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1.3E: Exercises for Section 1.3

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In exercises 1 - 4, use sigma notation to write each expressions as an infinite series.

1) 1+12+13+14+

Answer
n=11n

2) 11+11+

3) 112+1314+...

Answer
n=1(1)n1n

4) sin1+sin12+sin13+sin14+

In exercises 5 - 8, compute the first four partial sums S1,,S4 for the series having nth term an starting with n=1 as follows.

5) an=n

Answer
1,3,6,10

6) an=1/n

7) an=sinnπ2

Answer
1,1,0,0

8) an=(1)n

In exercises 9 - 12, compute the general term an of the series with the given partial sum Sn. If the sequence of partial sums converges, find its limit S.

9) Sn=11n,n2

Answer
an=SnSn1=1n11n. Since S=limnSn=limn(11n)=1, the series converges to S=1.

10) Sn=n(n+1)2,n1

11) Sn=n,n2

Answer
an=SnSn1=nn1=1n1+n.
The series diverges because the partial sums are unbounded.
That is, limnSn=limnn=.

12) Sn=2n+22n,n1

For each series in exercises 13 - 16, use the sequence of partial sums to determine whether the series converges or diverges.

13) n=1nn+2

Answer
S1=1/3,
S2=1/3+2/4>1/3+1/3=2/3,
S3=1/3+2/4+3/5>3(1/3)=1.
In general Sk>k/3, so the series diverges.
Note that the nth Term Test for Divergence could also be used to prove that this series diverges.

14) n=1(1(1)n))

15) n=11(n+1)(n+2) (Hint: Use a partial fraction decomposition like that for n=11n(n+1).)

Answer

S1=1/(23)=1/6=2/31/2,

S2=1/(23)+1/(34)=2/12+1/12=1/4=3/41/2,

S3=1/(23)+1/(34)+1/(45)=10/60+5/60+3/60=3/10=4/51/2,

S4=1/(23)+1/(34)+1/(45)+1/(56)=10/60+5/60+3/60+2/60=1/3=5/61/2.

The pattern is Sk=k+1k+212.
Then limnSn=limn(k+1k+212)=12, so the series converges to 1/2.

16) n=112n+1 (Hint: Follow the reasoning for n=11n.)

Suppose that n=1an=1, that n=1bn=1, that a1=2, and b1=3. Use this information to find the sum of the indicated series in exercises 17 - 20.

17) n=1(an+bn)

Answer
n=1(an+bn)=n=1an+n=1bn=1+(1)=0

18) n=1(an2bn)

19) n=2(anbn)

Answer
n=2(anbn)=n=2ann=2bn=(n=1ana1)(n=1bnb1)=(12)(1(3))=12=3

20) n=1(3an+14bn+1)

In exercises 21 - 26, state whether the given series converges or diverges and explain why.

21) n=11n+1000 (Hint: Rewrite using a change of index.)

Answer
The series diverges, n=10011n

22) n=11n+1080 (Hint: Rewrite using a change of index.)

23) 1+110+1100+11000+

Answer
This is a convergent geometric series, since r=110<1

24) 1+eπ+e2π2+e3π3+

25) 1+πe2+π2e4+π3e6+π4e8+

Answer
This is a convergent geometric series, since r=π/e2<1

26) 1π3+π29π327+

For each an in exercises 27 - 30, write its sum as a geometric series of the form n=1arn. State whether the series converges and if it does, find the exact value of its sum.

27) a1=1 and anan+1=5 for n1.

Answer
n=15(1/5)n, converges to 5/6

28) a1=2 and anan+1=1/2 for n1.

29) a1=10 and anan+1=10 for n1.

Answer
n=1100(1/10)n, converges to 1009

30) a1=110 and an/an+1=10 for n1.

In exercises 31 - 34, use the identity 11y=n=0yn (which is true for |y|<1) to express each function as a geometric series in the indicated term.

31) x1+x in x

Answer
xn=0(x)n=n=1(1)n1xn

32) x1x3/2 in x

33) 11+sin2x in sinx

Answer
n=0(1)nsin2n(x)

34) sec2x in sinx

In exercises 35 - 38, evaluate the telescoping series or state whether the series diverges.

35) n=121/n21/(n+1)

Answer
Sk=221/(k+1)1 as k.

36) n=11n131(n+1)13

37) n=1(nn+1)

Answer
Sk=1k+1 diverges

38) n=1(sinnsin(n+1))

Express each series in exercises 39 - 42 as a telescoping sum and evaluate its nth partial sum.

39) n=1ln(nn+1)

Answer
n=1[lnnln(n+1)],
Sk=ln(k+1)

40) n=12n+1(n2+n)2 (Hint: Factor denominator and use partial fractions.)

41) n=2ln(1+1n)(lnn)ln(n+1)

Answer
an=1lnn1ln(n+1) and Sk=1ln(2)1ln(k+1)1ln(2)

42) n=1(n+2)n(n+1)2n+1 (Hint: Look at 1/(n2n).

A general telescoping series is one in which all but the first few terms cancel out after summing a given number of successive terms.

43) Let an=f(n)2f(n+1)+f(n+2), in which f(n)0 as n. Find n=1an.

Answer
n=1an=f(1)f(2)

44) an=f(n)f(n+1)f(n+2)+f(n+3), in which f(n)0 as n. Find n=1an.

45) Suppose that an=c0f(n)+c1f(n+1)+c2f(n+2)+c3f(n+3)+c4f(n+4), where f(n)0 as n. Find a condition on the coefficients c0,,c4 that make this a general telescoping series.

Answer
c0+c1+c2+c3+c4=0

46) Evaluate n=11n(n+1)(n+2) (Hint: 1n(n+1)(n+2)=12n1n+1+12(n+2))

47) Evaluate n=22n3n.

Answer
2n31=1n12n+1n+1,
Sn=(11+1/3)+(1/22/3+1/4)+(1/32/4+1/5)+(1/42/5+1/6)+=1/2

48) Find a formula for n=11n(n+N) where N is a positive integer.

49) [T] Define a sequence tk=k1n=1(1/k)lnk. Use the graph of 1/x to verify that tk is increasing. Plot tk for k=1100 and state whether it appears that the sequence converges.

Answer

tk converges to 0.57721tk is a sum of rectangles of height 1/k over the interval [k,k+1] which lie above the graph of 1/x.

CNX_Calc_Figure_09_02_206.jpeg

50) [T] Suppose that N equal uniform rectangular blocks are stacked one on top of the other, allowing for some overhang. Archimedes’ law of the lever implies that the stack of N blocks is stable as long as the center of mass of the top (N1) blocks lies at the edge of the bottom block. Let x denote the position of the edge of the bottom block, and think of its position as relative to the center of the next-to-bottom block. This implies that (N1)x=(12x) or x=1/(2N). Use this expression to compute the maximum overhang (the position of the edge of the top block over the edge of the bottom block.) See the following figure.

CNX_Calc_Figure_09_02_201.jpeg

Each of the following infinite series converges to the given multiple of π or 1/π.

In each case, find the minimum value of N such that the Nth partial sum of the series accurately approximates the left-hand side to the given number of decimal places, and give the desired approximate value. Up to 15 decimals place, π=3.141592653589793....

51) [T] π=3+n=1n2nn!2(2n)!, error <0.0001

Answer
N=22,
SN=6.1415

52) [T] π2=k=0k!(2k+1)!!=k=02kk!2(2k+1)!, error <104

53) [T] 98012π=49801k=0(4k)!(1103+26390k)(k!)43964k, error <1012

Answer
N=3,
SN=1.559877597243667...

54) [T] 112π=k=0(1)k(6k)!(13591409+545140134k)(3k)!(k!)36403203k+3/2, error <1015

55) [T] A fair coin is one that has probability 1/2 of coming up heads when flipped.

a. What is the probability that a fair coin will come up tails n times in a row?

b. Find the probability that a coin comes up heads for the first time after an even number of coin flips.

Answer
a. The probability of any given ordered sequence of outcomes for n coin flips is 1/2n.
b. The probability of coming up heads for the first time on the n th flip is the probability of the sequence TTTH which is 1/2n. The probability of coming up heads for the first time on an even flip is n=11/22n or 1/3.

56) [T] Find the probability that a fair coin is flipped a multiple of three times before coming up heads.

57) [T] Find the probability that a fair coin will come up heads for the second time after an even number of flips.

Answer
5/9

58) [T] Find a series that expresses the probability that a fair coin will come up heads for the second time on a multiple of three flips.

59) [T] The expected number of times that a fair coin will come up heads is defined as the sum over n=1,2, of n times the probability that the coin will come up heads exactly n times in a row, or n2n+1. Compute the expected number of consecutive times that a fair coin will come up heads.

Answer
E=n=1n2n+1=1, as can be shown using summation by parts

60) [T] A person deposits $10 at the beginning of each quarter into a bank account that earns 4 annual interest compounded quarterly (four times a year).

a. Show that the interest accumulated after n quarters is $10(1.01n+110.01n).

b. Find the first eight terms of the sequence.

c. How much interest has accumulated after 2 years?

61) [T] Suppose that the amount of a drug in a patient’s system diminishes by a multiplicative factor r<1 each hour. Suppose that a new dose is administered every N hours. Find an expression that gives the amount A(n) in the patient’s system after n hours for each n in terms of the dosage d and the ratio r. (Hint: Write n=mN+k, where 0k<N, and sum over values from the different doses administered.)

Answer
The part of the first dose after n hours is drn, the part of the second dose is drnN, and, in general, the part remaining of the mth dose is drnmN, so A(n)=ml=0drnlN=ml=0drk+(ml)N=mq=0drk+qN=drkmq=0rNq=drk1r(m+1)N1rN,wheren=k+mN.

62) [T] A certain drug is effective for an average patient only if there is at least 1 mg per kg in the patient’s system, while it is safe only if there is at most 2 mg per kg in an average patient’s system. Suppose that the amount in a patient’s system diminishes by a multiplicative factor of 0.9 each hour after a dose is administered. Find the maximum interval N of hours between doses, and corresponding dose range d (in mg/kg) for this N that will enable use of the drug to be both safe and effective in the long term.

63) Suppose that an0 is a sequence of numbers. Explain why the sequence of partial sums of an is increasing.

Answer
SN+1=aN+1+SNSN

64) [T] Suppose that an is a sequence of positive numbers and the sequence Sn of partial sums of an is bounded above. Explain why n=1an converges. Does the conclusion remain true if we remove the hypothesis an0?

65) [T] Suppose that a1=S1=1 and that, for given numbers S>1 and 0<k<1, one defines an+1=k(SSn) and Sn+1=an+1+Sn. Does Sn converge? If so, to what? (Hint: First argue that Sn<S for all n and Sn is increasing.)

Answer
Since S>1,a2>0, and since k<1,S2=1+a2<1+(S1)=S. If Sn>S for some n, then there is a smallest n. For this n,S>Sn1, so Sn=Sn1+k(SSn1)=kS+(1k)Sn1<S, a contradiction. Thus Sn<S and an+1>0 for all n, so Sn is increasing and bounded by S. Let S=limSn. If S<S, then δ=k(SS)>0, but we can find n such that SSn<δ/2, which implies that Sn+1=Sn+k(SSn)>S+δ/2, contradicting that Sn is increasing to S. Thus SnS.

66) [T] A version of von Bertalanffy growth can be used to estimate the age of an individual in a homogeneous species from its length if the annual increase in year n+1 satisfies an+1=k(SSn), with Sn as the length at year n,S as a limiting length, and k as a relative growth constant. If S1=3,S=9, and k=1/2, numerically estimate the smallest value of n such that Sn8. Note that Sn+1=Sn+an+1. Find the corresponding n when k=1/4.

67) [T] Suppose that n=1an is a convergent series of positive terms. Explain why limNn=N+1an=0.

Answer
Let Sk=kn=1an and SkL. Then Sk eventually becomes arbitrarily close to L, which means that LSN=n=N+1an becomes arbitrarily small as N.

68) [T] Find the length of the dashed zig-zag path in the following figure.

CNX_Calc_Figure_08_02_202.jpeg

69) [T] Find the total length of the dashed path in the following figure.

CNX_Calc_Figure_09_02_203.jpeg

Answer
L=(1+12)n=112n=32.

70) [T] The Sierpinski triangle is obtained from a triangle by deleting the middle fourth as indicated in the first step, by deleting the middle fourths of the remaining three congruent triangles in the second step, and in general deleting the middle fourths of the remaining triangles in each successive step. Assuming that the original triangle is shown in the figure, find the areas of the remaining parts of the original triangle after N steps and find the total length of all of the boundary triangles after N steps.

CNX_Calc_Figure_08_02_204.jpeg

71) [T] The Sierpinski gasket is obtained by dividing the unit square into nine equal sub-squares, removing the middle square, then doing the same at each stage to the remaining sub-squares. The figure shows the remaining set after four iterations. Compute the total area removed after N stages, and compute the length the total perimeter of the remaining set after N stages.

CNX_Calc_Figure_09_02_205.jpeg

Answer
At stage one a square of area 1/9 is removed, at stage 2 one removes 8 squares of area 1/92, at stage three one removes 82 squares of area 1/93, and so on. The total removed area after N stages is N1n=08N9N+1=181(8/9)N18/91 as N. The total perimeter is 4+4n=08N3N+1.

This page titled 1.3E: Exercises for Section 1.3 is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform.

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