$$ Alternatively, $$sum_{n=1}^infty {3^nover 5^{n+1}}={1over5} imes{{3over5}over 1-{3over5}}={3over10}$$
Additional Exercises
1. Evaluate $$sum_{n=5}^{infty}(-{4over7})^n$$
Solution:
Geometric series: $$sum_{n=5}^{infty}(-{4over7})^n={(-{4over7})^5over 1-(-{4over7})}=-{1024over26411}$$
2. Test $$sum_{n=2}^{infty}-8cdot({6over11})^n$$
Solution:
$$sum_{n=2}^{infty}-8cdot({6over11})^n=-8cdotsum_{n=2}^{infty}({6over11})^n$$ which is a geometric series and $r < 1$, therefore it converges.
3. Evaluate $$sum_{i=2}^{infty}{12over 9i^2+21i+10}$$
Solution:
$$sum_{i=2}^{infty}{12over 9i^2+21i+10}=sum_{i=2}^{infty}{12over (3i+5) (3i+2)}$$ $$=sum_{i=2}^{infty}4cdot({1over 3i+2}-{1over 3i+5})=4 imes{1over8}={1over2}$$
4. Test $$sum_{m=3}^{infty}{(7m+5)cdot(m-8)over(5m+4)cdot(5m-7)}$$
Solution:
By $n^{ ext{th}}$ test: $$lim_{m oinfty}{(7m+5)cdot(m-8)over(5m+4)cdot(5m-7)}={7over25} eq0$$ Thus it diverges.
5. Test $$sum_{n=0}^{infty}{5over 7n+42}$$
Solution:
$$sum_{n=0}^{infty}{5over 7n+42}={5over7}cdotsum_{n=0}^{infty}{1over n+6}$$ which is a harmonic series. Thus it diverges.
6. Test $$sum_{n=5}^{infty}{2(n^2+2)over 7^n}$$
Solution:
Comparison test: $$sum_{n=5}^{infty}{2(n^2+2)over 7^n}=sum_{n=5}^{infty}{2n^2+4over 7^n}leqsum_{n=5}^{infty}{2^nover 7^n}, ext{when} ngeq7$$ And $$sum_{n=5}^{infty}{2^nover 7^n}=sum_{n=5}^{infty}({2over7})^n$$ is a geometric series which is convergent. Thus $$sum_{n=5}^{infty}{2(n^2+2)over 7^n}$$ is convergent, too.