Sequences CheatSheet

Transcription

1 Sequenes CheatSheet Intro This heatsheet ontains everything you should know about real sequenes. It s not meant to be exhaustive, but it ontains more material than the textbook. Definitions and properties In this first setion we define the notion of sequene of real numbers and the notion of subsequene. Definition.. A sequene of real numbers a n is a funtion whose domain is the natural numbers and range is ontained in the real numbers: N a R n a n Definition.. A sequene a n is alled inreasing if a i < a i+ for all i N. It s alled dereasing if a i > a i+ for all i N. It s alled monotoni if it s either inreasing or dereasing. Definition.3. Let p be an inreasing sequene of natural numbers (it s range is ontained in N). Let a n be a sequene, we all a subsequene of a n the omposition a pn : N p N a R k p k a pk Example.4. For example let p k = k and a n is any sequene; the omposition a pk = a k is the subsequene of all the even members of a n. If p k = k+, the omposition a k+ is the subsequene of all the odd members. Definition.5. A sequene a n is bounded from above if there is a real number L suh that a n L for all n N. It is bounded from below if there is a real L suh that a n L for all n N. It s bounded if it is bounded from above and below. 3 Convergene A real sequene a n is onvergent to a real number L if you an make a n arbitrarily lose to L by taking n large enough. Definition 3.. We say that a n onverges to L and we denote it with a n = L if ɛ R, ɛ > 0 there exists n(ɛ) N suh that n > n(ɛ), a n L < ɛ.

2 is. In this definition ɛ measures how lose we are to L and n(ɛ) tells us how large large enough Example 3.. Let a n be the sequene n p where p is a positive real number. Pik any number ɛ > 0. The ondition: n p < ɛ is fulfilled if: n > ɛ p The number ɛ p is positive but not natural, so we put n(ɛ) = ɛ p This hoie of n(ɛ) proves that a n onverges to zero. The writing x denotes the eiling funtion whih maps x to the smallest integer whih is greater than or equal to x. A real sequene a n is divergent if you an make a n arbitrarily large (positive or negative) by taking n large enough. Definition 3.3. We say that a n is divergent and denote it with: a n = + n + if C R there exists n(c) N suh that n > n(c), a n > C. Example 3.4. Let a n be the sequene ln (n). Let C be any positive real number. The ondition ln (n) > C is satisfied if we pik n > e C beause the exponential is an inreasing funtion. We prove that a n is divergent by hoosing n(c) = e C. A sequene might be neither onvergent nor divergent; for example a n = ( ) n is ertainly not divergent but it doesn t onverge to any number either. In the next setion we shall see a method to prove that a sequene is not onvergent. 4 Tehniques Sometimes a sequene an be a real funtion in disguise. In that ase the it of the sequene is the it of the funtion, if this it does exist. Proposition 4.. Let f(x) be a funtion of one real variable whose domain ontains the interval [, + ). Let a n be the sequene f(n) where n is any natural number. If: f(x) = L x + where L is a real number or also ±, then the sequene a n has the same it: a n = L If the funtion f(x) doesn t have a it we annot onlude anything about the sequene: it might or might not onverge. Consider for instane a n = sin (πn). The funtion f(x) = sin (πx) doesn t have a it but a n = sin(πn) = 0 for all natural numbers n, so a n onverges to zero. There are sequenes that do not ome from a funtion of one real variable or they ome from some non elementary funtion. For instane p n does not ome from a real funtion if p is negative. The sequene n! omes from a real funtion, the Gamma funtion, but unfortunately this funtion is not an elementary one. In all these ases we might need some speifi tehnique to study the onvergene.

3 Proposition 4.. A sequene a n onverges to L R if and only if every subsequene onverges to L. Example 4.3. The sequene a n = ( ) n is not onvergent. The subsequene a n onverges to and the subsequene a n+ onverges to. If a n were onvergent to some number L all its subsequenes should onverge to the same L, so a n is not onvergent. If two subsequenes onverge to the same it, in general we annot onlude that the sequene is onvergent. However the subsequene of even members and the subsequene of odd members are speial: if they are both onvergent and they onverge to the same it, then the original sequene is onvergent as well: Proposition 4.4. Let a n be a real sequene. If the subsequene a n onverges to a real number L and the subsequene a n+ onverges to the same number L, then a n onverges to L as well. Proof. Choose a positive ɛ R. Sine a n onverges to L, there is p(ɛ) suh that n > p(ɛ), a n L < ɛ. There is also q(ɛ) suh that n > q(ɛ), a n+ L < ɛ. Set n(ɛ) equal to max(p(ɛ), q(ɛ)+ ). For all n > n(ɛ) we are either in the odd or the even ase so a n L < ɛ, whih proves the onvergene of a n to L. This proposition an be used even when the two subsequenes go to infinity. With the following two propositions we an ompare sequenes both when they are onvergent and divergent. Proposition 4.5 (Squeeze theorem). Let a n, b n, n be three sequenes of real numbers. If there is n 0 N suh that n n 0, a n b n n and both a n and n onverge to L R then also b n onverges to L. Proof. Aording to the definition of it for every ɛ > 0 there exists n ɛ suh that for every n n ɛ : and there exists m ɛ suh that if n m ɛ : L ɛ < a n < L + ɛ L ɛ < n < L + ɛ If we now hoose p = max n ɛ, m ɛ, n 0 all the inequalities are true simultaneously for every n p and we an dedue: L ɛ < a n b n n < L + ɛ whih proves that b n onverges to L. Proposition 4.6. Let a n, b n be two sequenes. If there is n 0 N suh that n n 0, a n b n and a n = + then also n b n = +. Not as popular as the root ratio test for series, but even sequenes have a root and ratio test that an be helpful in ertain alulations: Proposition 4.7. Let a n be a sequene of real numbers suh that n a n onverges to L. If L < the sequene onverges to zero, if L > the sequene is divergent, if L = the test is inonlusive. Proposition 4.8. Let a n be a sequene of real numbers suh that a n+ a n onverges to L. If L < the sequene onverges to zero, if L > the sequene is divergent, if L = the test is inonlusive. 3

4 The two tests are perfetly equivalent; if one is inonlusive the other one is also inonlusive. You an hoose the most onvenient one in a speifi alulation. For example: determine for whih real numbers a the following sequene is onvergent: a n = n!an n n When dealing with a fatorial, the ratio test is always the best bet: a n+ a n = (n + )a (n + )( + n )n The it of the sequene is a e. If a > e the sequene diverges, if a < e the sequene onverges to zero. If a = e the test is inonlusive. It an be shown by finer tehniques that the ase a = e is atually divergent. proof of prop If L < we an hoose ɛ > 0 suh that L + ɛ <. Aording to the definition of it there exists n ɛ N suh that if n n ɛ we have: or equivalently: L ɛ < n a n < L + ɛ < 0 a n < (L + ɛ) n Sine L + ɛ < the sequene (L + ɛ) n onverges to zero and aording to the squeeze theorem a n onverges to zero as well. If the absolute value onverges to zero, even a n onverges to zero. If L > we an hoose ɛ > 0 suh that L ɛ >. If n n ɛ we have: (L ɛ) n a n and sine (L ɛ) n diverges to +, the sequenes a n and a n are also divergent. proof of prop If L < we an hoose ɛ > 0 suh that L + ɛ <. Aording to the definition of it there exists n ɛ N suh that if n n ɛ we have: a n+ < L + ɛ < a n If n N ɛ we an apply the inequality repeatedly and obtain: a n (L + ɛ) n nɛ a nɛ 0 a n < (L + ɛ) n Sine L + ɛ < the sequene (L + ɛ) n nɛ onverges to zero and aording to the squeeze theorem a n onverges to zero as well. If the absolute value onverges to zero, even a n onverges to zero. If L > we an hoose ɛ > 0 suh that L ɛ >. If n n ɛ we have: (L ɛ) a n+ Applying repeatedly we get: a n (L ɛ) n nɛ a nɛ a n and sine (L ɛ) n nɛ diverges to +, the sequenes a n and a n are also divergent. 4

5 The following proposition an be helpful when nothing else works. Proposition 4.9. A real sequene a n whih is bounded from above and inreasing is onvergent. A sequene whih is bounded from below and dereasing is onvergent. This statement is a onsequene of the least upper bound property of the real numbers. We introdue here some notation and definitions and we will need to explain this property. Without entering into formal details, we say that a set X is ordered if the symbol is defined on it, and for any two elements x, y X we an always say whether x y or y x. Definition 4.0. A subset A of an ordered set X is bounded from above if there exists x X suh that x a for every a A. We all x an upper bound of A. A subset A of an ordered set X is bounded from below if there exists x X suh that x a for every a A. We all x a lower bound of A. Definition 4.. The supremum of a bounded from above set A, is the smallest upper bound of A (it might not exist!). We denote it with sup(a). The infimum of a bounded from below set A, is the greatest upper bound of A (it might not exist!). We denote it with inf(a). If α is the supremum of A we an express that it is the smallest upper bound in the following way:. α a for every a A. For every positive number ɛ, the number α ɛ is not an upper bound, whih means that there exists a ɛ A suh that a ɛ > α ɛ. Definition 4.. An ordered set X has the least upper bound property if for every subset A X whih is bounded from above, there exists α X whih is the supremum of A. Theorem 4.3. The real numbers have the least upper bound property. The proof of this theorem is not obvious and it s founded on the rigorous definition of the real numbers that we have aurately omitted. Remark 4.4. The rational numbers for instane don t have the least upper bound property. The set A = {p Q p < } Q is learly bounded from above but it doesn t have a smallest upper bound. Suh a number would be whih is not rational. Remark 4.5. If a set has the least upper bound property then it s also true that every bounded from below subset has an infimum. proof of proposition 4.9. Denote with R the range of the sequene a n. Sine the sequene is bounded from above, the set R is also bounded from above. Sine the real numbers have the least upper bound property there is L R whih is the supremum of R. For every ɛ > 0, the number L ɛ is not a supremum so there exists n ɛ N suh that L ɛ < a nɛ. Sine the sequene is inreasing, for every n n ɛ we have: L ɛ < a nɛ a n < L + ɛ whih proves that L is atually the it of the sequene. The proof in the deresing bounded from below ase, is exatly the same. 5

6 5 Indution Mathematial indution is not a tehnique speifially meant to study the onvergene of a sequene, nonetheless it an be very helpful in this kind of senario. We denote with P n a statement involving a natural number n. For a given number n the statement an be true or false. Proposition 5.. Let P = {P n ; n N} be a set of statements involving a natural number n. Suppose that the following onditions are satisfied:. the statement P is true;. P n is true implies that P n+ is true as well; then the statement P n is true for every natural number n. We an replae the first ondition with: P n0 is true; in this ase the proposition hanges to: the statement P n is true for every n n 0. Example 5.. We an prove by indution that: P n : n i = i= n(n + ) is true for all natural numbers.. If n = the statement is trivially true.. Suppose that P n is true. n+ i = i= n i + (n + ) = i= n(n + ) + (n + ) = n + n + 3 = (n + )(n + ) This proves that: is true. P n+ : n+ i = i= (n + )(n + ) Aording to the proposition, the statement P n is true for every natural number. 6 Fundamental its of sequenes The following its should be onsidered fundamental, and you don t have to retrieve them eah time you do a alulation: np = + if p > 0 if p = 0 0 if p < 0 6

7 pn = 0 if < p < if p = + if p > does not exist if p n n = ( + n = e n) n! n n = 0 e n n! = 0 n p e n = 0 ln(n) n p = 0 if p > 0 The number p in all the previous formulas is a real number. The last four its ompare how fast the funtions (n n, n!, e n, n p, ln(n)) go to infinity. The result an be read as: n n is the fastest, followed by the fatorial, followed by the exponential, followed by any power funtion; and the logarithm is the slowest. 7 Examples of reursive sequenes A sequene an be defined reursively; this means that we hoose the first element in the sequene and we define any other element as a funtion of its predeessor: { a = a n+ = f(a n ) where f( ) is a funtion of one real variable. Sometimes it s possible to make a reursive sequene expliit, but it s usually harder to find the expliit sequene than studying the onvergene diretly. Example 7.. Consider the following reursive sequene: { a = The n-th term of the sequene looks like this: a n = a n+ = + a n } {{ } n+ times 7

8 First of all we notie that this sequene is inreasing. We an prove it by indution: a = < a = Assume that a n a n+ and try to prove that a n+ a n+ : This implies that: a n = a n+ a n+ = a n+ a n+ a n+ and this is satisfied if a n+ a n+ a n+. Now we prove by indution that the sequene is bounded above by 3. The first element is less then 3. Suppose that a n 3. This implies: a n+ = + a n 4 < 3 Aording to proposition 4.9 the sequene is onvergent. Sine the sequene is onvergent every subsequene must onverge to the same it, in partiular a n+ has the same it as a n : L = a n+ = + an = + L This is satisfied if L = + L, and sine L must be positive L = + 5 whih is the golden ratio of. Example 7.. Consider the degree equation x bx = 0 () where b, are integers and for simpliity we assume that they are stritly positive. Sine x = 0 is not a solution we an divide everything by x and obtain the following: x = b + x This equation expresses x as a funtion of itself ; we an take this funtion and plug it into the right hand side: x = b + b + x and we an repeat this same proedure indefinitely: x = b + b + b + b + b +... This representation of the solution of the degree two equation is known as ontinued fration and an be written as a reursive sequene: { a = a n+ = We an prove that this sequene is onvergent and it onverges to the solution of the equation minus b (with this kind of representation we are missing the first b in the ontinued fration). It s easy to 8 b+a n

9 prove that the sequene is bounded from above by and from below by b+, but the sequene is quite learly not monotoni (just alulate the first three members). At a loser look the sequene turns out to be osillating: the subsequene of the even members is inreasing, and the subsequene of the odd members is dereasing (this is left as an exerise to the reader). Sine a n and a n+ are bounded and monotoni they are both onvergent. A priori they might onverge to different its! We an expliitly alulate these its as in the previous example: and taking the it of both the two members: L = a n+ = b + b + a n a n+ = b + b + L This implies that the it L satisfies the following equation: L + bl = 0 The positive solution of this equation is preisely the solution of () minus b. With the same method we an prove that the it of the odd members satisfy the same equation. Sine the subsequenes of even and odd members onverge to the same it L, aording to proposition 4.4 we dedue that a n onverges to L. In the speial ase b = =, the solution to () is: x = and this is another representation of the golden ratio. 8 Stolz-Cesàro This last setion is devoted to a tehnique, known as Stolz-Cesàro theorem, whih is somehow the equivalent of l Hôpital for sequenes. Theorem 8. (Stolz-Cesàro). Let a n be a sequene of real numbers and b n an inreasing sequene suh that b n = +. If the it: then the sequene an b n is also onvergent to L: The number L an also be ±. a n+ a n = L b n+ b n a n = L b n 9

10 Proof. By definition of it, for every ɛ > 0 there exists n ɛ N suh that for every n n ɛ we have: L ɛ < a n+ a n b n+ b n < L + ɛ Sine b n is inreasing the number b n+ b n is always positive so we an multiply the inequality: (L ɛ)(b n+ b n ) < a n+ a n < (L + ɛ)(b n+ b n ) If k n ɛ, we an sum the three members of the inequality from n ɛ up to k. Sine the inequalities hold for eah one of the summands, they also hold for the summation: k k k (L ɛ) (b n+ b n ) < (a n+ a n ) < (L + ɛ) ((b n+ b n )) n=n ɛ n=n ɛ n=n ɛ In eah one of these summations, only the first and the last element don t anel out, so we get: (L ɛ)(b k+ b nɛ ) < a k+ a nɛ < (L + ɛ)(b k+ b nɛ ) If k is large enough we an assume that b k+ is positive, sine k + b k = +, so we an divide everything by b k+ : ( (L ɛ) b ) n ɛ + a n ɛ < a ( k+ < (L + ɛ) b ) n ɛ + a n ɛ b k+ b k+ b k+ b k+ b k+ The number n ɛ is fixed, and b k+ goes to + when k is large; for this reason both anɛ b k+ onverge to zero. We an find k ɛ N suh that for all k k ɛ we have: and also: Using these last two inequalities we get: ɛ < a n ɛ b k+ (L + ɛ) b n ɛ b k+ < ɛ ɛ < a n ɛ b k+ (L ɛ) b n ɛ b k+ < ɛ L ɛ < a k+ b k+ < L + ɛ and bnɛ b k+ whih proves that: a k+ = L k + b k+ Example 8.. Calulate the it: k + k n k n k+ Aording to Stolz-Cesàro the it is equal to the following: (n + ) k (n + ) k+ = nk+ k+ i=0 (n + ) k ) = n i nk+ ( k+ i 0 (n + ) k k ( k+ ) = i=0 i n i

11 = ( + n ) k k ( k+ ) = i=0 i n i k ( k+ k ) = k + If k the denominator is not an inreasing sequene so the method annot be applied. We will prove later on with some other tehnique that for k = we have: Example 8.3. Calulate the it: n = + n k= k log n log n Aording to Stolz-Cesàro the it is equal to the following: ) n+ log (n + ) + log n n+ ( log + n = ( ) = log (n + ) log n log + n + n ( ) n log + n ( ) = 0 n log + n We don t know if the numerator is onvergent or not, but this alulation shows that it annot diverge faster than log n. It also shows that n k= k goes to infinity as fast as log n.