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Timestamp: 2019-04-20 04:40:58+00:00

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number, e. You will need both of these later.
A sequence is a set of terms (or numbers) arranged in a definite order.
(i) 3, 7, 11, 15, . . .
(ii) 4, 9, 16, 25, . . .
This sequence can be rewritten as 22 , 32 , 42 , 52 , . . . The next term is 62 , or 36.
The dots(. . . ) indicate that the sequence continues indefinitely – it is an infinite sequence.
A sequence such as 3, 6, 9, 12 (stopping after a finite number of terms) is a finite sequence.
(i) 4, 9, 16, 25, . . . The formula for the nth term is un = (n + 1)2 .
(ii) un = 2n + 3. The sequence given by this formula is: 5, 7, 9, 11, . . .
(iii) un = 2n + n. The sequence is: 3, 6, 11, 20, . . .
(i) Suppose we know that: un = un−1 + 7n and u1 = 1.
find the whole sequence : 1, 15, 36, 64, . . .
The sequence defined by this formula is: 1, 1, 2, 3, 5, 8, 13, . . .
3 + 6 + 9 + 12 + 15.
(a) 2, 5, 8, 11, . . .
(d) 36, 18, 9, 4.5, . . .
(b) 0.25, 0.75, 1.25, 1.75, 2.25, . . .
(e) 1, −2, 3, −4, 5, . . .
(c) 5, −1, −7, . . .
the first five terms of the sequence.
the value of u4 .
SERIES AND LIMITS .3. SEQUENCES.
is −3. The common difference is 2.1: Some Arithmetic Sequences (i) 1. d.1. is 3. .3. The common difference is −6. Arithmetic and Geometric Sequences 2. 2. . a. the formula for the nth term is: un = a + (n − 1)d Examples 2. 5. pairing terms: Sn = a + (a + d) + (a + 2d) + · · · + (a + (n − 1)d) Sn = (a + (n − 1)d) + (a + (n − 2)d) + (a + (n − 3)d) + · · · + a 2Sn = (2a + (n − 1)d) + (2a + (n − 1)d) + (2a + (n − 1)d) + · · · + (2a + (n − 1)d) = n (2a + (n − 1)d) Dividing by 2 gives the formula. Arithmetic Series When the terms in an arithmetic sequence are summed. So the 10th term is: 5 + 9 × 7 = 68. the sum of the first n terms is: n Sn = (2a + (n − 1)d) 2 We can check that the formula works: S5 = 5 2 (2 × 3 + 4 × 3) = 45 2.2. (ii) 13. (ii) If an arithmetic sequence has u10 = 24 and u11 = 27. 7. . Arithmetic Sequences An arithmetic sequence is one in which each term can be obtained by adding a fixed number (called the common difference) to the previous term. 3. We can calculate it directly: S5 = 3 + 7 + 11 + 15 + 19 = 45 But there is a general formula: If an arithmetic sequence has first term a and common difference d. then add them together. . 7. we obtain an arithmetic series. . −5.2: Arithmetic Sequences (i) What is the 10th term of the arithmetic sequence 5. . ? In this sequence a = 5 and d = 7. what is the first term? The common difference.   39 2. Suppose we want to find the sum of the first 5 terms of the arithmetic sequence with first term 3 and common difference 4. 1. Using the formula for the 11th term: 27 = a + 10 × 3 Hence the first term. . 12. . . . 19. To Prove the Formula for an Arithmetic Series Write down the series in order and in reverse order. In an arithmetic sequence with first term a and common difference d. Examples 2.
and u22 = 108. +1. .7159 (ii) Which terms of the sequence are greater than 20? The nth term is given by un = 2 × (1. . −8. . 48. . the formula for the nth term is: un = arn−1 Examples 2. find the values of a and d for the arithmetic sequences: (a) 4. with a = 2 and r = 1. −11. (c) −7. . −10.1)9 = 4. .1)n−1 > log10 10 (n − 1) log10 1. . 2. . 8.1. .4. . . .1 So all terms from the 26th onwards are greater than 20. (c) Now use the formula to calculate the sum of 21+19+17+. 7.2: Arithmetic Sequences and Series (1) Using the notation above. −15. (b) −1. 1.1. . (2) Find the nth term in the following arithmetic sequences: (a) 44. Geometric Sequences A geometric sequence is one in which each term can be obtained by multiplying the previous term by a fixed number. SERIES AND LIMITS Exercises 3. what is the first term? (4) Use the formula for an arithmetic series to calculate the sum of the first 8 terms of the arithmetic sequence with first term 1 and common difference 10. 27. −11. .5.1)n−1 > 20 (1. 13. 10. 1.1 > 1 1 n > + 1 = 25.3: Geometric Sequences (i) 1 2 . −7. . . (i) What is the 10th term? Applying the formula. SEQUENCES. In a geometric sequence with first term a and common ratio r.2 log10 1. . 3. 2. Each term is double the previous one. (b) −3.5. 19. section 5): log10 (1. (b) Use the formula for an arithmetic series to calculate 21 + 19 + 17 + 15 + 13. . called the common ratio. 17. 4. (6) Is the sequence un = −4n + 2 arithmetic? If so. (ii) 81. 46.4: Consider the geometric sequence with first term 2 and common ratio 1. 15. . Examples 2. .1)n−1 . . 13. The common ratio is 13 .40 3. . . The common ratio is 2. . . . u10 = 2 × (1. what is the common difference? 2. . 9. (5) (a) Find the values of a and d for the arithmetic sequence: 21. It exceeds 20 if: 2 × (1. (3) If an arithmetic sequence has u20 = 100.1)n−1 > 10 Taking logs of both sides (see chapter 1. 5.
4. .5 2.8. 9. what are the values of a and r for the sequence: 4.5)9 There is a general formula: For a geometric sequence with first term a and common ratio r. −8. 1. 20. SERIES AND LIMITS 41 2.3 Geometric Series. . rSn = ar + ar2 + ar3 + . .5 + 0.75 + · · · + 3 × (0. . . . Simplify your answer as much as possible.5)10 ) = 5.5 + 0. 0.5. To Prove the Formula for a Geometric Series Write down the series and then multiply it by r: Sn = a + ar + ar2 + ar3 + . . (6) Find the sum of the first n terms of the geometric sequence: 20.25. 64. .994 1 − 0. (3) In the sequence 1. . • Jacques §3.25. . 0. . which is the first term term greater than 1000? (4) (a) Using the notation above. 16.3.5. ? (b) Use the formula for a geometric series to calculate: 4 + 2 + 1 + 0. . (5) Find the sum of the first 10 terms of the geometric sequence: 4. . (7) Use the formula for a geometric series to show that: 1 + x x2 x3 16 − x4 + + = 2 4 8 16 − 8x Further reading and exercises • For more practice with arithmetic and geometric sequences and series. 4. . 40. 0. . Subtract the second equation from the first: + arn−1 + arn−1 + arn Sn − rSn = a − arn a(1 − rn ) =⇒ Sn = 1−r Exercises 3. . .3: Geometric Sequences and Series (1) Find the 8th term and the nth term in the geometric sequence: 5. . 27. . the sum of the first n terms is: a(1 − rn ) Sn = 1−r So the answer is: S50 = 3(1 − (0. 10. SEQUENCES. you could use an A-level pure maths textbook. (2) Find the 15th term and the nth term in the geometric sequence: −2. 2. 16.6. .5: S10 = 3 + 1. . Geometric Series Suppose we want to find the sum of the first 10 terms of the geometric sequence with first term 3 and common ratio 0. . . 3. .
SERIES AND LIMITS . SEQUENCES.42 3.
42. so that you have £561. work through Jacques Chapter 3. section 5): t log10 1. (ii) How long will you have to wait to double your initial investment? The initial amount will have doubled when: 500 × (1. in practice. In this case we say that the bank pays an equivalent annual rate of i.80 in total.06)t = 1000 =⇒ (1. then after one year you have a total amount y1 : y1 = P (1 + i) after two years: y2 = (P (1 + i)) (1 + i) = P (1 + i)2 and after t years: yt = P (1 + i)t This is a geometric sequence with common ratio (1 + i).0610 = £895. Suppose the bank now pays interest m times a year at a rate of mi .1 .  3.06)per annum. and the interest is paid at the end of each year. Interval of Compounding In the previous section we assumed that interest was paid annually.1 More generally. 0. After two years. Economic Application: Interest Rates.80. financial institutions often pay interest more frequently. at a fixed interest rate of 6% (that is.06)t = 2 Taking logs of both sides (see chapter 1. However. if you invest an amount P (the “principal”) and interest is paid annually at interest rate i. After 1 year you would have:   i m P 1+ m and after t years:   i mt P 1+ m 1If you are not confident with calculations involving percentages. At the end of one year you receive an interest payment of 0.8957 log10 1. 3.1: If you save £500 at a fixed interest rate of 6% paid annually: (i) How much will you have after 10 years? Using the formula above.06 = log10 2 log10 2 t = = 11. you receive an interest payment of 0. and so on.1.06 × 530 = £31. Savings and Loans  Suppose that you invest £500 at the bank. y10 = 500 × 1.06 × 500 = £30. perhaps quarterly or even monthly. which is added to your account.06 So you will have to wait 12 years. We call the time period between interest payments the interval of compounding. Examples 3. so you have £530.
17% Note: When we say. at a fixed interest rate i (compounded annually).5.05) − 1) = £2641.00512 = 1.243% in one year. Banks often describe their savings accounts in terms of the APR.024 = £1082. We can use the formula from section 2. if you saved £200 at the beginning of each year for 10 years.08 = £1080.08 5×12 1000 × 1 + = £1489. gives the same yield. if compounded annually.06 12 = 1. Regular Savings Suppose that you invest an amount A at the beginning of every year. The yield is the same if:   0.2. (iii) How much will you have at the end of 5 years if the bank pays interest monthly? Using the formula with m = 12 and t = 5.06 12t t P (1 + i) = P 1 + 12   0.3. for example. 3. at 5% interest. it is the rate of interest that. At the end of t years. what is the APR? 12t If you invested an amount P for t years. for example. The sum is:   A(1 + i) 1 − (1 + i)t A(1 + i) St = = (1 + i)t − 1 1 − (1 + i) i So. 10 then you would accumulate 200 1.2: You invest £1000 for two years in the bank. not the APR.06 . or APR. Examples 3. and so on.3: Annual Percentage Rate If a bank pays interest monthly at an equivalent annual rate of 6%. SEQUENCES.05 0. whereas if you 12 invested at interest rate i compounded annually.43.36. The total amount that you will have at the end of t years is: St = A(1 + i)t + A(1 + i)t−1 + A(1 + i)t−2 + · · · + A(1 + i)2 + A(1 + i) = A(1 + i) + A(1 + i)2 + A(1 + i)3 + · · · + A(1 + i)t−1 + A(1 + i)t This is the sum of the first t terms of a geometric sequence with first term A(1 + i). you can see that if the bank pays interest quarterly and the equivalent annual rate is 8%. This rate is known as the Annual Percentage Rate. you would have P (1 + i)t .85 12 From this example.0617 So the APR is 6. (i) How much will you have at the end of one year if the bank pays interest annually? You will have: 1000 × 1. (ii) How much will you have at the end of one year if the bank pays interest quarterly? Using the formula above with m = 4. SERIES AND LIMITS 43 Examples 3. . “the annual interest rate is 3%” or “the interest rate is 3% per annum” we normally mean the equivalent annual rate. the amount you invested in the second year will be worth A(1 + i)t−1 . which pays interest at an equivalent annual rate of 8%.0617 (1 + i) = 1+ 12 =⇒ i = 0.05 ((1. Note that you are better off (for a given equivalent annual rate) if the interval of compounding is shorter. you would have P 1 + 0. so that customers do not need to do calculations involving the interval of compounding. you will have 1000 × 1. then your investment grows by 8. you will have:   0. and common ratio (1 + i). the amount you invested at the beginning of the first year will be worth A(1 + i)t .
Using the formula from section 2. + y The right-hand side of this equation is the sum of t terms of a geometric sequence with first term y and common ratio (1 + i) (in reverse order). SERIES AND LIMITS 3. at an interest rate of 9%.2.1 and 3. Exercises 3. Xt must be zero: =⇒ L(1 + i)t = y(1 + i)t−1 + y(1 + i)t−2 + . Paying Back a Loan If you borrow an amount L. . what is the APR? (4) If you save £10 at the beginning of each year for 20 years. • Anthony & Biggs Chapter 4 .4: Interest Rates.) (1) Suppose that you save £300 at a fixed interest rate of 4% per annum. (a) How much would you have after 4 years if interest were paid annually? (b) How much would you have after 10 years if interest were compounded monthly? (2) If you invest £20 at 15% interest. Savings. to be paid back in annual repayments over t years.3. and the interest rate is i. (a) What should the repayments be if the loan is to be repaid in 25 years? (b) Find a formula for the repayments if the repayment period is T years. interest will have been added to the loan.44 3. After repaying y you will owe: X1 = L(1 + i) − y At the end of two years you will owe: X2 = (L(1 + i) − y) (1 + i) − y = L(1 + i)2 − y(1 + i) − y At the end of three years: X3 = L(1 + i)3 − y(1 + i)2 − y(1 + i) − y and at the end of t years: Xt = L(1 + i)t − y(1 + i)t−1 − y(1 + i)t−2 − . how much will you have at the end of 20 years? (5) Suppose you take out a loan of £100000. . Further reading and exercises • Jacques §3. to be repaid in regular annual repayments. and Loans (Assume annual compounding unless otherwise specified. SEQUENCES. how long will it be before you have £100? (3) If a bank pays interest daily at an equivalent annual rate of 5%. how much do you need to repay each year? Let the annual repayment be y. . At the end of the first year. and the annual interest rate is 5%. . − y But if you are to pay off the loan in t years.5:  y (1 + i)t − 1 t L(1 + i) = i Li(1 + i)t =⇒ y = ((1 + i)t − 1) This is the amount that you need to repay each year.
If you accepted the £1000 today. A friend of yours has the winning ticket. if your friend will sell the ticket for less than £4535. . you would have £1000(1 + i) in a year’s time. and saved it at interest rate i. Or. SERIES AND LIMITS   45  4. How much would you be prepared to pay to buy the ticket. you would be indifferent between (a) taking out a loan of £4535. Another way of looking at this is to consider what cash sum now would be equivalent to a gift of a gift of £1050 in one year’s time. and (b) doing nothing.1: Present Value and Investment (i) The prize in a lottery is £5000. the future value of (a) is £1080. and repaying the loan after 2 years when you receive the prize. We could say: Future value of (a) = 1000(1 + i) Future value of (b) = 1050 You should accept the gift that has higher future value. the Present Value of an amount A to be received in t years’ time is: A P = (1 + i)t The present value is also known as the Present Discounted Value.15. For example. Examples 4. if the interest rate is 8%. and (b) saving your money at 5%. Present Value and Investment  Would you prefer to receive (a) a gift of £1000 today. Either way. or (b) a gift of £1050 in one year’s time? Your decision (assuming you do not have a desperate need for some immediate cash) will depend on the interest rate. you would be indifferent between (a) paying this for the ticket.05)2 This is the maximum amount you should pay. you should buy it.15.15 (1.3. but the prize will be paid in two years’ time. it would be better to take (b). If you have £4535. if you are able to borrow and save at an interest rate of 5%? The present value of the ticket is: P = 5000 = 4535. buying the ticket. so you should accept that.15. if you don’t have any money at the moment. payments received in the future are worth less – we “discount” them at the interest rate i. An amount P received now would be equivalent to an amount £1050 in one year’s time if: P (1 + i) = 1050 1050 =⇒ P = 1+i We say that: The Present Value of “£1050 in one year’s time” is 1050 1+i More generally: If the annual interest rate is i. SEQUENCES. But if the interest rate is less than 5%.
however.90 + 3247.04 .5: P = −10000 + = −10000 +    1 10 1.04 1. . we can calculate the present value of an annuity: PV = = = A A A A + + + ··· + 1 + i (1 + i)2 (1 + i)3 (1 + i)N   N  1 A 1+i 1 − 1+i   1 1 − 1+i   N  1 A 1 − 1+i i The present value tells you the price you would be prepared to pay for the asset.1. without thinking about exactly how the money to buy the ticket is to be obtained. If the interest rate is 4%. should you take it? We can calculate the present value of the investment opportunity by adding up the present values of all the amounts paid out and received: P = −10000 + 1000 1000 1000 1000 5000 + + + ··· + + 2 3 10 1. The first term is 1000 1 1. Using the formula for a geometric series. This does rely. for an initial outlay of £10000.0411 + 3247.04 1000 1. SERIES AND LIMITS We can see from this example that Present Value is a powerful concept: a single calculation of the PV enables you to answer the question. (ii) An investment opportunity promises you a payment of £1000 at the end of each of the next 10 years. on the assumption that you can borrow and save at the same interest rate. SEQUENCES.04 + 1 1 − 1.90 0.04 1. Using the formula from section 2.04    1 10 1000 1 − 1.04 1. and a capital sum of £5000 at the end of the 11th year. 4.46 3.0411 In the middle of this expression we have (again) a geometric series.04   ! 1 10 = −10000 + 25000 1 − + 3247.90 = −10000 + 11358.04 1− 5000 1.04 1.80 The present value of the opportunity is positive (or equivalently.04 and the common ratio is 1. Annuities An annuity is a financial asset which pays you an amount A each year for N years.80 = £1358.90 1. the present value of the return is greater than the initial outlay): you should take it.04 = −10000 + 8110.
What price will you be prepared to accept? (3) The useful life of a bus is five years. What is the present value of the gift (a) if the interest rate is 3% (b) if the interest rate is 10%? (2) (a) How much would you pay for an annuity that pays £20 a year for 10 years. Operating the bus brings annual profits of £10000.4 • Anthony & Biggs Chapter 4 • Varian also discusses Present Value and has more economic examples.5: Present Value and Investment (1) On your 18th birthday. then after receiving the third payment. . In the first year.3. What is the value of a new bus if the interest rate is 6%? (4) An investment project requires an initial outlay of £2400. if the interest rate is 5%? (b) You buy it. your parents promise you a gift of £500 when you are 21. thereafter operating costs increase by £500 a year. (a) What is the maximum length of time for which the project should operate? (b) Should it be undertaken if the interest rate is 5%? (c) Should it be undertaken if the interest rate is 10%? Further reading and exercises • Jacques §3. SERIES AND LIMITS 47 Exercises 3. operating costs are £600. SEQUENCES. and can generate revenue of £2000 per year. you consider selling the annuity.
. . (iii) un = n1 The terms of this sequence get smaller and smaller: 1. We can see that it converges: lim un = 4 n→∞ (ii) un = (−1)n This sequence is −1.9999. . This is a geometric sequence with common ratio 2. We say that “the limit of the sequence as n tends to infinity is zero” or “the sequence converges to zero” or: lim un = 0 n→∞ Examples 5.000000954 we can see that as n gets larger.99.1: Limits of Sequences (i) un = 4 − (0. +1.1. 16. .999. by n3 . so: lim un = lim n→∞ n→∞ 2+ 3 1 n ! = 2 3 . 14 . un gets closer and closer to zero. . 12 . 3n3 A useful trick is to divide the numerator and the denominator by the highest power of n. SEQUENCES.000977 u20 = = 0. that is. . 32. Then: ! 2 + n1 un = 3 and we know that 1 n → 0. It has no limit. . . It converges to zero: lim 1 n→∞ n =0 (iv) 2. +1. The terms get bigger and bigger. 4. . SERIES AND LIMITS   5. Limits   5. .9.48 3. 3. −1. 3.1)n The sequence is: 3. 3. 15 . .5 u10 = ( 12 )10 ( 12 )20 = 0. −1. 13 . 8. The Limit of a Sequence If we write down some of the terms of the geometric sequence: un = ( 12 )n : u1 = ( 21 )1 = 0. +1. . It diverges: un → ∞ as n → ∞ (v) un = 2n3 + n2 .
(1) un = ( 13 )n (2) un = −5 + ( 14 )n (3) un = (− 13 )n (4) un = 7 − ( 25 )n (5) un = 10 n3 (6) un = (1. . SEQUENCES. r →∞ 5. find the limit. n→∞ n • If r > 1.2. Using the sigma notation this can be written: ∞ X i=1  1 i−1 2 =2 . SERIES AND LIMITS 49 Exercises 3. 12 .6: Say whether each if the following sequences converges or diverges as n → ∞.5: Sn n 1 − 12 a(1 − rn ) =2 1− = = 1−r 1 − 12 n−1 = 2 − 12   1 n 2 As the number of terms gets larger and larger.3. 14 .2)n 10 n3 (7) un = 25n + (8) un = 7n2 +5n n2 From examples like these we can deduce some general results that are worth remembering: 1 =0 n→∞ n 1 1 • Similarly lim 2 = 0 and lim 3 = 0 etc n→∞ n n→∞ n • lim lim rn = 0 • If |r| < 1. their sum gets closer and closer to 2: lim Sn = 2 n→∞ Equivalently. If it converges. . We can find the sum of the first n terms: n−1 Sn = 1 + 12 + 14 + 18 + · · · + 12 using the formula from section 2. It is a geometric sequence with first term a = 1 and common ratio r = 21 . . Infinite Geometric Series Consider the sequence: 1. we can write this as: 1+ 1 2 + 1 4 + 1 8 + ··· = 2 So we have found the sum of an infinite number of terms to be a finite number. 18 .
SEQUENCES. the present value tells you the price you would be prepared to pay for the asset. If the interest rate is i.2: Find the sum to infinity of the following series: (i) 2 − 1 + 12 − 14 + 18 . Economic Application: Perpetuities In section 4. so it converges. . (assuming 0 < x < 0.3. We know 0 < r < 1. (Another way to get this result is to let = . so it diverges: the infinite sum sum does not exist. Using the formula above: a 2 4 S∞ = = 3 = 1−r 3 2 (ii) x + 2x2 + 4x3 + 8x4 + . rn → 0 so the series converges: lim Sn = n→∞ a 1−r or equivalently: ∞ X ari−1 = i=1 a 1−r But note that if |r| > 1 the terms of the series get bigger and bigger. The sum of the first n terms is: Sn = a(1 − rn ) 1−r As n → ∞. This is a geometric series.. . Even an asset that pays out forever has a finite price.5) This is a geometric series with a = x and r = 2x. The common ratio is of an infinite series: PV 1 1+i .. the present value of a perpetuity is: PV = A A A + + + . provided that |r| < 1. with a = 2 and r = − 12 . It converges because |r| < 1. The formula for the sum to infinity gives: a x S∞ = = 1−r 1 − 2x 5. Examples 5. 2 1 + i (1 + i) (1 + i)3 This is an infinite geometric series.50 3. Using the formula for the sum A 1+i = 1−  1 1+i  A i Again. SERIES AND LIMITS or (a little more neatly): ∞ X  1 i 2 =2 i=0 The same procedure works for any geometric series with common ratio r.1 we calculated the present value of an annuity – an asset that pays you an amount A each year for a fixed number of years. A perpetuity is an asset that pays you an amount A each year forever. . .
and Perpetuities (1) Evaluate the following infinite sums: 2 3 4 (a) 13 + 13 + 13 + 13 + . .2 + 0.3. . (b) 1 + 0. . refer to an A-level pure maths textbook. what is the present value of the firm? Further reading and exercises • Anthony & Biggs: §3.7: Infinite Series. • For more on limits of sequences. SERIES AND LIMITS 51 N → ∞ in the formula for the present value of an annuity that we obtained earlier. . .04 + 0. .) Exercises 3. (d) (e) 1 2 1 + 14 − 18 + 16 − . What assumption is needed here? x x x (2) If the interest rate is 4%. If the interest rate is 5%..008 + 0. and infinite sums. • Varian has more on financial assets including perpetuities. .. (c) 1 − 2 3 ∞ X  1 r 2 r=3 (f) ∞ X xr assuming |x| < 1.0016 + . SEQUENCES. (Why is this assumption necessary?) r=0 2y 2 4y 3 8y 4 + 2 + 3 + . . and works out the present value of a perpetuity in a different way. and then to rise by 2% each year after that (forever). . what is the present value of: (g) (a) an annuity that pays £100 each year for 20 years? (b) a perpetuity that pays £100 each year forever? How will the value of each asset have changed after 10 years? (3) A firm’s profits are expected to be £1000 this year. 4 7 + 23 + 23 + .3 discusses limits briefly.
As n → ∞. As the interval of compounding get shorter. from section 3. SEQUENCES...01)100 = (1. .8: Verify (approximately.705 = 2. The Number e   If we evaluate the numbers in the sequence:   1 n un = 1 + n we get: u1 = 2. .594 = 2.718281828459 .52 3. that if interest is paid m times a year at equivalent annual rate i. n = e2 . interest is compounded almost continuously. . or daily (m = 365).0001)10000 = (1.001)1000 = (1. weekly (m = 52). . SERIES AND LIMITS   6. then the return after t years from investing an inital amount P is:   i mt P 1+ m Interest might be paid quarterly (m = 4). So:   1 n lim 1 + = e (≈ 2. using a calculator) that lim 1 + n→∞ n Hint: Most calculators have a button that evaluates ex for any number x. . Or it could be paid even more frequently – every hour.71828) n→∞ n e is important in calculus (as we will see later) and arises in many economic applications. 6.717 = 2.1. Economic Application: Continuous Compounding Remember.1. the return after t years on an initial amount P is: P eit .25.71814 = 2. We can generalise this result to:  r n For any value of r. .370. monthly (m = 12). for example: u10 u100 u1000 u10000 u100000 = (1.1)10 = (1. This is an irrational number (see Chapter 1) known simply as e. u3 = (1 + 13 )3 = 2. . we can apply our result above to say that:   i m lim 1 + = ei m→∞ m and so: If interest is compounded continuously at rate i. u2 = (1 + 12 )2 = 2. every second . un gets closer and closer to a limit of 2. lim 1 + = er n→∞ n  2 Exercises 3. As m → ∞. For some higher values of n we have.71826 .00001)100000 = 2.
when solving economic problems we often simplify by assuming continuous compounding. We can see from these examples that with continuous compounding the APR is little different from the interest rate. an amount P invested for one year yields: P e0. So “Ae−it now” and “A after t years”.3.3. the present value of an amount A received in t years is: P = Ae−it Continuous compounding is particularly useful because it allows us to calculate the present value when t is not a whole number of years.4. e0. With continuous compounding. and interest is compounded countinuously: (a) How much will you have after 1 year? (b) How much will you have after 5 years? (c) What is the APR? (3) You expect to receive a gift of £100 on your next birthday.05127)P So the APR is 5. .1: If interest is compounded continuously. what is the APR if: (i) the interest rate is 5%? Applying the formula. If the interest rate is 5%.2.08329. are worth the same. SEQUENCES. the interest rate is 5%. note that if you have an amount Ae−it now. (ii) the interest rate is 8%? Similarly.2 and §7. To see where the formula comes from.9: e (1) Express the following in terms of e: (a) limn→∞ (1 + n1 )n (b) limn→∞ (1 + n5 )n (c) limn→∞ (1 + 1 n 2n ) (2) If you invest £100. 6. SERIES AND LIMITS 53 Examples 6.05127P = (1 + 0. when we showed that the present value of an amount A received in t years time A is (1+i) t . you will then have Ae−it eit = A. Exercises 3.329%. because it avoids the messy calculations for the interval of compounding. Present Value with Continuous Compounding In section 4. and you save it for t years with continuous compounding. if the interest rate is i.127%. So. • Jacques §2. what is the present value of the gift (a) six months before your birthday (b) 2 days before your birthday? Further reading and exercises • Anthony & Biggs: §7.08 = 1. we were assuming annual compounding of interest.05 = 1. so the APR is 8.
95 Exercises 3.389 to 3 decimal places) Exercises 3.75 (5) S10 = 1398100 n (6) Sn = 25(1 − 15 )  n−2 = 25 − 15  (7) a = 1.6: (1) un → 0 as n → ∞ (2) un → −5 as n → ∞ (3) un → 0 as n → ∞ (4) un → 7 as n → ∞ (5) un → 0 as n → ∞ (6) un → ∞ as n → ∞ 2This Version of Workbook Chapter 3: August 22.09 after ten years.13% i P = 100e− 2 = £97.25 (e) −6 (2) −6.g. (b) £2500.+(2n+1) (7) (a) n (b) 4n − 3 (8) There are several possibilities. so d = −4 Exercises 3. 26 (3) 4. un = −2 − 4(n − 1).03. 2 (a) r Pr=1 (b) P7r=2 3r n 2 (c) r=4 r 540.26 T (b) y = 5000(1. 6.53 2i (b) P = 100e− 365 = £99.75 (c) −13 (d) 2.73 (3) £42123 (4) (a) 3 years. Same value in ten years. (b) Yes PV=£95. 333.25 (2) t = 12 (3) 5. P25e.66 (2) (a) £154.96 (b) 447. SERIES AND LIMITS Solutions to Exercises in Chapter 3 Exercises 3. . .7: (1) (a) (b) (c) (d) (e) (f) 1 2 5 4 2 3 18 19 1 4 1 1−x If |x| ≥ 1 sequence diverges 2y 2 (g) x−2y assuming |2y| < |x| (2) (a) £1359. r = x2 .57 (b) £375.97 . 5 × 2n−1 −32768. (−2)n 8th (a) a = 4.05 −1 Exercises 3. (3) £33. 4 (4) u4 = −2 (5) 30 1 1 (6) (a) 1 + 14 + 19 + 16 + 25 (b) 1 + 2 + 4 + 8 (c) 1+3+5+. SEQUENCES. n = 4 16−x4 ⇒ S4 = 16−8x Exercises 3.05) T 1.13 £128. 4. 4.33 Exercises 3.13% (4) S20 = £557.9: (1) (a) (b) (c) (2) (a) (b) (c) (3) (a) e e5 1 e2 £105.40 5.65 (5) (a) £8024.43 (b) £115.54 3.5 (2) (a) un = 44 + 2(n − 1) = 42 + 2n (b) un = −3 − 4(n − 1) = −(4n + 7) (3) u1 = 24 (4) 288 (5) (a) a = 21 d = −2 (b) S5 = 85 (c) S11 = 121 (6) Yes.1: (1) (a) 14 (b) 2. 2003 (7) un → ∞ as n → ∞ (8) un → 7 as n → ∞ Exercises 3.5: (1) (a) £457.4: (1) (a) 350. r = 12 (b) S5 = 7.19 (c) No PV=−£82. 4.2: (1) (a) a = 4 d = 3 (b) a = −1 d = 3 (c) a = −7 d = −1.8: (1) (e2 = 7.3: (1) (2) (3) (4) Exercises 3. Is only worth £811.
SEQUENCES.3. SERIES AND LIMITS  55 .
P 2 (2) Write out the series: n−1 r=0 (2r − 1) without using sigma notation.75)n Longer Questions (1) Carol (an economics student) is considering two possible careers. 12.75 + 0. (5) If you invest £500 at a fixed interest rate of 3% per annum. and that her choice is made on graduation day. how much will you have after 4 years: (a) if interest is paid annually? (b) if interest is paid monthly? What is the APR in this case? (c) if interest is compounded continuously? If interest is paid annually. (a) If she decides to be an acrobat: (i) How much will she earn in the 3rd year of her career? (ii) How much will she earn in the nth year? (iii) What will be her total career earnings? (b) If she decides to be a beekeeper: (i) What will be her total career earnings? (ii) In which year will her annual earnings first exceed what she would have earned as an acrobat? (c) She knows that what matters for her choice of career is the present value of her earnings. .2.8. when will your savings exceed £600? (6) If the interest rate is 5% per annum. 15. (c) 0. she will earn £30000 in the first year. of: 5n2 + 4n + 3 (a) 3(1 + (0.25+· · ·+(0. 5. and can expect her earnings to increase at 1% per annum thereafter.8. as n → ∞. 27. 0.5+0.2)n ) (b) 2n2 + 1 (c) 0. . . (3) For each of the following series. . 3. . As a beekeeper. but the subsequent increase will be 5% per annum. work out how many terms there are and hence find the sum: (a) 3+4+5+· · ·+20 (b) 1+0. (Assume that earnings are received at the end of each year. Worksheet 3: Sequences. .) If she decides to be an acrobat: . . . (b) 1. As an acrobat. and Limits.5625 + · · · + (0. she will earn only £20000 in the first year. 8. 64. showing the first four terms and the last two terms.5)n−1 (c) 5+10+20+· · ·+5×2n (4) Express the series 3 + 7 + 11 + · · · + (4n − 1) + (4n + 3) using sigma notation. . Series. The rate of interest is i. what is the present value of: (a) An annuity that pays £100 a year for 20 years? (b) A perpetuity that pays £50 a year? (7) Find the limit. 10. the Economics of Finance  Quick Questions (1) What is the nth term of each of the following sequences: (a) 20. She plans to work for 40 years.2.
The computer he wants cost £1000. The insurance premium on the computer will start at £40 for the first year. what value of computer would the employer agree to purchase for Bill (assuming the insurance. SERIES AND LIMITS (i) What is the present value of her first year’s earnings? (ii) What is the present value of the nth year’s earnings? (iii) What is the total present value of her career earnings? (d) Which career should she choose if the interest rate is 3%? (e) Which career should she choose if the interest rate is 15%? (f) Explain these results. and decline by £5 per year throughout the life of the computer. (2) Bill is due to start a four year degree course financed by his employer and he feels the need to have his own computer. repair and resale schedules remain unchanged)? (b) Bill has another idea. and repair bills at the end.56 3. (a) Bill offers to forgo £360 per annum from his grant if his employer purchases a computer for him and meets all insurance and repair bills. SEQUENCES. He still wants the £1000 computer. while repair bills start at £50 in the computer’s first year. and increase by 50% per annum thereafter. Again. How should the employer respond in this case? . The computer will be sold at the end of the degree course and the proceeds paid to the employer.) The resale value for computers is given by the following table: Resale value at end of year Year Year Year Year 1 2 3 4 and onwards 75% of initial cost 60% of initial cost 20% of initial cost £100 The interest rate is 10% per annum. Should the employer agree to the scheme? If not. the employer would meet all bills and receive the proceeds from the sale of both computers. (The grant and the insurance premium are paid at the beginning of each year. but suggests that it be replaced after two years with a new one. and Bill would forgo £360 per annum from his grant.

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