Queen Mary University of London ECN115 Mathematical Methods PDF

ECN115 Mathematical Methods in Economics and Finance Evgenii Safonov 2024-2025. LECTURE 3 1 Quiz I ◁ Online Quiz I will be on Monday, 14 October ◁ Based on the material of first 2 lectures and homeworks (3rd week’s material is...

ECN115 Mathematical Methods in Economics and Finance Evgenii Safonov 2024-2025. LECTURE 3 1 Quiz I ◁ Online Quiz I will be on Monday, 14 October ◁ Based on the material of first 2 lectures and homeworks (3rd week’s material is not included) ◁ 7.5% of the total grade ◁ 1 hour since you start the quiz; you can start at any time on 14 October (to be on the safe side, start before 23:00) ◁ Work alone; there are several variants of each question and each student gets a random variant of each question ◁ 10 questions (3 of them harder than the rest) ◁ Each question is 10 marks ◁ 3 types of questions: either write down an answer (number), or choose one of the options, or choose all answers that are correct 2 Outline ◁ Exponential and logarithmic functions ◁ Trigonometric functions ◁ Sequences and their limits ◁ Arithmetic properties of limits 3 Exponential functions Definition (informal). For a given a, b ∈ R, b > 0 define function f : R → R as follows: Q, then f (x) = b 1. If ax ∈ ax 2. Otherwise if ax ̸∈ Q, f (x) is the value that makes the graph of the function f on the (x, y ) plane a continuous (!?) line 4 Exponential functions Definition (informal). For a given a, b ∈ R, b > 0 define function f : R → R as follows: Q, then f (x) = b 1. If ax ∈ ax 2. Otherwise if ax ̸∈ Q, f (x) is the value that makes the graph of the function f on the (x, y ) plane a continuous (!?) line Special case. When b = 2.71828... = e (Euler’s number), we write f (x) = e ax = exp(ax) 4 Exponential functions 6 5 4x 4 exp(x) 2x 3 20.5 x 2 1x 2-x 1 -3 -2 -1 0 1 2 3 5 Logarithmic functions Definition. Given number b ∈ (0, 1) ∪ (1, ∞) (that is, b > 0, b ̸= 1), define R function logb : (0, ∞) → as follows: logb (x) = y such that b y = x 6 Logarithmic functions Definition. Given number b ∈ (0, 1) ∪ (1, ∞) (that is, b > 0, b ̸= 1), define R function logb : (0, ∞) → as follows: logb (x) = y such that b y = x Remarks ◁ We assume that such number exists. We might actually prove it (later) ◁ We assume that such number is unique. We might actually prove it ◁ The domain consists of strictly positive real numbers, since b y > 0 for b>0 Special case. When b = 2.71828... = e (Euler’s number), we write f (x) = log(x) = ln(x) So, the notation is ln(x) = log(x) = loge (x) 6 Logarithmic functions 6 4 log10 (x) 2 log(x) log2 (x) 0.5 1.0 1.5 2.0 2.5 log1.1 (x) -2 log0.9 (x) -4 log0.5 (x) -6 7 Algebraic properties of the exponential and logarithm functions b logb (x) = x logb (b x ) = x (b x )y = b xy logb (x y ) = y · logb (x) bx · by = b x+y logb (xy ) = logb (x) + logb (y ) b logb (c)·x = cx logb (x) = logb (c) · logc (x) 8 Monotonicity of the common functions ◁ f : (0, ∞) → R given by f (x) = x a is: ✓ Strictly increasing when a > 0; ✓ Constant when a = 0 (in this case, f (x) = 1); 1 ✓ Strictly decreasing when a < 0, since f (x) = x −a. 9 Monotonicity of the common functions ◁ f : (0, ∞) → R given by f (x) = x a is: ✓ Strictly increasing when a > 0; ✓ Constant when a = 0 (in this case, f (x) = 1); 1 ✓ Strictly decreasing when a < 0, since f (x) = x −a. ◁ f : R → R given by f (x) = b ax with b > 0 is: ✓ Strictly increasing when b a > 1; ✓ Constant when b a = 1 (in this case, f (x) = 1); ✓ Strictly decreasing when b a < 1. It is convenient to denote c = b a , then f (x) = (b a )x = c x. 9 Monotonicity of the common functions ◁ f : (0, ∞) → R given by f (x) = x a is: ✓ Strictly increasing when a > 0; ✓ Constant when a = 0 (in this case, f (x) = 1); 1 ✓ Strictly decreasing when a < 0, since f (x) = x −a. ◁ f : R → R given by f (x) = b ax with b > 0 is: ✓ Strictly increasing when b a > 1; ✓ Constant when b a = 1 (in this case, f (x) = 1); ✓ Strictly decreasing when b a < 1. It is convenient to denote c = b a , then f (x) = (b a )x = c x. ◁ f : (0, ∞) → R given by f (x) = log (x) with b > 0, b ̸= 1 is: b ✓ Strictly increasing when b > 1; ✓ Strictly decreasing when 0 < b < 1. 9 Outline ◁ Exponential and logarithmic functions ◁ Trigonometric functions ◁ Sequences and their limits ◁ Arithmetic properties of limits 10 Trigonometric functions 11 Trigonometric functions—sinus Definition. Define function sin : R → R as follows: 1. If x ∈ [0, 2𝜋), then sin(x) is the vertical coordinate of the point corresponding to x radians of anticlockwise rotation from the positive horizontal axis in a unit circle centered at the origin 2. If x ̸∈ [0, 2𝜋), then sin(x) = sin(x − 2𝜋n) for some n ∈ Z such that x − 2𝜋n ∈ [0, 2𝜋) where 𝜋 = 3.14159... is the ratio of circle’s circumference to its diameter. The function sin is called sinus 12 Trigonometric functions—cosinus Definition. Define function cos : R → R as follows: 1. If x ∈ [0, 2𝜋), then cos(x) is the horizontal coordinate of the point corresponding to x radians of anticlockwise rotation from the positive horizontal axis in a unit circle centered at the origin 2. If x ̸∈ [0, 2𝜋), then cos(x) = cos(x − 2𝜋n) for some n ∈ Z such that x − 2𝜋n ∈ [0, 2𝜋) where 𝜋 = 3.14159... is the ratio of circle’s circumference to its diameter. The function sin is called cosinus 13 Trigonometric functions—tangent R R {︂ }︂ 5𝜋 3𝜋 𝜋 𝜋 3𝜋 5𝜋 Definition. The function tan : ∖..., − , − , − , , , ,... → 2 2 2 2 2 2 by sin(x) tan(x) = cos(x) where 𝜋 = 3.14159... is the ratio of circle’s circumference to its diameter. The function tan is called tangent 14 Trigonometric functions—tangent R R {︂ }︂ 5𝜋 3𝜋 𝜋 𝜋 3𝜋 5𝜋 Definition. The function tan : ∖..., − , − , − , , , ,... → 2 2 2 2 2 2 by sin(x) tan(x) = cos(x) where 𝜋 = 3.14159... is the ratio of circle’s circumference to its diameter. The function tan is called tangent Remarks 𝜋 Z ◁ Tangent is not defined for x = + 𝜋n for n ∈ , since the denominator in 2 the formula above assumes value zero at these points 14 Trigonometric functions—tangent R R {︂ }︂ 5𝜋 3𝜋 𝜋 𝜋 3𝜋 5𝜋 Definition. The function tan : ∖..., − , − , − , , , ,... → 2 2 2 2 2 2 by sin(x) tan(x) = cos(x) where 𝜋 = 3.14159... is the ratio of circle’s circumference to its diameter. The function tan is called tangent Remarks 𝜋 Z ◁ Tangent is not defined for x = + 𝜋n for n ∈ , since the denominator in 2 the formula above assumes value zero at these points ◁ We have tan(x + 2𝜋n) = tan(x) for n ∈ Z ◁ People also use cotan : R∖ {..., −3𝜋, −2𝜋 − 𝜋, 0, 𝜋, 2𝜋, 3𝜋,...} → R given cos(x) by cotan(x) = sin(x) 14 Trigonometric functions—properties and graphs Basic properties: ◁ sin2 (x) + cos2 (x) = 1 ◁ cos(x) = sin(x + 𝜋/2) ◁ sin(x + 𝜋) = − sin(x), cos(x + 𝜋) = − cos(x) ◁ sin(0) = 0, sin(𝜋/2) = 1, sin(𝜋) = 0, sin(3𝜋/2) = −1 ◁ cos(0) = 1, cos(𝜋/2) = 0, cos(𝜋) = −1, cos(3𝜋/2) = 0 4 1.0 0.8 2 0.6 sin(x) cos(x) -6 -4 -2 2 4 6 0.4 tan(x) -2 0.2 0.0 0.0 0.2 0.4 0.6 0.8 1.0 -4 15 Outline ◁ Exponential and logarithmic functions ◁ Trigonometric functions ◁ Sequences and their limits ◁ Arithmetic properties of limits 16 Sequences Definition. A sequence is an infinite ordered list x = (x1 , x2 , x3 ,...), where N xn = x1 , x2 , x3 ,... are indexed by n ∈ N A more formal definition. A sequence x a function x : → R with the domain equal to the set of natural numbers. We denote by xn = x(n) Examples: ◁ Writing H for heads and T for tails, a sequence of coin flip results might appear as (T , T , T , T , H, T , T , T , H, H,...) ◁ Beginning in 1886, the sequence of world chess champions is (Steinitz, Lasker, Capablanca, Alekhine,... , Carlsen, Liren,...) ◁ A macroeconomic series might include 4 numbers for each period: xt = (output(t), employment(t), debt(t), inflation(t)) and t = 1, 2, 3,... denotes the time period 17 Numerical sequences Definition. A sequence x = (x1 , x2 , x3 ,...) with each xk ∈ R is called a numerical sequence Examples: ◁ The sequence a = (7, 7, 7, 7, 7, 7, 7, 7, 7, 7,...) is constant. ◁ The n-th element of the sequence b = (3, 3.1, 3.14, 3.141,...) represents the first n digits of the decimal expression of the number 𝜋 ◁ The sequence c = (−0.5, 0.11, 0.3596, 0.365881,...) represents f (bn ), for f (x) = x 2 − 9.5 ◁ The sequence d = (0, 0.69, 1.09, 1.38, 1.60, 1.79, 1.94,...) represents the first two digits of the decimal expression of ln(n) Notations. All of the following may denote a sequence (note that the index need not be n; people use various letters such as i, j, k, l, m as indexes): x, x1 , x2 , x3 ,..., (x1 , x2 , x3 ,...), (xn ), (xn )n∈N , (xn )N n=1 18 What happens with an when n grows arbitrary large? ◁ Consider an = 2−n. Thus, the sequence is (︂ )︂ 1 1 1 1 1 1 1 1 1 1 1 (an ) = , , , , , , , , , , ,... 2 4 8 16 32 64 128 256 512 1024 2048 ◁ When n grows larger, an stays positive, but becomes smaller and smaller, and closer and closer to zero ◁ Thus, the number 0 is a good description of what happens “at infinity” ◁ Would our sequence, instead, be (︂ )︂ 1 1 1 1 1 1 1 1 1 1 1 (bn ) = , , , , , , , , , , ,... 2 4 8 16 32 64 128 256 256 256 256 then the number 1/256 would better describe its behavior at infinity ◁ As it turns out, the behavior “at infinity” cannot be fully described by the properties of the first elements of the sequence. Thus, we need a clever definition of the behavior at infinity 19 Limits of sequences Definition. Number x ∈ R is a limit of sequence (an ) if for any real number 𝜖 > 0, there exists a natural number N such that for all n > N, |an − x| < 𝜖. In this case, we say that sequence (an ) has a finite limit, or that it converges 20 Limits of sequences Definition. Number x ∈ R is a limit of sequence (an ) if for any real number 𝜖 > 0, there exists a natural number N such that for all n > N, |an − x| < 𝜖. In this case, we say that sequence (an ) has a finite limit, or that it converges Thus, x is a limit of an if ∀𝜖 > 0, ∃N ∈ N: ∀n > N, |an − x| < 𝜖 20 Limits of sequences Definition. Number x ∈ R is a limit of sequence (an ) if for any real number 𝜖 > 0, there exists a natural number N such that for all n > N, |an − x| < 𝜖. In this case, we say that sequence (an ) has a finite limit, or that it converges Thus, x is a limit of an if ∀𝜖 > 0, ∃N ∈ N: ∀n > N, |an − x| < 𝜖 Example 1. Consider an = 2−n. Thus, the sequence is (︂ )︂ 1 1 1 1 1 1 1 1 1 1 1 (an ) = , , , , , , , , , , ,... 2 4 8 16 32 64 128 256 512 1024 2048 20 Limits of sequences Definition. Number x ∈ R is a limit of sequence (an ) if for any real number 𝜖 > 0, there exists a natural number N such that for all n > N, |an − x| < 𝜖. In this case, we say that sequence (an ) has a finite limit, or that it converges Thus, x is a limit of an if ∀𝜖 > 0, ∃N ∈ N: ∀n > N, |an − x| < 𝜖 Example 1. Consider an = 2−n. Thus, the sequence is (︂ )︂ 1 1 1 1 1 1 1 1 1 1 1 (an ) = , , , , , , , , , , ,... 2 4 8 16 32 64 128 256 512 1024 2048 Let us see if x = 0 is its limit. Note that |an − x| = |2−n − 0| = 2−n 20 Limits of sequences: illustration Example 1. Consider an = 2−n. Thus, the sequence is (︂ )︂ 1 1 1 1 1 1 1 1 1 1 1 (an ) = , , , , , , , , , , ,... 2 4 8 16 32 64 128 256 512 1024 2048 Let us see if x = 0 is its limit. Note that |an − x| = |2−n − 0| = 2−n ◁ Let 𝜖 = 1, then |an − x| < 𝜖 for all n. We might take N = 1 21 Limits of sequences: illustration Example 1. Consider an = 2−n. Thus, the sequence is ⎛ ⎞ ⎜1 1 1 1 1 1 1 1 1 1 1 ⎟ ⎝ 2 , 4 , 8 , 16 , 32 , 64 , 128 , 256 , 512 , 1024 , 2048 ,...⎠ (an ) = ⎜ ⎟ ⏟ ⏞ n>3 for 𝜖=0.1 Let us see if x = 0 is its limit. Note that |an − x| = |2−n − 0| = 2−n ◁ Let 𝜖 = 1, then |an − x| < 𝜖 for all n. We might take N = 1 ◁ Let 𝜖 = 0.1, then |an − x| < 𝜖 starting from a4 = 1/16. Take N = 3 21 Limits of sequences: illustration Example 1. Consider an = 2−n. Thus, the sequence is ⎛ ⎞ ⎜1 1 1 1 1 1 1 1 1 1 1 ⎟ ⎝ 2 , 4 , 8 , 16 , 32 , 64 , 128 , 256 , 512 , 1024 , 2048 ,...⎠ (an ) = ⎜ ⎟ ⏟ ⏞ n>6 for 𝜖=0.01 Let us see if x = 0 is its limit. Note that |an − x| = |2−n − 0| = 2−n ◁ Let 𝜖 = 1, then |an − x| < 𝜖 for all n. We might take N = 1 ◁ Let 𝜖 = 0.1, then |an − x| < 𝜖 starting from a4 = 1/16. Take N = 3 ◁ Let 𝜖 = 0.01, then |an − x| < 𝜖 starting from a7 = 1/128. Take N = 6 21 Limits of sequences: illustration Example 1. Consider an = 2−n. Thus, the sequence is ⎛ ⎞ ⎜1 1 1 1 1 1 1 1 1 1 1 ⎟ ⎝ 2 , 4 , 8 , 16 , 32 , 64 , 128 , 256 , 512 , 1024 , 2048 ,...⎠ (an ) = ⎜ ⎟ ⏟ ⏞ n>8 for 𝜖=0.003 Let us see if x = 0 is its limit. Note that |an − x| = |2−n − 0| = 2−n ◁ Let 𝜖 = 1, then |an − x| < 𝜖 for all n. We might take N = 1 ◁ Let 𝜖 = 0.1, then |an − x| < 𝜖 starting from a4 = 1/16. Take N = 3 ◁ Let 𝜖 = 0.01, then |an − x| < 𝜖 starting from a7 = 1/128. Take N = 6 ◁ Let 𝜖 = 0.003, then |an − x| < 𝜖 starting from a9 = 1/512. Take N = 8 21 Limits of sequences: illustration Example 1. Consider an = 2−n. Thus, the sequence is (︂ )︂ 1 1 1 1 1 1 1 1 1 1 1 (an ) = , , , , , , , , , , ,... 2 4 8 16 32 64 128 256 512 1024 2048 Let us see if x = 0 is its finite limit. Note that |an − x| = |2−n − 0| = 2−n Remark. Note that N = 100 works for 𝜖 = 1, 0.1, 0.01, 0.003 as well. It does not matter which particular N to choose for which 𝜖. The important thing is that for ANY 𝜖 > 0 there is SOME N such that |an − x| < 𝜖 when n > N In words, people sometimes say that |an − x| < 𝜖 for large enough n, meaning the same statement 22 Limits of sequences—example 1 Example 1. Consider an = 2−n. Thus, the sequence is (︂ )︂ 1 1 1 1 1 1 1 1 1 1 1 (an ) = , , , , , , , , , , ,... 2 4 8 16 32 64 128 256 512 1024 2048 Let us see if x = 0 is its finite limit. Note that |an − x| = |2−n − 0| = 2−n Formal proof (there are many proofs) ◁ Consider arbitrary 𝜖 > 0 ◁ Since 𝜖 > 0, there is some natural K such that 𝜖 > 0.1K = 0.0...0 ⏟ 1⏞ , K −th digit otherwise 𝜖 = 0.000.... = 0 ◁ Take N = 4K. Since 2−4 = 1/16 < 0.1, then aN = a4K = 2−4K = (2−4 )K < 0.1K ≤ 𝜖 ◁ Since ak+1 = ak /2, and ak > 0 for any k, then an < aN < 𝜖 for any n > N, which we wanted to prove 23 Limits of sequences—example 2 Example 2. Consider an = 1 + 1/n. Thus, the sequence is 3 4 5 6 7 2, , , , , ,... 2 3 4 5 6 Let us show that x = 1 is a limit of the sequence (an ) ◁ Consider arbitrary 𝜖 > 0. (Note that we need to prove that the statement holds for ANY 𝜖 > 0) ◁ Let N = ⌈1/𝜖⌉ be the first natural number not less than 1/𝜖. (Note that it is enough to find ONE N that works for a given 𝜖) ◁ Consider arbitrary n > N. (Here, we again need to show that the inequality holds for ANY n > N, given N that we have picked) Then |an − 1| = |1/n| = 1/n < 1/N ≤ 𝜖 ◁ Therefore, |an − 1| < 𝜖 for all n > N, which we wanted to prove 24 Limits of sequences—when there is no limit Example 3. Consider the sequence {︃ 1 if n is odd 1, 0, 1, 0, 1, 0, 1, 0, 1, 0,... thus, an = 0 if n is even Let us show that sequence (an ) does not have a (finite) limit ◁ Towards a contradiction, assume that x is a limit of (an ) (We should consider ARBITRARY x). ◁ Consider 𝜖 = 1/4. (Enough to find SOME 𝜖 > 0 to get a contradiction) ◁ Then there should exists N ∈ N such that for all n > N, |an − x| < 𝜖 (We should consider ARBITRARY N) ◁ Note that one of the numbers aN+1 or aN+2 is zero, and another is one, hence it should be that |1 − x| < 𝜖 = 1/4 and |0 − x| < 𝜖 = 1/4 (here, we used the fact that N + 1, N + 2 > N)., then: 1 = |1 − 0| ≤ |1 − x| + |x − 0| < 1/4 + 1/4 = 1/2 ⏟ ⏞ by triangle inequality (HW-1) ◁ We get 1 < 1/2, contradiction 25 There cannot be many limits of a sequence Theorem L3.1. A sequence of real numbers has at most one finite limit. Proof (by contradiction). 1) Towards a contradiction, assume x and y are two distinct limits of (an ) |x − y | 2) Consider 𝜖 = >0 2 3) Since x is a limit of (an ), there is N1 : for all n > N1 , |an − x| < 𝜖 4) Since y is a limit of (an ), there is N2 : for all n > N2 , |an − y | < 𝜖 5) Take n = max{N1 , N2 } + 1, then |an − x| < 𝜖 and |an − y | < 𝜖, then 2𝜖 = |x − y | ≤ |x − an | + |an − y | < 𝜖 + 𝜖 = 2𝜖 ⏟ ⏞ by triangle inequality 6) Hence, 2𝜖 < 2𝜖, in contradiction 26 Notations ◁ Since there can be at most one finite limit of a sequence, when it exists, it is convenient to have a notation for it. ◁ Let sequence (an ) have a finite limit x. We denote it by x = lim an or by an −→ x n→∞ ◁ The notion of limit can be generalized to infinite limits (next slide). ◁ A generalization of Theorem L3.1. says that a sequence of real numbers has at most one finite or infinite limit. 27 Infinite limits Definition. We say that a sequence (an ) tends to infinity if for any number M there is N ∈ N such that for all n > N, an > M. In this case, we write lim an = ∞ an −→ ∞ n→∞ Definition. We say that a sequence (an ) tends to minus infinity if for any number M there is N ∈ N such that for all n > N, an < M. In this case, we write lim an = −∞ an −→ −∞ n→∞ ◁ an = n, then lim an = ∞ n→∞ 2 ◁ an = −n , then lim an = −∞ n→∞ ◁ (an ) = (1, 0, 2, 0, 3, 0, 4, 0,...) does not have either finite or infinite limit ◁ an = ln(n), then lim an = ∞ n→∞ 28 Outline ◁ Exponential and logarithmic functions ◁ Trigonometric functions ◁ Sequences and their limits ◁ Arithmetic properties of limits 29 Calculating limits ◁ Proving that x is a limit of (an ) can be lengthy ◁ It is useful to calculate limits without addressing the definition each time ◁ For instance, what is a limit (and does it exist?) of the following sequence? 3n2 + 4n + 2 an = 5n2 − 1 ◁ We can use the arithmetic properties of the limits to calculate the limits of such sequences 30 Limit of a sum of sequences Theorem L3.2. If sequences (an ) and (bn ) have finite limits, then the sequence (an + bn ) also has a finite limit; moreover, lim (an + bn ) = lim an + lim bn n→∞ n→∞ n→∞ 31 Limit of a sum of sequences Theorem L3.2. If sequences (an ) and (bn ) have finite limits, then the sequence (an + bn ) also has a finite limit; moreover, lim (an + bn ) = lim an + lim bn n→∞ n→∞ n→∞ Proof. Let a = lim an , b = lim bn n→∞ n→∞ 1) Consider arbitrary 𝜖 > 0 2) Since an → a, there is N1 such that for all n > N1 , |an − a| < 𝜖/2 3) Since bn → b, there is N2 such that for all n > N2 , |bn − b| < 𝜖/2 4) Take N = max{N1 , N2 }. Then N ≥ N1 , N2. Hence, for all n > N, |an − a| < 𝜖/2 and |bn − a| < 𝜖/2. Therefore, for all n > N 𝜖 𝜖 |an + bn − (a + b)| = |(an − a) + (bn − b)| ≤ |an − a| + |bn − b|< + = 𝜖 ⏟ ⏞ 2 2 by triangle inequality Thus, we’ve proved that a + b is the limit of an + bn 31 Limit of a product of sequences Theorem L3.3. If sequences (an ) and (bn ) have finite limits, then the sequence (an · bn ) also has a finite limit; moreover, lim (an · bn ) = lim an · lim bn n→∞ n→∞ n→∞ Proof. Omitted 32 limit of a ratio of sequences Theorem L3.4. If sequences (an ) and (bn ) have finite limits, and lim bn ̸= 0, n→∞ an then the sequence ( ) also has a finite limit; moreover, bn (︂ an )︂ lim an n→∞ lim = n→∞ bn lim bn n→∞ Proof. Omitted 33 Calculating limits—examples Example 4. Let k be a natural number. Then 1 lim = 0 n→∞ nk ◁ We have already established it for (k = 1); i.e. lim (1/n) = 0 n→∞ ◁ When k = 2, we have: (︂ )︂ 1 1 1 1 1 lim = lim · = lim · lim = 0·0 = 0 n→∞ n2 n→∞ n n n→∞ n n→∞ n ◁ Applying mathematical induction, we get (︂ )︂ (︂ )︂ (︂ )︂ 1 1 1 1 1 lim k+1 = lim · = lim · lim = 0·0 = 0 n→∞ n n→∞ nk n n→∞ nk n→∞ n c Note also that lim = 0 for any constant c (can you show it?) n→∞ nk 34 Calculating limits—examples 3n2 + 4n + 2 Example 5. Consider an = 5n2 − 1 ◁ It is tempting to use our fraction rule the numerator limit of the numerator limit of = the denominator limit of the denominator ◁ However, both the numerator and the denominator does not have finite limits1 , so the theorem is not applied ◁ A trick helps: let us divide the numerator and the denominator by n2 , then 3 + 4/n + 2/n2 bn an = = 5 − 1/n2 cn where bn = 3 + 4/n + 2/n2 and cn = 5 − 1/n2 1 In fact, both the numerator and the denominator have infinite limits here. 35 Calculating limits—examples ◁ We can calculate the limit of bn = 3 + 4/n + 2/n2 by interchanging summation and multiplication with taking the limits (using Theorems L3.2 and L3.3): (︁ )︁ lim 3+4/n+2/n2 = lim 3+ lim (4/n)+ lim (2/n2 ) = 3+0+0 = 3 n→∞ n→∞ n→∞ n→∞ ◁ Similarly for cn = 5 − 1/n2 (︁ )︁ lim 5 − 1/n2 = lim 5 + lim (−1/n2 ) = 5 + 0 = 5 n→∞ n→∞ n→∞ ◁ Altogether (using also Theorem L3.4), (︁ )︁ 2 3n + 4n + 2 3 + 4/n + 2/n 2 lim 3 + 4/n + 2/n2 3 n→∞ lim = lim = (︁ )︁ = n→∞ 5n2 − 1 n→∞ 5 − 1/n2 lim 5 − 1/n 2 5 n→∞ 36 Infinite sums Definition. Given a sequence a, an infinite (from one side) sum of its elements is defined as follows: ∞ (︃ n )︃ ∑︁ ∑︁ aj = lim aj n→∞ j=1 j=1 the infinite sum in the left hand side exists when the limit on the right hand side exists. n ∑︁ Put it differently, consider a new sequence bn defined as bn = aj , then ∞ j=1 ∑︁ aj = lim bn n→∞ j=1 37 Infinite sums—example ∞ ∑︁ Example 6. Consider q j , where 0 < |q| < 1. Using Homework 2 and the j=0 definition of the infinite sum, we get: ∞ n 1 − q n+1 ∑︁ ∑︁ (︂ )︂ q j = lim q j = lim j=0 n→∞ j=0 n→∞ 1−q when the corresponding limit exists. Let us show that lim q n = 0 n→∞ ◁ Consider arbitrary 𝜖 > 0 ◁ Since 0 < |q| < 1, then 𝜒 = |q|−1 − 1 > 0 ◁ Using the Binomial formula, (︀ −1 )︀n |q| = (1 + 𝜒)n = 1 + n𝜒 +... > n𝜒 1 ◁ Take N > , then for all n > N, 𝜒·𝜖 1 1 |q n − 0| = |q n | = |q|n < <