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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Review
Determine whether the series converges.
1. $displaystylesum_{n=0}^{infty}{nover n^2+4}$
Solution:
When $n > 4$ we have $${nover n^2+4} > {nover n^2+n} = {1over n+1} o ext{diverge}$$ By $p$-series test and comparison test, it diverges.
2. $displaystyle{1over1cdot2}+{1over3cdot4}+{1over5cdot6} +{1over7cdot8}+cdotscdots$
Solution: $$s=sum_{n=0}^{infty}{1over (2n+1)(2n+2)}$$ and $${1over (2n+1)(2n+2)} = {1over 4n^2+6n+2} < {1over n^2} o ext{converge}$$ By $p$-series test and comparison test, it converges.
3. $displaystylesum_{n=0}^{infty}{nover(n^2+4)^2}$
Solution: $${nover(n^2+4)^2} < {nover (n^2)^2} = {1over n^3} o ext{converge}$$ By $p$-series test and comparison test, it converges.
4. $displaystylesum_{n=0}^{infty}{n!over8^n}$
Solution: $$lim_{n oinfty}a_{n+1}/a_n={(n+1)!over8^{n+1}}cdot{8^nover n!}={n+1over8} oinfty$$ By ratio test, it diverges.
5. $displaystyle1-{3over4}+{5over8}-{7over12}+{9over16}+ cdotscdots$
Solution: $$s=1+sum_{n=1}^{infty}(-1)^n{2n+1over4n}$$ and $$lim_{n oinfty}{2n+1over4n}={1over2} eq0$$ Thus it diverges.
6. $displaystylesum_{n=0}^{infty}{1oversqrt{n^2+4}}$
Solution: $$lim_{n oinfty}{1oversqrt{n^2+4}}ig/{1over n} = lim_{n oinfty} {noversqrt{n^2+4}}=1 > 0$$ Since ${1over n}$ is harmonic series which is divergent, by limit comparison test, the original series is divergent.
7. $displaystylesum_{n=0}^{infty}{sin^3nover n^2}$
Solution: $${sin^3nover n^2}leq{1over n^2} o ext{converge}$$ By $p$-series test and comparison test, it converges.
8. $displaystylesum_{n=0}^{infty}{nover e^n}$
Solution: $$lim_{n oinfty}a_{n+1}/a_n=lim_{n oinfty}{n+1over e^{n+1}}cdot{e^nover n}={1over e} < 1$$ By ratio test, it converges.
9. $displaystylesum_{n=0}^{infty}{n!over1cdot3cdot5cdots(2n-1)}$
Solution: $$lim_{n oinfty}a_{n+1}/a_n=lim_{n oinfty}{(n+1)!over1cdot3cdot5cdots(2n+1)}cdot{1cdot3cdot5cdot(2n-1)over n!}$$ $$=lim_{n oinfty}{n+1over (2n+1)cdot n}=0 < 1$$ By ratio test, it converges.
10. $displaystylesum_{n=1}^{infty}{1over nsqrt{n}}$
Solution: $$int_{1}^{infty}{dxover xsqrt{x}}=int_{1}^{infty}x^{-{3over2}}dx=-2x^{-{1over2}}Big|_{1}^{infty}=2 o ext{converge}$$ By integral test, it converges.
11. $displaystyle{1over2cdot3cdot4}+{2over3cdot4cdot5} +{3over4cdot5cdot6}+{4over5cdot6cdot7}+cdotscdots$
Solution: $$s=sum_{n=0}^{infty}{n+1over(n+2)(n+3)(n+4)}$$ and $${n+1over(n+2)(n+3)(n+4)} < {n+1over(n+1)(n+1)(n+1)} = {1over(n+1)^2} < {1over n^2} o ext{converge}$$ By $p$-series test and comparison test, it converges.
12. $displaystylesum_{n=1}^{infty}{1cdot3cdot5cdots(2n-1)over(2n)!}$
Solution: $$lim_{n oinfty}a_{n+1}/a_n=lim_{n oinfty}{1cdot3cdot5cdots(2n+1)over(2n+2)!} cdot {(2n)!over1cdot3cdot5cdots(2n-1)}$$ $$=lim_{n oinfty}{2n+1over(2n+2)(2n+1)}=0 < 1$$ By ratio test, it converges.
13. $displaystylesum_{n=0}^{infty}{6^nover n!}$
Solution: $$lim_{n oinfty}a_{n+1}/a_n=lim_{n oinfty} {6^{n+1}over(n+1)!} cdot {n!over6^n}lim_{n oinfty}{6over n+1}=0 < 1$$ By ratio test, it converges.
14. $displaystylesum_{n=1}^{infty}{(-1)^{n-1}oversqrt{n}}$
Solution: $$lim_{n oinfty}{1oversqrt{n}}=0$$ and it is decreasing. By alternating series test, it converges (conditionally).
15. $displaystylesum_{n=1}^{infty}{2^n3^{n-1}over n!}$
Solution: $$lim_{n oinfty}a_{n+1}/a_n=lim_{n oinfty}{2^{n+1} 3^{n}over(n+1)!}cdot {n!over2^n3^{n-1}}=lim_{n oinfty}{6over n+1}=0 < 1$$ By ratio test, it converges.
16. $displaystyle1+{5^2over2^2}+{5^4over(2cdot4)^2}+{5^6over(2cdot4 cdot6)^2}+{5^8over(2cdot4cdot6cdot8)^2}+cdotscdots$
Solution: $$s=sum_{n=0}^{infty}{5^{2n}over4^{n}cdot(n!)^2}$$ and $$lim_{n oinfty}a_{n+1}/a_n=lim_{n oinfty}{25^{n+1} over4^{n+1}cdot((n+1)!)^2}cdot{4^ncdot(n!)^2over25^n}$$ $$=lim_{n oinfty}{25over4(n+1)^2}=0 < 1$$ By ratio test, it converges.
17. $displaystylesum_{n=1}^{infty}sin{1over n}$
Solution: $$lim_{n oinfty}{sin{1over n}over{1over n}}=1 > 0$$ Since ${1over n}$ is harmonic series which is divergent, by limit comparison test, the original series is divergent. Find the interval and radius of convergence; you need not check the endpoints of the intervals.
18. $displaystylesum_{n=0}^{infty}{2^nover n!}x^n$
Solution:
By ratio test, we have $$lim_{n oinfty}{2^{n+1}over(n+1)!}cdot{n!over2^n}cdotig|{x^{n+1}over x^{n}}ig|=lim_{n oinfty}{2|x|over n+1}=0Rightarrow xin(-infty, infty)$$
19. $displaystylesum_{n=0}^{infty}{x^nover1+3^n}$
Solution:
By ratio test, we have $$lim_{n oinfty} {1+3^nover1+3^{n+1}}cdotig|{x^{n+1}over x^{n}}ig|={1over3}|x| < 1Rightarrow xin(-3, 3)$$
20. $displaystylesum_{n=1}^{infty}{x^nover ncdot3^n}$
Solution:
By ratio test, we have $$lim_{n oinfty}{ncdot3^nover(n+1)cdot3^{n+1}}cdot|x|={1over3}|x| < 1Rightarrow xin(-3, 3)$$
21. $displaystyle x+{1over2}cdot{x^3over3}+{1cdot3over2cdot4}cdot {x^5over5}+{1cdot3cdot5over2cdot4cdot6}cdot{x^7over7} +cdots$
Solution: $$s=sum_{n=0}^{infty}{(2n)!ig/2^ncdot n!over2^ncdot n!}cdot {x^{2n+1}over2n+1}=sum_{n=0}^{infty}{(2n)!over 4^ncdot(n!)^2}cdot {x^{2n+1}over2n+1}$$ and by ratio test, we have $$lim_{n oinfty}{(2n+2)!over4^{n+1}cdot((n+1)!)^2}cdot{1over2n+3} cdot {4^ncdot(n!)^2over(2n)!}cdot(2n+1)cdotig|{x^{2n+3}over x^{2n+1}}ig|$$ $$=lim_{n oinfty}{(2n+2)(2n+1)cdot (2n+1)over 4(n+1)^2cdot(2n+3)}cdot x^2=x^2 < 1Rightarrow xin(-1, 1)$$
22. $displaystylesum_{n=1}^{infty}{n!over n^2}x^n$
Solution:
By ratio test, we have $$lim_{n oinfty}{(n+1)!over(n+1)^2}cdot{n^2over n!}cdot|x|=lim_{n oinfty}(n+1)cdot|x| oinfty$$ Thus $R=0$ and it converges only when $x=0$.
23. $displaystylesum_{n=1}^{infty}{(-1)^nover n^2cdot3^n}x^{2n}$
Solution:
By ratio test, we have $$lim_{n oinfty} {n^2cdot3^nover(n+1)^2cdot3^{n+1}}cdot {x^{2n+2}over x^{2n}}={x^2over3} < 1Rightarrow xin(-sqrt3, sqrt3)$$
24. $displaystylesum_{n=0}^{infty}{(x-1)^nover n!}$
Solution: $$lim_{n oinfty}{n!over(n+1)!}cdot|x-1|=0Rightarrow xin(-infty, infty)$$
Find a series for each function, using the formula for Maclaurin series and algebraic manipulation as appropriate.
25. $2^x$
Solution: $$f(0)=2^xig|_{x=0}=1$$ $$f'(0)=2^xcdotlog2ig|_{x=0}=log2$$ $$f''(0)=2^xcdot(log2)^2ig|_{x=0}=(log2)^2$$ $$cdotscdotscdots$$ Thus $$2^x=1+{log2over1!}x+{(log2)^2over2!}x^2+cdots =sum_{n=0}^{infty}{(log2)^nover n!}x^n$$
26. $log(1+x)$
Solution: $$f(0)=log(1+x)ig|_{x=0}=0$$ $$f'(0)={1over1+x}ig|_{x=0}=1$$ $$f''(0)={-1over(1+x)^2}ig|_{x=0}=-1=-1!$$ $$f'''(0)={2over(1+x)^3}ig|_{x=0}=2=2!$$ $$f^{(4)}(0)={-6over(1+x)^4}ig|_{x=0}=-6=-3!$$ $$cdotscdotscdots$$ Thus $$log(1+x)=x-{1!over2!}x^2+{2!over3!}x^3-{3!over4!}x^4+cdots =sum_{n=0}^{infty}{(-1)^nover n+1}x^{n+1}$$
27. $displaystylelogleft({1+xover1-x} ight)$
Solution: $$log(1+x)=sum_{n=0}^{infty}{(-1)^nover n+1}x^{n+1} = x - {1over2}x^2+{1over3}x^3-{1over4}x^4+cdots$$ $$log(1-x)=sum_{n=0}^{infty}{(-1)^nover n+1}(-x)^{n+1} = sum_{n=0}^{infty}{-1over n+1}x^{n+1}=-x - {1over2}x^2-{1over3}x^3-{1over4}x^4+cdots$$ Thus $$logig({1+xover1-x}ig)=log(1+x)-log(1-x)$$ $$=2x+{2over3}x^3+{2over5}x^5+cdots =sum_{n=0}^{infty}{2over 2n+1}x^{2n+1}$$
28. $sqrt{1+x}$
Solution: $$f(0)=sqrt{1+x}ig|_{x=0}=1$$ $$f'(0)={1over2}(1+x)^{-{1over2}}ig|_{x=0}={1over2}$$ $$f''(0)=-{1over4}(1+x)^{-{3over2}}ig|_{x=0}=-{1over4}$$ $$f'''(0)={3over8}(1+x)^{-{5over2}}ig|_{x=0}={3over8}$$ $$f^{(4)}(0)=-{15over16}(1+x)^{-{7over2}}ig|_{x=0}=-{15over16}$$ $$cdotscdotscdots$$ Thus $$sqrt{1+x}=1+{1over2cdot1!}x-{1over4cdot2!}x^2+{3over8cdot3!}x^3 -{15over16cdot4!}x^4+cdots$$ $$=1+{xover2}+sum_{n=2}^{infty}{(-1)^{n+1}cdot1cdot3cdots(2n-3)over 2^ncdot n!}x^n$$
29. $displaystyle{1over1+x^2}$
Solution: $${1over 1-x}=sum_{n=0}^{infty}x^nRightarrow {1over1+x} = sum_{n=0}^{infty}(-x)^n=sum_{n=0}^{infty}(-1)^ncdot x^n$$ $$Rightarrow {1over1+x^2}=sum_{n=0}^{infty}(-1)^ncdot x^{2n}$$
30. $arctan(x)$
Solution: $$int{1over1+x^2}dx=arctan(x)$$ $$Rightarrow arctan(x)=intsum_{n=0}^{infty}(-1)^ncdot x^{2n}dx = sum_{n=0}^{infty}{(-1)^nover2n+1}cdot x^{2n+1}$$
31. Use the answer to the previous problem to discover a series for a well-known mathematical constant $pi$.
Solution: $${piover4}=arctan1=sum_{n=0}^{infty}{(-1)^nover2n+1}Rightarrow pi=sum_{n=0}^{infty}(-1)^ncdot{4over2n+1}$$
Final
1. To say that the sequence $a_n$ converges to $L$ means what? In other words, what is the definition of the statement $displaystylelim_{n oinfty}a_n=L$?
Solution:
For every positive real number $varepsilon > 0$ there exists an $Ninmathbf{N}$ so that whenever $n geq N$, we have $|a_n-L| < varepsilon$.
2. To say that $displaystylesum_{k=4}^infty a_k = L$ means what? In order words, what does it mean to say that the ``value'' of a series is $L$?
Solution:
The sequence of partial sums $s_n = displaystylesum_{k=4}^n a_k$ converges to $L$.
3. Which of the following could be the initial terms of a monotonic sequence?
A. $3,quad 1,quad -5,quad -7,quad -10,quad -6,quadldots$
B. $3,quad -2,quad -4,quad -7,quad -13,quad -18,quadldots$
C. $3,quad 0,quad 3,quad 2,quad -4,quad -5,quadldots$
D. $3,quad 8,quad 4,quad 2,quad -4,quad -10,quadldots$
E. $3,quad -2,quad -5,quad -4,quad -6,quad -8,quadldots$
Solution:
A monotonic sequence is a sequence which is either increasing, decreasing, non-increasing, or non-decreasing. So B is correct.
4. Consider the sequence $b_m=-4m^2-m-2$. Is the sequence bounded above? Bounded below?
Solution: $$-4m^2-m-2=-4(m+{1over8})^2-{31over16} leq -{31over16}$$ Bounded above, but not bounded below.
5. Evaluate $displaystylesum_{j=3}^infty left(-displaystylefrac{3}{10} ight)^{j}$.
Solution:
This is geometric series, $displaystylesum_{n=k}^{infty}r^n={r^kover1-r}$, where $|r| < 1$. Thus $$sum_{j=3}^infty left(-frac{3}{10} ight)^{j}={(-{3over10})^3over1-(-{3over10})}=-frac{27}{1300}$$
6. Does the series $displaystylesum_{n=7}^infty left( frac{n^{3.0} - 3 \, n^{2}}{n^{5.0} + 3 \, n^{3}} ight)$ converge or diverge?
Solution: $$frac{n^{3.0} - 3 \, n^{2}}{n^{5.0} + 3 \, n^{3}} < {n^3over n^5+3n^3} < {n^3over n^5} = {1over n^2} o ext{converge}$$ By $p$-series test and comparison test, it converges.
7. Evaluate $displaystylesum_{m=4}^inftyfrac{20}{16 \, m^{2} + 40 \, m + 21}$.
Solution: $$frac{20}{16 \, m^{2} + 40 \, m + 21}={20over(4m+3)(4m+7)} =5cdot{(4m+7)-(4m+3)over(4m+3)(4m+7)}=5({1over4m+3}-{1over4m+7})$$ Thus $$sum_{m=4}^inftyfrac{20}{16 \, m^{2} + 40 \, m + 21}=5cdot({1over19}-{1over23}+{1over23}-{1over27}+cdots) ={5over19}-lim_{m oinfty}{1over4m+7}={5over19}$$
8. Does the series $displaystylesum_{m=0}^infty left(frac{7}{5 \, m + 20} ight)$ converge or diverge?
Solution: $$frac{7}{5m + 20} > {5over5m+20} ={1over m+4} o ext{diverge}$$ By $p$-series test and comparison test, it diverges.
9. Suppose $(a_n)$ is a sequence involving both positive and negative numbers, and suppose that the series $displaystylesum_{n=1}^{infty}|a_n|$ converges. What can be known for certain about the series $displaystylesum_{n=1}^{infty}a_n$?
Solution:
Since $$0leq a_n+|a_n| < 2|a_n| o ext{converge}$$ by comparison test we have $displaystylesum_{n=1}^{infty}(a_n+|a_n|)$ converges, and hence $$displaystylesum_{n=1}^{infty}(a_n+|a_n|)-sum_{n=1}^{infty}|a_n|=displaystylesum_{n=1}^{infty}a_n$$ converges. This shows that "absolute convergence implies convergence".
10. Does the series $displaystylesum_{i=5}^infty left( frac{left(-1 ight)^{i}}{i^{1.16}} ight)$ converge or diverge?
Solution:
This is $p$-series where $p=1.16 > 1$, thus it converges absolutely.
11. Does the series $displaystylesum_{n=7}^infty left( frac{10^{n} n!}{left(2 \, n ight)^{n}} ight)$ converge or diverge?
Solution: $$lim_{n oinfty}{a_{n+1}over a_n}=lim_{n oinfty} {10^{n+1}(n+1)!over(2n+2)^{n+1}}cdot{(2n)^nover10^ncdot n!}$$ $$=lim_{n oinfty}{10(n+1)over2n+2}cdotleft({2nover2n+2} ight)^n = {5over e} > 1$$ By ratio test, it diverges.
12. Does the series $displaystylesum_{n=7}^infty left( left(frac{7 \, n + 3}{5 \, n + 9} ight)^{n} ight)$ converge or diverge?
Solution: $$lim_{n oinfty}sqrt[n]{a_n}=lim_{n oinfty}{7n+3over5n+9}={7over5} > 1$$ By root test, it diverges.
13. For which real numbers $x$ does the series $$sum_{n=1}^infty left(frac{2 \, left(-frac{4}{7} ight)^{n}cdot x^{n}}{5 \, n} ight)$$ converge?
Solution:
By ratio test, we have $$lim_{n oinfty}{|a_{n+1}over |a_n|}cdot|x| = lim_{n oinfty}{({4over7})^{n+1}over5(n+1)}cdot{5nover({4over7})^n}cdot|x| = {4over7}cdot|x| < 1Rightarrow -{7over4} < x < {7over4}$$ When $x=displaystyle-{7over4}$, $displaystylesum_{n=1}^{infty}{2over5n}$ diverges; When $x=displaystyle{7over4}$, $displaystylesum_{n=1}^{infty}{(-1)^ncdot2over5n}$ is an alternating series and it is convergent. Thus the interval of convergence is $displaystyle(-{7over4}, {7over4}]$.
14. What is the radius of convergence of the series $displaystylesum_{k=4}^infty left( frac{2 \, {left(k + 1 ight)} x^{k}}{6^{k} + k + 4} ight)$?
Solution:
By ratio test, we have $$lim_{k oinfty}{2(k+2)over6^{k+1}+k+1+4}cdot{6^k+k+4over2(k+1)}cdot|x| ={1over6}|x| < 1 Rightarrow -6 < x < 6$$ Thus the radius is $R=6$.
15. Suppose $displaystylesum_{n=1}^infty b_n = frac{x}{{left(x^{2} - 1 ight)}^{2}}$. What is the expression of $b_n$?
Solution: $$f(x)=int{xover(x^2-1)^2}dx={1over2}int{d(x^2-1)over(x^2-1)^2} $$ $$=-{1over2}cdot{1over x^2-1}={1over2}cdot{1over1-x^2}={1over2}cdotsum_{n=0}^{infty}x^{2n}$$ Thus $$F(x)=f'(x)=sum_{n=1}^{infty}ncdot x^{2n-1}=sum_{n=1}^{infty}b_nRightarrow b_n=n x^{2n-1}$$
16. What are the first few terms of the Taylor series for $f(x) = e^{left(e^{x} - 1 ight)} - 1$ centered around the point $a=0$?
Solution: $$f(0)=e^{left(e^x-1 ight)}-1ig|_{x=0}=0$$ $$f'(0)=e^{left(e^x-1 ight)}cdot e^xig|_{x=0}=1$$ $$f''(0)=e^{left(e^x-1 ight)}cdot e^{2x}+e^{left(e^x-1 ight)}cdot e^xig|_{x=0}=2$$ $$f'''(0)=e^{left(e^x-1 ight)}cdot e^{3x}+2e^{left(e^x-1 ight)}cdot e^{2x}+e^{left(e^x-1 ight)}cdot e^{2x}+e^{left(e^x-1 ight)}cdot e^xig|_{x=0}=5$$ $$cdotscdotscdots$$ Thus $$e^{left(e^{x} - 1 ight)} - 1=0+x+x^2+{5over6}x^3+cdotscdots$$
17. By finding the Taylor series around $x=1$, rewrite the polynomial $p(x)=-3x^3-x^2+x-1$ as a polynomial in $(x-1)$.
Solution: $$p(1)=-3x^3-x^2+x-1ig|_{x=1}=-4$$ $$p'(1)=-9x^2-2x+1ig|_{x=1}=-10$$ $$p''(1)=-18x-2ig|_{x=1}=-20$$ $$p'''(1)=-18ig|_{x=1}=-18$$ Thus $$p(x)=-4-10(x-1)-10{(x-1)}^2-3{(x-1)}^3$$
18. By considering Taylor series, evaluate $$lim_{x o 0} frac{{left(displaystylefrac{cosleft(x ight) - 1}{x} + sinleft(x ight) ight)}^{2}}{logleft(x + 1 ight) anleft(x ight)}.$$
Solution: $$cos(x)=1-{x^2over2}+{x^4over24}+O(x^6)Rightarrow {cos(x)-1over x}=-{1over2}x+{1over24}x^3+O(x^5)$$ $$sin(x)=x-{1over6}x^3+O(x^5)$$ $$log(x+1)=x-{1over2}x^2+{1over3}x^3-{1over4}x^4+O(x^5)$$ $$ an(x)=x+{1over3}x^3+O(x^5)$$ Thus $$lim_{x o 0} frac{{left(displaystylefrac{cosleft(x ight) - 1}{x} + sinleft(x ight) ight)}^{2}}{logleft(x + 1 ight) anleft(x ight)} = lim_{x o0}{displaystyleleft({1over2}x-{1over8}x^3+O(x^5) ight)^2over displaystyle x^2-{1over2}x^3+O(x^4)}={1over4}$$
19. Let $f(x)=cos(x^3)$. By considering the Taylor series for $f$ around 0, compute $f^{(48)}(0)$.
Solution:
Since $$cos(x)=sum_{n=0}^{infty}{(-1)^nover (2n)!}x^{2n}Rightarrow cos(x^3)=sum_{n=0}^{infty}{(-1)^nover (2n)!}x^{6n}$$ And hence we have $${f^{(48)}(0)over48!}={1over16!}Rightarrow f^{(48)}(0)={48!over16!}$$
20. You may remember that $$frac{d}{dx} arctan x = frac{1}{1+x^2}.$$ By computing some terms of the (alternating!) Taylor series for arctangent, approximate $arctan left(displaystylefrac{1}{3} ight)$ to within ${1over100}$.
Solution: $$arctan(x)=int{1over1+x^2}dx=int{1over1-(-x^2)}dx$$ $$=intsum_{n=0}^{infty}(-x^2)^ndx=intsum_{n=0}^{infty}(-1)^ncdot x^{2n}dx=sum_{n=0}^{infty}{(-1)^nover2n+1}x^{2n+1}$$ So $$arctanleft({1over3} ight)=sum_{n=0}^{infty}{(-1)^nover2n+1}cdotleft({1over3} ight)^{2n+1}$$ We hope to find a term that $|a_n| leq displaystyle{1over100}$. We can write the first few terms: $$arctanleft({1over3} ight)={1over3}-{1over81} +{1over1215}+cdots$$ Thus the first two terms are enough and the result is $displaystyle{1over3}-{1over81}={26over81}$ .