Sequences, Series, and Limits; the Economics of Finance, Exams of Economics

Download Sequences, Series, and Limits; the Economics of Finance and more Exams Economics in PDF only on Docsity! CHAPTER 3 Sequences, Series, and Limits; the Economics of Finance If you have done A-level maths you will have studied Sequences and Series (in particular Arithmetic and Geometric ones) before; if not you will need to work carefully through the first two sections of this chapter. Sequences and series arise in many economic applications, such as the economics of finance and investment. Also, they help you to understand the concept of a limit and the significance of the natural number, e. You will need both of these later. —./—  1. Sequences and Series 1.1. Sequences A sequence is a set of terms (or numbers) arranged in a definite order. Examples 1.1: Sequences (i) 3, 7, 11, 15, . . . In this sequence each term is obtained by adding 4 to the previous term. So the next term would be 19. (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. Suppose we write u1 for the first term of a sequence, u2 for the second and so on. There may be a formula for un, the nth term: Examples 1.2: The nth term of a 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, . . . Or there may be a formula that enables you to work out the terms of a sequence from the preceding one(s), called a recurrence relation: Examples 1.3: Recurrence Relations (i) Suppose we know that: un = un−1 + 7n and u1 = 1. Then we can work out that u2 = 1 + 7 × 2 = 15, u3 = 15 + 7 × 3 = 36, and so on, to find the whole sequence : 1, 15, 36, 64, . . . 37 38 3. SEQUENCES, SERIES AND LIMITS (ii) un = un−1 + un−2, u1 = 1, u2 = 1 The sequence defined by this formula is: 1, 1, 2, 3, 5, 8, 13, . . . 1.2. Series A series is formed when the terms of a sequence are added together. The Greek letter Σ (pronounced “sigma”) is used to denote “the sum of”: n∑ r=1 ur means u1 + u2 + · · ·+ un Examples 1.4: Series (i) In the sequence 3, 6, 9, 12, . . . , the sum of the first five terms is the series: 3 + 6 + 9 + 12 + 15. (ii) 6∑ r=1 (2r + 3) = 5 + 7 + 9 + 11 + 13 + 15 (iii) k∑ r=5 1 r2 = 1 25 + 1 36 + 1 49 + · · ·+ 1 k2 Exercises 3.1: Sequences and Series (1) Find the next term in each of the following sequences: (a) 2, 5, 8, 11, . . . (b) 0.25, 0.75, 1.25, 1.75, 2.25, . . . (c) 5,−1,−7, . . . (d) 36, 18, 9, 4.5, . . . (e) 1,−2, 3,−4, 5, . . . (2) Find the 2nd, 4th and 6th terms in the sequence given by: un = n2 − 10 (3) If un = un−1 2 + 2 and u1 = 4 write down the first five terms of the sequence. (4) If un = u2 n−1 + 3un−1 and u3 = −2, find the value of u4. (5) Find the value of ∑4 r=1 3r (6) Write out the following sums without using sigma notation: (a) ∑5 r=1 1 r2 (b) ∑3 i=0 2i (c) ∑n j=0(2j + 1) (7) In the series ∑n−1 i=0 (4i+1), (a) how many terms are there? (b) what is the formula for the last term? (8) Express using the Σ notation: (a) 12 + 22 + 32 + ... + 252 (c) 16 + 25 + 36 + 49 + · · ·+ n2 (b) 6 + 9 + 12 + · · ·+ 21 Further reading and exercises • For more practice with simple sequences and series, you could use an A-level pure maths textbook. 3. SEQUENCES, SERIES AND LIMITS 41 2.5. 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 0.5: S10 = 3 + 1.5 + 0.75 + · · ·+ 3× (0.5)9 There is a general formula: For a geometric sequence with first term a and common ratio r, the sum of the first n terms is: Sn = a(1− rn) 1− r So the answer is: S50 = 3(1− (0.5)10) 1− 0.5 = 5.994 2.6. To Prove the Formula for a Geometric Series Write down the series and then multiply it by r: Sn = a + ar + ar2 + ar3 + . . . + arn−1 rSn = ar + ar2 + ar3 + . . . + arn−1 + arn Subtract the second equation from the first: Sn − rSn = a− arn =⇒ Sn = a(1− rn) 1− r Exercises 3.3: Geometric Sequences and Series (1) Find the 8th term and the nth term in the geometric sequence: 5, 10, 20, 40, . . . (2) Find the 15th term and the nth term in the geometric sequence: −2, 4,−8, 16, . . . (3) In the sequence 1, 3, 9, 27, . . . , which is the first term term greater than 1000? (4) (a) Using the notation above, what are the values of a and r for the sequence: 4, 2, 1, 0.5, 0.25, . . . ? (b) Use the formula for a geometric series to calculate: 4 + 2 + 1 + 0.5 + 0.25. (5) Find the sum of the first 10 terms of the geometric sequence: 4, 16, 64, . . . (6) Find the sum of the first n terms of the geometric sequence: 20, 4, 0.8, . . . Simplify your answer as much as possible. (7) Use the formula for a geometric series to show that: 1 + x 2 + x2 4 + x3 8 = 16− x4 16− 8x Further reading and exercises • For more practice with arithmetic and geometric sequences and series, you could use an A-level pure maths textbook. • Jacques §3.3 Geometric Series. 42 3. SEQUENCES, SERIES AND LIMITS   3. Economic Application: Interest Rates, Savings and Loans Suppose that you invest £500 at the bank, at a fixed interest rate of 6% (that is, 0.06) per annum, and the interest is paid at the end of each year. At the end of one year you receive an interest payment of 0.06× 500 = £30, which is added to your account, so you have £530. After two years, you receive an interest payment of 0.06 × 530 = £31.80, so that you have £561.80 in total, and so on.1 More generally, if you invest an amount P (the “principal”) and interest is paid annually at interest rate i, 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). Examples 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, y10 = 500× 1.0610 = £895.42. (ii) How long will you have to wait to double your initial investment? The initial amount will have doubled when: 500× (1.06)t = 1000 =⇒ (1.06)t = 2 Taking logs of both sides (see chapter 1, section 5): t log10 1.06 = log10 2 t = log10 2 log10 1.06 = 11.8957 So you will have to wait 12 years. 3.1. Interval of Compounding In the previous section we assumed that interest was paid annually. However, in practice, financial institutions often pay interest more frequently, perhaps quarterly or even monthly. We call the time period between interest payments the interval of compounding. Suppose the bank has a nominal (that is, stated) interest rate i, but pays interest m times a year at a rate of i m . After 1 year you would have: P ( 1 + i m )m and after t years: P ( 1 + i m )mt 1If you are not confident with calculations involving percentages, work through Jacques Chapter 3.1 3. SEQUENCES, SERIES AND LIMITS 43 Examples 3.2: You invest £1000 for two years in the bank, which pays interest at a nominal rate of 8%. (i) How much will you have at the end of one year if the bank pays interest annually? You will have: 1000× 1.08 = £1080. (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, you will have 1000× 1.024 = £1082.43. Note that you are better off (for a given nominal rate) if the interval of compounding is shorter. (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, you will have: 1000× ( 1 + 0.08 12 )5×12 = £1489.85 From this example, you can see that if the bank pays interest quarterly and the nominal rate is 8%, then your investment actually grows by 8.243% in one year. The effective annual in- terest rate is 8.243%. In the UK this rate is known as the Annual Equivalent Rate (AER) (or sometimes the Annual Percentage Rate (APR)). Banks often describe their savings accounts in terms of the AER, so that customers do not need to do calculations involving the interval of compounding. If the nominal interest rate is i, and interest is paid m times a year, an investment P grows to P (1 + i/m)m in one year. So the formula for the Annual Equivalent Rate is: AER = ( 1 + i m )m − 1 Examples 3.3: Annual Equivalent Rate If the nominal interest rate is 6% and the bank pays interest monthly, what is the AER? AER = ( 1 + 0.06 12 )12 − 1 = 0.0617 The Annual Equivalent Rate is 6.17%. 3.2. Regular Savings Suppose that you invest an amount A at the beginning of every year, at a fixed interest rate i (compounded annually). At the end of t years, the amount you invested at the beginning of the first year will be worth A(1 + i)t, the amount you invested in the second year will be worth A(1 + i)t−1, and so on. 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), and common ratio (1 + i). We can use the formula from section 2.5. The sum is: St = A(1 + i) ( 1− (1 + i)t ) 1− (1 + i) = A(1 + i) i ( (1 + i)t − 1 ) So, for example, if you saved £200 at the beginning of each year for 10 years, at 5% interest, then you would accumulate 2001.05 0.05((1.05)10 − 1) = £2641.36. 46 3. SEQUENCES, 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, without thinking about ex- actly how the money to buy the ticket is to be obtained. This does rely, however, on the assumption that you can borrow and save at the same interest rate. (ii) An investment opportunity promises you a payment of £1000 at the end of each of the next 10 years, and a capital sum of £5000 at the end of the 11th year, for an initial outlay of £10000. 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 1.04 + 1000 1.042 + 1000 1.043 + · · ·+ 1000 1.0410 + 5000 1.0411 In the middle of this expression we have (again) a geometric series. The first term is 1000 1.04 and the common ratio is 1 1.04 . Using the formula from section 2.5: P = −10000 + 1000 1.04 ( 1− ( 1 1.04 )10) 1− 1 1.04 + 5000 1.0411 = −10000 + 1000 ( 1− ( 1 1.04 )10) 0.04 + 3247.90 = −10000 + 25000 ( 1− ( 1 1.04 )10 ) + 3247.90 = −10000 + 8110.90 + 3247.90 = −10000 + 11358.80 = £1358.80 The present value of the opportunity is positive (or equivalently, the present value of the return is greater than the initial outlay): you should take it. 4.1. Annuities An annuity is a financial asset which pays you an amount A each year for N years. Using the formula for a geometric series, we can calculate the present value of an annuity: PV = A 1 + i + A (1 + i)2 + A (1 + i)3 + · · ·+ A (1 + i)N = A 1+i ( 1− ( 1 1+i )N ) 1− ( 1 1+i ) = A ( 1− ( 1 1+i )N ) i The present value tells you the price you would be prepared to pay for the asset. 3. SEQUENCES, SERIES AND LIMITS 47 Exercises 3.5: Present Value and Investment (1) On your 18th birthday, your parents promise you a gift of £500 when you are 21. 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, if the interest rate is 5%? (b) You buy it, then after receiving the third payment, you consider selling the annuity. What price will you be prepared to accept? (3) The useful life of a bus is five years. Operating the bus brings annual profits of £10000. What is the value of a new bus if the interest rate is 6%? (4) An investment project requires an initial outlay of £2400, and can generate revenue of £2000 per year. In the first year, operating costs are £600; 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 §3.4 • Anthony & Biggs Chapter 4 • Varian also discusses Present Value and has more economic examples. 48 3. SEQUENCES, SERIES AND LIMITS    5. Limits 5.1. The Limit of a Sequence If we write down some of the terms of the geometric sequence: un = (1 2)n: u1 = (1 2)1 = 0.5 u10 = (1 2)10 = 0.000977 u20 = (1 2)20 = 0.000000954 we can see that as n gets larger, un gets closer and closer to zero. We say that “the limit of the sequence as n tends to infinity is zero” or “the sequence converges to zero” or: lim n→∞ un = 0 Examples 5.1: Limits of Sequences (i) un = 4− (0.1)n The sequence is: 3.9, 3.99, 3.999, 3.9999, . . . We can see that it converges: lim n→∞ un = 4 (ii) un = (−1)n This sequence is −1,+1,−1,+1,−1,+1, . . . It has no limit. (iii) un = 1 n The terms of this sequence get smaller and smaller: 1, 1 2 , 1 3 , 1 4 , 1 5 , . . . It converges to zero: lim n→∞ 1 n = 0 (iv) 2, 4, 8, 16, 32, . . . This is a geometric sequence with common ratio 2. The terms get bigger and bigger. It diverges: un →∞ as n →∞ (v) un = 2n3 + n2 3n3 . A useful trick is to divide the numerator and the denominator by the highest power of n; that is, by n3. Then: un = ( 2 + 1 n 3 ) and we know that 1 n → 0, so: lim n→∞ un = lim n→∞ ( 2 + 1 n 3 ) = 2 3 3. SEQUENCES, SERIES AND LIMITS 51 N →∞ in the formula for the present value of an annuity that we obtained earlier.) Exercises 3.7: Infinite Series, and Perpetuities (1) Evaluate the following infinite sums: (a) 1 3 + ( 1 3 )2 + ( 1 3 )3 + ( 1 3 )4 + . . . (b) 1 + 0.2 + 0.04 + 0.008 + 0.0016 + . . . (c) 1− 1 2 + 1 4 − 1 8 + 1 16 − . . . (d) 2 3 + ( 2 3 )4 + ( 2 3 )7 + . . . (e) ∞∑ r=3 ( 1 2 )r (f) ∞∑ r=0 xr assuming |x| < 1. (Why is this assumption necessary?) (g) 2y2 x + 4y3 x2 + 8y4 x3 + . . . What assumption is needed here? (2) If the interest rate is 4%, what is the present value of: (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, and then to rise by 2% each year after that (forever). If the interest rate is 5%, what is the present value of the firm? Further reading and exercises • Anthony & Biggs: §3.3 discusses limits briefly. • Varian has more on financial assets including perpetuities, and works out the present value of a perpetuity in a different way. • For more on limits of sequences, and infinite sums, refer to an A-level pure maths textbook. 52 3. SEQUENCES, SERIES AND LIMITS    6. The Number e If we evaluate the numbers in the sequence: un = ( 1 + 1 n )n we get: u1 = 2, u2 = (1 + 1 2)2 = 2.25, u3 = (1 + 1 3)3 = 2.370, . . . For some higher values of n we have, for example: u10 = (1.1)10 = 2.594 u100 = (1.01)100 = 2.705 u1000 = (1.001)1000 = 2.717 u10000 = (1.0001)10000 = 2.71814 u100000 = (1.00001)100000 = 2.71826 . . . As n → ∞, un gets closer and closer to a limit of 2.718281828459 . . . This is an irrational number (see Chapter 1) known simply as e. So: lim n→∞ ( 1 + 1 n )n = e (≈ 2.71828) e is important in calculus (as we will see later) and arises in many economic applications. We can generalise this result to: For any value of r, lim n→∞ ( 1 + r n )n = er Exercises 3.8: Verify (approximately, using a calculator) that lim n→∞ ( 1 + 2 n )n = e2. Hint : Most calculators have a button that evaluates ex for any number x. 6.1. Economic Application: Continuous Compounding Remember, from section 3.1, that if interest is paid m times a year and the nominal rate is i, then the return after t years from investing an inital amount P is: P ( 1 + i m )mt Interest might be paid quarterly (m = 4), monthly (m = 12), weekly (m = 52), or daily (m = 365). Or it could be paid even more frequently – every hour, every second . . . As the interval of compounding get shorter, interest is compounded almost continuously. As m →∞, we can apply our result above to say that: lim m→∞ ( 1 + i m )m = ei and so: If interest is compounded continuously at rate i, the return after t years on an initial amount P is: Peit 3. SEQUENCES, SERIES AND LIMITS 53 Examples 6.1: If interest is compounded continuously, what is the AER if: (i) the interest rate is 5%? Applying the formula, an amount P invested for one year yields: Pe0.05 = 1.05127P = (1 + 0.05127)P So the AER is 5.127%. (ii) the interest rate is 8%? Similarly, e0.08 = 1.08329, so the AER is 8.329%. We can see from these examples that with continuous compounding the AER is little dif- ferent from the interest rate. So, when solving economic problems we often simplify by assuming continuous compounding, because it avoids the messy calculations for the interval of compounding. 6.2. Present Value with Continuous Compounding In section 4, when we showed that the present value of an amount A received in t years time is A (1+i)t , we were assuming annual compounding of interest. With continuous compounding, if the interest rate is i, 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. To see where the formula comes from, note that if you have an amount Ae−it now, and you save it for t years with continuous compounding, you will then have Ae−iteit = A. So “Ae−it now” and “A after t years”, are worth the same. Exercises 3.9: e (1) Express the following in terms of e: (a) limn→∞(1 + 1 n)n (b) limn→∞(1 + 5 n)n (c) limn→∞(1 + 1 2n)n (2) If you invest £100, the interest rate is 5%, and interest is compounded countinu- ously: (a) How much will you have after 1 year? (b) How much will you have after 5 years? (c) What is the AER? (3) You expect to receive a gift of £100 on your next birthday. If the interest rate is 5%, 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: §7.2 and §7.3. • Jacques §2.4.