Mathematical induction. Limits of sequences

LATEST EDITION APRIL 7, 205 AT 5:46 Mathematical induction. Limits of sequences Proofreading of English by Laurence Weinstock Table des matières Mathematical induction 2. Domino effect or chain reaction..................... 2.2 Significance of mathematical induction................ 2.3 Axiom of induction............................ 3.4 Bernoulli s inequality........................... 4.5 Application to sequences......................... 4.6 Situations leading to an erroneous conclusion............. 5 2 Limit of a sequence 6 2. Finite it................................. 6 2.2 Infinite it................................ 6 2.3 Limit by comparison and on an interval................ 7 2.4 Operations on its........................... 8 2.4. Limit of a sum........................... 8 2.4.2 Limit of a product........................ 8 2.4.3 Limit of a quotient........................ 9 2.5 Limit of a geometric sequence...................... 0 2.6 Convergence of a monotonic sequence................. 2.6. Upper-bounded, lower-bounded and bounded sequences. 2.6.2 Convergence theorems...................... 2.7 The method of Heron of Alexandria (st century)........... 3 PAUL MILAN TERMINALE S

TABLE DES MATIÈRES Mathematical induction. Domino effect or chain reaction Mathematical induction is similar to the domino effect. It is analogous to a row of dominoes spaced regularly :? d 0 d d 2 d n d n+ The first domino falls. Primer If the nth domino falls, it knocks over the (n+)th. Propagation The first domino d 0 falls. This primes the reaction. The dominoes are close enough together so that if one domino d n falls the following d n+ also falls. There is therefore propagation along of the row of dominoes. So it can be concluded that all dominoes in the row fall one after the other. Let s transpose this domino effect into a mathematical property. Let (u n ) be the sequence defined by : u 0 = 0, 3 and n N, u n+ = 2 u n+ 2 Consider the property (P) : n N, 0 < u n < The first domino falls : u 0 = 0, 3 then 0 < u 0 <. The property is primed. If a domino falls, the following also falls : If 0 < u n < 0 < 2 u n < 2 2 < 2 u n+ 2 <. We then have 0 < 2 < u n+ <. The property is hereditary. As the first domino has fallen and the others fall by propagation, all the dominoes fall and therefore the property is satisfied for any natural integer..2 Significance of mathematical induction Consider the sequence (u n ) defined by : u 0 = 0 and n N, u n+ = 2u n + We would like to have a formula to explicitly compute u n as a function of n. At first, there is no obvious formula to be seen. In such a situation, calculating the first terms of the sequence is interesting as it often allows you to see a pattern emerge. Let s calculate the first few terms : u = 2u 0 + = (2 ) u 2 = 2u + = 3 (2 2 ) u 3 = 2u 2 + = 7 (2 3 ) u 4 = 2u 3 + = 5 (2 4 ) u 5 = 2u 4 + = 3 (2 5 ) PAUL MILAN 2 TERMINALE S

. MATHEMATICAL INDUCTION The terms of (u n ) seem to follow a simple law : by adding to each term, the successive power of 2 is obtained. From this observation, the following conjecture can be put forward : n N, u n = 2 n A conjecture is not a proof (i.e. not necessarily a true affirmation, some conjectures sometimes prove to be false...). It is only the statement of a property resulting from a number of observations. How can this conjecture be confirmed? Note (P) the property, defined by : n N, u n = 2 n Let s assume for any arbitrary n that the property (P) is true : u n = 2 n Then we have : u n+ = 2u n + = 2(2 n )+ = 2 n+ This is the property (P) for the index n+. If the property is true for an arbitrary index n (induction hypothesis) then the property is also true for the following index n+. This is called the induction step. We have checked that the property is true for the indices 0,, 2, 3, 4, 5. This is called the basis step. But with the induction step, it will also be true for the index n = 6, then for the index n = 7 etc. The property is therefore true for all n..3 Axiom of induction Definition : Consider the property (P n ) defined on N. If the property is satisfied for the first index 0 or n 0 : basis step and if the induction step is true from index 0 or n 0, i.e. : n 0 or n n 0 then P n P n+ then the property is true for all n from index 0 or n 0 Note : Mathematical induction is like the domino effect : If a domino falls, then the next one falls. If the first falls then the second falls, then the third, then the fourth... Conclusion : if the first domino falls, then they all fall. Mathematical induction has two steps : Prove the basis step Prove the induction step If we can carry out these two steps then the property is proven for all natural numbers. It must be ensured that the two conditions "basis step" and "induction step" are proven. As a matter of fact, if one of the two conditions is not met, then an erroneous conclusion is made, as shown by two examples in paragraph.6. PAUL MILAN 3 TERMINALE S

TABLE DES MATIÈRES.4 Bernoulli s inequality Theorem : Let a be a strictly positive real number. We then have n N, (+ a) n +na OPK Organized Presentation of Knowledge Prove this inequality by mathematical induction. Basis step : (+ a) 0 = and +0a =, then (+ a) 0 +0 a. The basis step is established. Induction step : Supposing that (+ a) n +na, let s show that (+ a) n+ +(n+)a By hypothesis : (+ a) n +na as + a > 0 and as a result : (+ a)(+ a) n (+ a)(+na) (+ a) n+ +na+a+na 2 The induction step is established. +(n+)a+na 2 +(n+)a By reason of both the basis and induction steps : n N, (+ a) n +na Note : For the induction step, the inequality is proven by "transitivity" :.5 Application to sequences a > b and b > c so a > c Let (u n ) be a sequence defined by : u 0 = and n N, u n+ = 2+u n a) Prove for all n, 0 < u n < 2 b) Prove that the sequence is strictly increasing. a) Let s prove by mathematical induction that u n is bounded. Basis step : We know that u 0 = so 0 < u 0 < 2. The basis step is established. Induction step : Supposing that 0 < u n < 2, show that 0 < u n+ < 2. 0 < u n < 2 2 < u n + 2 < 4 As the square root function is increasing on R +, 2 < un + 2 < 2 0 < 2 < u n+ < 2 The induction step is established. Therefore, by reason of both the basis and induction steps, we can conclude that : n N, 0 < u n < 2. PAUL MILAN 4 TERMINALE S

. MATHEMATICAL INDUCTION b) Let s show by mathematical induction that the sequence (u n ) is increasing. Basis step : we know that u = 3 so u > u 0. The basis step is established. Induction step : Supposing u n+ > u n, show that u n+2 > u n+. u n+ > u n u n+ + 2 < u n + 2 As the square root function is increasing on R +, un+ + 2 > u n + 2 u n+2 < u n+ The induction step is established. By reason of both the basis and induction steps, the sequence (u n ) is increasing..6 Situations leading to an erroneous conclusion Situation : Only the induction step established. Consider the property : n N, 3 divides 2 n Induction step : Supposing that 3 divides 2 n, show that 3 divides 2 n+. If 3 divides 2 n, then there exists a natural number k such that : 2 n = 3k By multiplying by 2 : 2 n+ = 2 3k = 3(2k). So therefore 3 divides 2 n+ Conclusion : The induction step is established but not the basis step, so it cannot be concluded that the property is true. That is fortunate, because this property is false! Situation 2 : The basis step is established up to a certain index. Consider the following property : n N, n 2 n+4 is a prime number. The basis step is established because for n = 0 we obtain 4 which is a prime number. But there is no mathematical induction even though P(n) is true up to n = 40. This can be seen with a table of prime numbers and a list of the first terms of the sequence (u n ) defined by u n = n 2 n + 4. n u n n u n n u n n u n 0 4 5 22 503 33 097 4 2 73 23 547 34 63 2 43 3 97 24 593 35 23 3 47 4 223 25 64 36 30 4 53 5 25 26 69 37 373 5 6 6 28 27 743 38 447 6 7 7 33 28 797 39 523 7 83 8 347 29 853 40 60 8 97 9 383 30 9 9 3 20 42 3 97 0 3 2 46 32 033 For n = 4, we have : property is false. 4 2 4+4 = 4 2 which is not a prime number. The Conclusion : The truth of a proposition for the first set of values does not make a general rule! PAUL MILAN 5 TERMINALE S

TABLE DES MATIÈRES 2 Limit of a sequence 2. Finite it Definition 2 : The sequence (u n ) is said to have a it l if, and only if, all open intervals containing l contain all the terms of the sequence from a certain index onwards. The it is denoted : u n = l and the sequence(u n ) is said to converge tol Note : If the it exists, it is unique (This is easily proven by contradiction). This definition conveys the idea of the accumulation of terms u n around l ] [.0.5 2.0 Consequently The sequences defined for all natural numbers n = 0 by : l u n = n, v n = n 2, w n = n 3, t n = n, have a it of 0 Algorithm : Determine the integer N from which u n is in an interval containing l. Consider (u n ) defined by : { u0 = 0, u n+ = 2u n ( u n ) This sequence converges tol = 0, 5. We want to find the integer N from which the terms of the sequence are in the open interval centered at 0,5 and with a radius of 0 3. The following program will display N, with a "while" loop. We then obtain : N = 5 and u 5 0, 5 = 3, 96 0 4 Variables: N : integer U : real number Inputs and initialization 0, U 0 N Processing while U 0, 5 0 3 do 2U( U) U N+ N end Output : Print N, U 0, 5 2.2 Infinite it Definition 3 : The sequence(u n ) is said to have a it+ (resp. ) if, and only if, every interval ]A;+ [ (resp. ] ; B[) contains all terms of the sequence from a certain index. The it is denoted : u n = + resp. u n = The sequence is said to diverge to + (resp. ) Note : This definition conveys the idea that the terms of the sequence will always exceed the number A, no matter how large. PAUL MILAN 6 TERMINALE S

2. LIMIT OF A SEQUENCE A sequence can have no it. For example : u n = ( 2) n. It is said that the sequence diverges. Consequently The sequences defined for all natural numbers by : u n = n, v n = n 2, w n = n 3, t n = n, have a it of + Algorithm : Determine the number N from which u n is greater than a given number A (increasing sequence). Consider the sequence (u n ) defined by : u 0 = 2 u n+ = 4 3 u n + We can show that this sequence is increasing and diverges to +. We want find the integer N from which u n is greater than 0 3 The following program allows you to find N, by means of a "while loop". We find that : N = 25 and U = 325, 83 Variables: N : integer U : real number Inputs and initialization 2 U 0 N Processing while U 0 3 do 4 3 U+ U N+ N end Output : Print N, U 2.3 Limit by comparison and on an interval Theorem 2 : Let (u n ), (v n ) and (w n ) be three sequences. If, from a certain index : ) "Squeeze" theorem v n u n w n and if 2) Comparison theorem u n v n and if u n w n and if v n = l and w n = l then v n = + then u n = + w n = then u n = u n = l OPK Proof : Only the proof of the comparison theorem is on the syllabus. We know that : v n = +, so for any real number A, there is an integer N such that if n > N then v n ]A;+ [ As u n > v n from the index p so if n > max(n, p) then u n ]A;+ [ Hence we have : Examples : u n = + Prove that the sequence (u n ) defined by : u n = sin n n+ converges. n N, n+ however sin n n+ n+ n+ = n+ = 0 PAUL MILAN 7 TERMINALE S

TABLE DES MATIÈRES So, according to the squeeze theorem : u n = 0 Show that the sequence (v n ) defined by : v n = n+sin n diverges to + n N however so according to the comparison theorem : 2.4 Operations on its n+sin n n n = + v n = + You are not required to prove the following theorems. It is fairly intuitive that the it of the sum is the sum of the its, and likewise that the it of the product (or quotient) is the product (or quotient) of the its. Only 4 cases have indeterminate forms. When faced with an indeterminate form, the best approach is to try to change the form of the sequence, or use the comparison or squeeze theorems or the theorem on monotonic sequences (see further on) in order to reach a conclusion. 2.4. Limit of a sum If (u n ) has a it l l l + + If (v n ) has a it l + + then (u n + v n ) has a it l+l + + I.F. Note : I.F. = Indeterminate form Examples : Determine the its of the following sequences : n N, u n = 3n++ 2 n 3n+ = + 2 n = 0 The sum is u n = + n N, v n = ( ) n + 5 3 n ( 3 ) n = 0 5 n = 5 The sum is v n = 5 n N, w n = n 2 n+2 n2 = + I.F. Another method n+2 = must be used 2.4.2 Limit of a product If (u n ) has a it l l = 0 0 If (v n ) has a it l then (u n v n ) has a it l l * I.F. * *Follow the usual rules on multiplying unlike or like signs PAUL MILAN 8 TERMINALE S

2. LIMIT OF A SEQUENCE Examples : Determine the its of the following sequences : a) n N, u n = n 2 n+2 = n 2 ( n + 2 n 2 ) n2 = + n + 2 n 2 = The product is u n = + n N, v n = (2 n) 3 n 3n = + 2 n = The product is v n = n N, w n = n+ (n2 + 3) n+ = 0 I.F. Expressed in n2 + 3 = + this form 2.4.3 Limit of a quotient If (u n ) has a it l l = 0 0 l If (v n ) has a it l = 0 0 () 0 l ( ) un l then has a it l * I.F. 0 * I.F. v n *Follow the usual rules on dividing like or unlike signs () 0 without changing sign Examples : Determine the its of the following sequences : a) n N, u n = 5 2n 2 + 5 = 5 2n2 + 3 = + The quotient is u n = 0 b) n N, v n = n n = 0, 5 n 0, 5n = 0 + The quotient is v n = 3 c) n N, w n = n2 + 3 n+ n+ = n + n Factoring out n n+ 3 n = + + n = The quotient is w n = + PAUL MILAN 9 TERMINALE S

TABLE DES MATIÈRES 2.5 Limit of a geometric sequence Theorem 3 : Let q be a real number. Consider the following sequences : If q > then qn = + If q = then qn = If < q < then qn = 0 If q then qn does not exist OPK Proof : Only the proof of the first it is on the syllabus. Bernoulli s inequality is proven by mathematical induction. Therefore for all a > 0 n N, (+ a) n +na Let q = + a so if a > 0 then q >. The inequality becomes : q n +na Seeing as a > 0 we have : +na = + According to the comparison theorem : qn = + Note : To prove the third it, let Q =, with 0 < q < therefore Q >. q By taking the it on each side of the equality we can conclude with the quotient of the its. { u0 = 2 Example : Consider the sequence (u n ) defined by : u n+ = 2u n + 5 Let us define the sequence (v n ) such that v n = u n + 5 ) Show that the sequence (v n ) is geometric 2) Express v n then u n in terms of n 3) Deduce the it of (u n ) ) We have to show that x N v n+ = qv n v n+ = u n+ + 5 = (2u n + 5)+5 = 2(u n + 5) = 2v n Therefore (v n ) is a geometric sequence with a common ratio of q = 2 and a st term of v 0 = u 0 + 5 = 7 2) It can therefore be deduced that : v n = v 0 q n = 7 2 n so u n = v n 5 = 7 2 n 5 3) According to the aforementioned theorem, 2 >, so 2n = + Using the sum and product of its, we can conclude that : u n = + PAUL MILAN 0 TERMINALE S

2. LIMIT OF A SEQUENCE 2.6 Convergence of a monotonic sequence 2.6. Upper-bounded, lower-bounded and bounded sequences Definition 4 : The sequence (u n ) is said to be bounded above if, and only if, there is a real number M such that : n N u n M The sequence (u n ) is said to be bounded below if, and only if, there is a real number m such that : n N u n m The sequence (u n ) is said to be is bounded if it has an upper bound and a lower bound. Example : Show that the sequence (u n ) defined on N by : u n = n+ + n+2 + + 2n is bounded on the interval n terms n+ + n+2 + + {}}{ 2n n + n + + n n+ + n+2 + + 2n n n n+ + n+2 + + 2n n+ + n+2 + + 2n n+ + n+2 + + 2n n+ + n+2 + + 2n 2 n terms {}}{ 2n + 2n + + 2n n 2n [ ] 2 ; So therefore : 2 u n 2.6.2 Convergence theorems Theorem 4 : Divergence If a sequence (u n ) is increasing and not bounded above then the sequence (u n ) diverges to +. If a sequence (u n ) is decreasing and not bounded below then the sequence (u n ) diverges to. PAUL MILAN TERMINALE S

TABLE DES MATIÈRES OPK Proof : Let (u n ) be an increasing sequence with no upper bound. (u n ) is not bounded above, so for all intervals ]A;+ [, N N such that : u N ]A;+ [ Seeing as (u n ) is increasing : n > N so u n > u N Hence : n > N and therefore u n ]A;+ [ So from a certain index, all the terms of the sequence are in the interval ]A;+ [. The sequence (u n ) diverges to +. Example : : Consider (u n ) defined by : u 0 = and u n+ = u n + 2n+3. It can easily be shown that the sequence (u n ) is increasing and by induction that u n n 2 (see exercise). So this sequence diverges to + The converse of this theorem is false, if a sequence diverges to +,it is not necessarily increasing. To prove this, let us consider two sequences that diverge to + and that are not monotonic : { u n = n+( ) n vn = n if n is even and v n = 2n if n is odd Theorem 5 : Convergence If a sequence (u n ) is increasing and bounded above then the sequence (u n ) converges. If a sequence (u n ) is decreasing and bounded below then the sequence (u n ) converges. Note : The proof of this theorem is not on the syllabus. This theorem allows us to show that a sequence converges to a it but does not give the value of this it. We can only say that if (u n ) is increasing and bounded above by M then l M and if (u n ) is decreasing and bounded below by m then l m Example : Consider (u n ) defined by : { u0 = 0 u n+ = 3u n + 4 ) Show that the sequence (u n ) is increasing and bounded above by 4. 2) Deduce that the sequence(u n ) converges. By using an algorithm we can conjecture that (u n ) converges to 4 and determine the integer N from which u n > 3, 99. PAUL MILAN 2 TERMINALE S

2. LIMIT OF A SEQUENCE ) Show by mathematical induction that the sequence (u n ) is increasing and bounded above by 4, i.e. : n N, 0 u n u n+ 4 Basis step : u 0 = 0 and u = 4 = 2, we therefore find that : 0 u 0 u 4 The basis step is established. Induction step : : Supposing that 0 u n u n+ 4, show that 0 u n+ u n+2 4. 0 u n u n+ 4 0 3u n 3u n+ 2 4 3u n + 4 3u n+ + 4 6 As the square root function is increasing on R + The induction step is established. 2 3u n + 4 3u n+ + 4 4 0 2 u n+ u n+2 4 By reason of the basis and induction steps, the sequence(u n ) is increasing and bounded above by 4. 2) (u n ) is increasing and bounded above by 4, according to the theorem of monotonic sequences, (u n ) is convergent. The algorithm shown opposite relates to the exercise in paragraph 2.. We want to find the index N from which u n > 3, 99, then u n 4 < 0 2 We find that : N = 7 and u 7 4 0, 007 thus u 7 3, 993 Variables: N : integer U : real number Inputs and initialization 0 U 0 N Processing while U 4 0 2 do 3U+ 4 U N+ N end Output : Print N, U 4 2.7 The method of Heron of Alexandria (st century) This is a method of calculating an approximation of a square root. It depends on having a first approximate value of A written a. Si a < A a > A A a > A A A a > A So : a < A < A a Similarly, we can show that if a > A then A a < A < a PAUL MILAN 3 TERMINALE S

TABLE DES MATIÈRES Algorithm : Knowing the value of a, we can determine an bounded interval containing A. The interval is then reduced by taking the average m of the values a and A. We then have : a m = 2 ( a+ A ) a The value of a is then replaced by the value of m and the process is repeated. u 0 = a Let us construct a sequence (u n ) defined on N by : u n+ = ) (u n + Aun 2 The sequence converges very quickly : after each iteration, the number of correct decimals is doubled! However, in the first century, nobody had ever heard of the decimal system or even the number zero. Calculating these fractions was certainly complicated! The process can be simplified with the following program : The first term is determined by searching for the nearest square A with a "while loop", that value is assigned to U, it takes N iterations to find u N. The approximate value of 43 with 2 iterations is : We then find : U = A = 43, N = 2 380 6 66 480 20,760 544 The absolute error is : U 43 5 0 6 Variables: A, N, I integers U : real number Inputs and initialization Input A, N 0 I while A > I 2 do I + I end I U Processing for I from to N do ( U+ A ) U 2 U end Output : Print U, U A PAUL MILAN 4 TERMINALE S