Sequences and series convergence – university Year 1 maths worksheet answers to download as PDF

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1. Answer to Question 1

   1. For a rational expression, divide top and bottom by the highest power \(n^2\): \(a_n=\dfrac{5-3/n}{2+7/n^2}\). Since \(1/n\to0\) and \(1/n^2\to0\), the limit is \(\dfrac52\).
   2. Both terms tend to infinity, so multiply by the conjugate: \(b_n=\dfrac{(n^2+n)-n^2}{\sqrt{n^2+n}+n}=\dfrac{n}{\sqrt{n^2+n}+n}=\dfrac1{\sqrt{1+1/n}+1}\). The denominator tends to \(2\), so the limit is \(\dfrac12\).
   3. Use the standard limit \(\left(1+\dfrac xn\right)^n\to e^x\) with \(x=2\): the limit is \(e^2\).
   4. Take logarithms: \(\ln d_n=\dfrac{\ln n}n\to0\), because \(\ln n\) grows more slowly than \(n\). Since the exponential is continuous, \(d_n=e^{\ln d_n}\to e^0=\) 1.

2. Answer to Question 2

   1. Put everything over the common denominator: \(a_n-3=\dfrac{3n+1-3(n+2)}{n+2}=\dfrac{-5}{n+2}\). Taking absolute values, \(|a_n-3|=\dfrac5{n+2}\).
   2. We need \(\dfrac5{n+2}\lt 0.01\), that is \(n+2>500\), or \(n>498\). So the smallest such value is \(N=498\): every term from \(a_{499}\) onwards is within \(0.01\) of 3.
   3. Let \(\varepsilon>0\) be given and choose an integer \(N\ge\dfrac5\varepsilon\). For every \(n>N\): \(|a_n-3|=\dfrac5{n+2}\lt \dfrac5n\lt \dfrac5N\le\varepsilon\). Since such an \(N\) exists for every \(\varepsilon>0\), this is exactly the definition of \(a_n\to3\).

3. Answer to Question 3

   1. This is a geometric series with first term \(a=5\) and ratio \(r=\frac23\). Since \(|r|\lt 1\), it converges to \(\dfrac a{1-r}=\dfrac5{1-2/3}=\) 15.
   2. Use partial fractions: \(\dfrac1{n(n+2)}=\dfrac12\left(\dfrac1n-\dfrac1{n+2}\right)\). In the partial sum most terms cancel (it telescopes), leaving \(\dfrac12\left(1+\dfrac12-\dfrac1{N+1}-\dfrac1{N+2}\right)\). As \(N\to\infty\) this tends to \(\dfrac34\).
   3. Geometric with first term \(0.9\) (the sum starts at \(n=1\)) and ratio \(0.9\): \(\dfrac{0.9}{1-0.9}=\) 9.
   4. \(0.\overline{27}=0.27+0.0027+\dots\), a geometric series with \(a=0.27\) and \(r=0.01\). Its sum is \(\dfrac{0.27}{1-0.01}=\dfrac{27}{99}=\) \(\dfrac3{11}\).

4. Answer to Question 4

   1. Since \(n^2+3n>n^2\), we have \(0\lt \dfrac1{n^2+3n}\lt \dfrac1{n^2}\). The series \(\sum\frac1{n^2}\) converges (\(p=2>1\)), so the given series converges by the comparison test.
   2. Compare with \(\frac1{\sqrt n}\): \(\dfrac{1/(\sqrt n+1)}{1/\sqrt n}=\dfrac{\sqrt n}{\sqrt n+1}\to1\), a finite non-zero limit. Since \(\sum\frac1{\sqrt n}\) diverges (\(p=\frac12\le1\)), the series diverges by limit comparison.
   3. For large \(n\) the term behaves like \(\frac n{n^3}=\frac1{n^2}\). Indeed \(\dfrac{(n+1)/(n^3+2)}{1/n^2}=\dfrac{n^3+n^2}{n^3+2}\to1\), and \(\sum\frac1{n^2}\) converges, so the series converges.
   4. For \(n\ge3\), \(\ln n>1\), so \(\dfrac{\ln n}n>\dfrac1n\). The harmonic series diverges, so this series diverges by comparison.

5. Answer to Question 5

   1. Factorials suggest the ratio test: \(\dfrac{a_{n+1}}{a_n}=\dfrac{3^{n+1}}{(n+1)!}\cdot\dfrac{n!}{3^n}=\dfrac3{n+1}\to0\lt 1\), so the series converges (its sum is \(e^3\)).
   2. \(\dfrac{a_{n+1}}{a_n}=\dfrac{(n+1)^2}{2^{n+1}}\cdot\dfrac{2^n}{n^2}=\dfrac12\left(\dfrac{n+1}n\right)^2\to\dfrac12\lt 1\): the series converges; the exponential beats the power.
   3. \(\dfrac{a_{n+1}}{a_n}=\dfrac{(n+1)!}{2^{n+1}}\cdot\dfrac{2^n}{n!}=\dfrac{n+1}2\to\infty\): the series diverges; in fact the terms do not even tend to 0.
   4. The term is an \(n\)th power, so use the root test: \(\sqrt[n]{a_n}=\dfrac{2n}{3n+1}\to\dfrac23\lt 1\). The series converges.

6. Answer to Question 6

   1. \(\frac1{\sqrt n}\) decreases to 0, so the alternating series test gives convergence. But \(\sum\frac1{\sqrt n}\) diverges (\(p=\frac12\)), so the convergence is conditional.
   2. \(\dfrac n{n^2+1}\) decreases to 0 for \(n\ge1\), so the series converges. \(\dfrac{n}{n^2+1}\sim\dfrac1n\), so the absolute series diverges by limit comparison: conditionally convergent.
   3. Ratio: \(\left|\dfrac{a_{n+1}}{a_n}\right|=\dfrac{n+1}n\cdot\dfrac{|x|}3\to\dfrac{|x|}3\), so \(R=3\). At \(x=\pm3\) the terms are \(n(\pm1)^n\), which do not tend to 0. The interval is \((-3,3)\).

7. Answer to Question 7

   1. \(1.125\times0.75=\) 0.84375 m.
   2. \(2+2(1.5+1.125+\dots)=2+2\times\dfrac{1.5}{1-0.75}=2+12=\) 14 m.
   3. Fall time from \(h_k=2(0.75)^k\) is \(\sqrt{2h_k/g}=t_0(\sqrt{0.75})^k\) with \(t_0=\sqrt{4/9.8}\approx0.639\) s.
   4. Each bounce has a rise and a fall of equal time: total \(=t_0+2t_0\dfrac{r}{1-r}\) with \(r=\sqrt{0.75}\approx0.866\), giving \(0.639(1+12.93)\approx\) 8.9 s.
   5. The terms shrink geometrically, so their sum converges, like Zeno's paradox. In reality the ball stops after a finite number of visible bounces.

8. Answer to Question 8

   1. \(a_2=0.3(100)+100=130\), \(a_3=0.3(130)+100=139\).
   2. By induction, \(a_{n+1}=0.3a_n+100\) adds one more power of 0.3. Geometric sum: \(a_n=\dfrac{100(1-0.3^n)}{0.7}\).
   3. \(a_5=\dfrac{100(1-0.00243)}{0.7}\approx\) 142.5 mg; limit \(\dfrac{100}{0.7}\approx\) 142.9 mg.
   4. Yes: the level never exceeds 142.9 mg. For a limit of 120 mg we need \(\dfrac D{0.7}=120\), so \(D=\) 84 mg.

9. Answer to Question 9

   1. The first payment (end of month 1) earns interest at the end of months 2 and 3, so it grows to \(200\times1.004^2\). The second payment earns interest once: \(200\times1.004\). The third has just been paid: \(200\). Adding, the balance is \(200(1.004^2+1.004+1)\).
   2. In the same way, \(B_n=200(1+1.004+\dots+1.004^{n-1})\), a geometric series with \(n\) terms, first term 1 and ratio \(1.004\). Its sum is \(\dfrac{1.004^n-1}{1.004-1}\), so \(B_n=200\dfrac{1.004^n-1}{0.004}\).
   3. \(B_{24}=200\times\dfrac{1.004^{24}-1}{0.004}\approx\) £5 027.42. She paid in \(24\times200=\)£4 800, so the interest earned is £227.42.
   4. We need \(B_n>10\,000\), that is \(1.004^n>1+\dfrac{10000\times0.004}{200}=1.2\). Taking logarithms, \(n>\dfrac{\ln1.2}{\ln1.004}\approx45.7\), so it takes 46 months.

10. Answer to Question 10

    1. In stage 1 Achilles runs the 100 m gap in 10 s; meanwhile the tortoise moves 10 m. In stage 2 he covers those 10 m in 1 s, while the tortoise moves 1 m, and so on. Each gap, and each time, is 10 times smaller than the previous one, so stage 4 takes 0.01 s.
    2. The times form a geometric series with \(a=10\) and \(r=0.1\): \(10+1+0.1+\dots=\dfrac{10}{1-0.1}=\dfrac{100}9\approx\) 11.1 s.
    3. Achilles catches up when the distances are equal: \(10t=100+t\), so \(9t=100\) and \(t=\dfrac{100}9\). This matches part (b): an infinite number of stages can take a finite time, which resolves the paradox.
    4. Distance \(=\) speed \(\times\) time \(=10\times\dfrac{100}9=\dfrac{1000}9\approx\) 111.1 m.

11. Answer to Question 11

    1. The number of sides is multiplied by 4 and the length divided by 3, so the perimeter is multiplied by \(\frac43>1\): it tends to infinity.
    2. \(\displaystyle\sum_{n\ge1}3\cdot4^{n-1}\dfrac{A_0}{9^n}=\dfrac{A_0}3\sum_{n\ge1}\left(\tfrac49\right)^{n-1}\).
    3. \(\dfrac{A_0}3\cdot\dfrac1{1-4/9}=\dfrac{A_0}3\cdot\dfrac95=\dfrac35A_0\). Total area \(A_0+\dfrac35A_0=\) \(\dfrac85A_0\).
    4. The boundary becomes infinitely wiggly, yet the shape stays inside a fixed circle, so its area is bounded: a first example of a fractal.

12. Answer to Question 12

    1. Add the next term of the harmonic series: \(H_4=H_3+\frac14=1.833+0.25=\) 2.083.
    2. The top book is completely beyond the edge when the overhang exceeds one book length: \(\dfrac12H_n>1\), so \(H_n>2\). From the table, \(H_3\lt 2\lt H_4\), so 4 books are needed.
    3. The overhang is \(\frac12H_n\), and the harmonic series diverges: \(H_n\) exceeds any bound when \(n\) is large enough. So any overhang, however large, can be reached in theory, although the number of books grows enormously.
    4. An overhang greater than 2 needs \(H_n>4\). Since \(H_{30}\approx3.995\lt 4\lt H_{31}\approx4.027\), we need 31 books. Because \(H_n\) grows like \(\ln n\), each extra book length of overhang needs about \(e^2\approx7.4\) times as many books.

13. Answer to Question 13

    1. \(S_6=1.6487196\) (true value \(1.6487213\)).
    2. The tail is \(\dfrac{0.5^{N+1}}{(N+1)!}\left(1+\dfrac{0.5}{N+2}+\dots\right)\) and each ratio is less than \(0.5\), so it is bounded by a geometric series.
    3. \(N=6\): bound \(\dfrac{0.5^7}{7!}\times2\approx3.1\times10^{-6}\), too big. \(N=7\): \(\dfrac{0.5^8}{8!}\times2\approx1.9\times10^{-7}\). So \(N=7\).
    4. The ratio test gives \(\dfrac{|x|}{k+1}\to0\) for every \(x\): the radius of convergence is infinite.

14. Answer to Question 14

    1. Apply the recurrence: \(a_1=0.8(500)+400=800\), \(a_2=0.8(800)+400=1040\) and \(a_3=0.8(1040)+400=1232\).
    2. If \(a_n\to L\), then also \(a_{n+1}\to L\). Taking limits in the recurrence, \(L=0.8L+400\), so \(0.2L=400\) and \(L=2000\).
    3. Let \(b_n=a_n-2000\). Then \(b_{n+1}=0.8a_n+400-2000=0.8(a_n-2000)=0.8b_n\), so \((b_n)\) is geometric with \(b_0=-1500\). Hence \(b_n=-1500(0.8)^n\) and \(a_n=2000-1500(0.8)^n\).
    4. Since \(1500(0.8)^n>0\), \(a_n\lt 2000\) for all \(n\). Also \(a_{n+1}-a_n=1500(0.8)^n(1-0.8)>0\), so the sequence is increasing. Being increasing and bounded above, it converges by the monotone convergence theorem.
    5. We need \(1500(0.8)^n\lt 100\), that is \(0.8^n\lt \frac1{15}\), so \(n>\dfrac{\ln15}{\ln1.25}\approx12.1\). The population first exceeds 1900 after 13 years.

15. Answer to Question 15

    1. Because \(\frac1{x^2}\) is decreasing, \(\frac1{n^2}\) lies between \(\int_n^{n+1}\) and \(\int_{n-1}^n\) of \(\frac1{x^2}\). Summing for \(n>N\) gives the inequalities.
    2. \(\int_a^\infty x^{-2}dx=\frac1a\).
    3. \(1.549768+\frac1{11}=\) 1.640677 \(\lt S\lt \) \(1.549768+0.1=\) 1.649768.
    4. \(\pi^2/6\approx1.644934\) lies in the interval of width \(0.009\). The error of \(S_N\) alone is about \(\frac1N\), so about 100 terms give the same accuracy; the integral bounds save a lot of work.

16. Answer to Question 16

    1. \(\displaystyle\int_0^x\dfrac{dt}{1-t}=-\ln(1-x)\) and \(\int_0^xt^n\,dt=\dfrac{x^{n+1}}{n+1}\).
    2. \(\ln(1+x)=x-\frac{x^2}2+\frac{x^3}3-\cdots\); at \(x=1\) we get \(\ln2\) (Abel's theorem justifies the endpoint).
    3. The error is at most the first omitted term \(\frac1{N+1}\), so we need \(N\ge\) 1000 terms.
    4. The tail after 10 terms is \(\displaystyle\sum_{n\ge11}\dfrac1{n2^n}\lt \dfrac1{11\cdot2^{11}}\cdot\dfrac1{1-1/2}\approx8.9\times10^{-5}\). Indeed \(S_{10}=0.693065\) and \(\ln2=0.693147\).
    5. The second: it converges geometrically, so 10 terms beat 1000 terms of the first.

17. Answer to Question 17

    1. \(P(X=4)=\frac16\left(\frac56\right)^3\approx\) 0.0965. The probabilities form a geometric series with \(a=\frac16\) and \(r=\frac56\), so their sum is \(\dfrac{1/6}{1-5/6}=1\), as it must be.
    2. With \(a_n=n\cdot\frac16\left(\frac56\right)^{n-1}\), \(\dfrac{a_{n+1}}{a_n}=\dfrac{n+1}n\cdot\dfrac56\to\dfrac56\lt 1\), so the series for \(E[X]\) converges by the ratio test.
    3. The geometric series \(\sum x^n=\frac1{1-x}\) has radius of convergence 1. Inside it, a power series may be differentiated term by term, so \(\sum nx^{n-1}=\dfrac d{dx}\dfrac1{1-x}=\dfrac1{(1-x)^2}\) for \(|x|\lt 1\).
    4. With \(x=\frac56\): \(E[X]=\dfrac16\cdot\dfrac1{(1/6)^2}=\) 6 rolls, which matches intuition.
    5. \(X>12\) means the first 12 rolls are not sixes: \(P(X>12)=\left(\frac56\right)^{12}\approx\) 0.112. About 1 time in 9 you wait more than twice the average.

18. Answer to Question 18

    1. \(\dfrac1{2k-1}-\dfrac1{2(2k-1)}-\dfrac1{4k}=\dfrac1{2(2k-1)}-\dfrac1{4k}\).
    2. Group \(k\) is \(\dfrac12\left(\dfrac1{2k-1}-\dfrac1{2k}\right)\), so the sum is half the alternating harmonic series: \(\frac12\ln2\approx0.3466\).
    3. The series is only conditionally convergent: the positive and negative parts both diverge, so the order of the terms matters.
    4. A conditionally convergent series can be rearranged to have any sum, or to diverge. For an absolutely convergent series like \(\sum\frac{(-1)^n}{n^2}\), every rearrangement has the same sum.

19. Answer to Question 19

    1. Substitute \(u=-x^2\) in \(e^u=\sum\dfrac{u^n}{n!}\): \(e^{-x^2}=\sum\dfrac{(-x^2)^n}{n!}=1-x^2+\dfrac{x^4}2-\dfrac{x^6}6+\cdots\). The exponential series converges for every \(u\), so \(R=\infty\).
    2. Inside the radius of convergence we may integrate term by term. Each term is \(\dfrac{(-1)^n}{n!}x^{2n}\), and \(\int_0^1x^{2n}dx=\dfrac1{2n+1}\), so \(I=\displaystyle\sum_{n\ge0}\dfrac{(-1)^n}{n!(2n+1)}\).
    3. The first five terms are \(1-\frac13+\frac1{10}-\frac1{42}+\frac1{216}\approx\) 0.747487.
    4. The series alternates and its terms decrease to 0, so the error is less than the first omitted term: \(\dfrac1{5!\times11}=\dfrac1{1320}\approx0.00076\).
    5. Actual error \(=0.747487-0.746824=0.00066\), which is less than \(0.00076\), as predicted.

20. Answer to Question 20

    1. The tail is \(\dfrac1{(n+1)!}\left(1+\dfrac1{n+2}+\dfrac1{(n+2)(n+3)}+\cdots\right)\). Each fraction in the bracket is at most the corresponding power of \(\frac1{n+1}\), so the tail is less than \(\dfrac1{(n+1)!}\cdot\dfrac1{1-\frac1{n+1}}=\dfrac1{(n+1)!}\cdot\dfrac{n+1}n=\dfrac1{n!\,n}\). It is positive because every term is positive.
    2. If \(e=\frac ab\), then \(b!\,e=a(b-1)!\) is an integer. Also \(\dfrac{b!}{k!}\) is an integer for every \(k\le b\). So \(x\), a difference of integers, is an integer.
    3. Apply (a) with \(n=b\) and multiply by \(b!\): \(0\lt x\lt \dfrac{b!}{b!\,b}=\dfrac1b\le1\).
    4. There is no integer strictly between 0 and 1, so the assumption \(e=\frac ab\) leads to a contradiction. Therefore \(e\) is irrational.