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Limits of Sequences of Real Numbers 2013 Sequences of Real Numbers Limits through Definitions The Squeeze Theorem Using the Squeeze Theorem Monotonous Sequences Index FAQ VIDEO and INTERNET SUPPORT FOR THIS LECTURE Explains the main points in THIS slide show: http://www.youtube.com/watch?v=yBE1WApSpV4 Examples: http://www.youtube.com/watch?v=hc64LUtPjP0 Theory through examples: http://archives.math.utk.edu/visual.calculus/6/sequences.3/index.html Index FAQ Sequences of Numbers Definition A sequence x1,x2 ,x3, is a rule that assigns, to each natural number n, the number xn. Examples Index 1 1 1 1 1, 2 , 4 , 8 , 2 1,1.4,1.41,1.414,1.4142, 3 1, 3,5, 7,9, FAQ Limits of Sequences Definition A finite number L is the limit of the sequence x1,x2 ,x3, if the numbers xn get arbitrarily close to the number L as the index n grows. If a sequence has a finite limit, then we say that the sequence is convergent or that it converges. Otherwise it diverges and is divergent. Examples Index 1 1 1 1 The sequence 1, , , , 2 4 8 0 and its limit is 0. converges FAQ 1 1 1 The sequence 1, , , , 2 4 8 0 and its limit is 0. Index converges FAQ Limits of Sequences 2 3 The sequence 1,1.4,1.41,1.414,1.4142, converges and its limit is 2. The sequence (1,-2,3,-4,…) diverges. Notation lim xn L n Index FAQ Computing Limits of Sequences (1) The limit of a sequence xn can be often computed by inserting n in the formula defining the general term xn . If this expression can be evaluated and the result is finite, then this finite value is the limit of the sequence. This usually requires a rewriting of the expression xn . Index FAQ Computing Limits of Sequences (1) Examples 1 2 1 1 1 1 The limit of the sequence 1, , , , n 1 is 0 because 2 4 8 2 1 inserting n to the formula xn n 1 one gets 0. 2 n2 1 The limit of the sequence 2 n 1 1 2 n2 1 n is 1 because rewriting 2 n 1 1 1 n2 1 and inserting n one gets 1. 1 n2 Index 0 FAQ Computing Limits of Sequences Examples continued 3 The limit of the sequence n 1 n n 1 n is 0 because of the rewriting n 1 n n 1 n n 1 n n 1 n 1 . n 1 n n 1 n Insert n to get the limit 0. Index FAQ Formal Definition of Limits of Sequences Definition A finite number L is the limit of the sequence x1,x2 ,x3, if 0 : n such that n n L xn . Example 1 0 since if 0 is given, then n n lim 1 1 1 0 if n n . n n Index FAQ Visualizing the formal definition of a sequence http://archives.math.utk.edu/visual.calcul us/6/sequences.3/index.html Index FAQ Immediate consequence of the formal definition of a sequence Theorem Proof Every convergent sequence is bounded. Suppose that lim xn=L . Take ϵ = 1 (any number works). Find N 1 so that whenever n > N1 we have xn within 1 of L. Then apart from the finite set { a1, a2, ... , aN} all the terms of the sequence are bounded by L+ 1 and L - 1. So an upper bound for the sequence is max {x1 , x2 , ... , xN , L+ 1 }. Similarly one can find a lower bound. Index FAQ The Limit of a Sequence is UNIQUE Theorem Proof The limit of a sequence is UNIQUE Indirectly, suppose, that a sequence would have 2 limits, L1 and L2. Than for a given ∃N 1 ∈N:∀n∈N:n>N 1 :|L1 −xn|<ϵ ∃N 2 ∈N:∀n∈N:n>N 2 :| L2 −xn|<ϵ if N=max{N 1 ,N 2 }, xn would be arbitrary close to L1 and arbitrary close to L2 at the same, it is impossible-this is the contradiction (Unless L1 =L2) Index FAQ Calculating limit using unique prop. Index FAQ Limit of Sums Theorem Assume that the limits lim xn x and lim y n y n are finite. Proof n Then lim xn y n x y . n Let 0 be given. We have to find a number n with the property n n xn y n x y . To that end observe that also Index 2 0. FAQ Limit of Sums Proof Hence there are numbers n1 and n2 such that n n1 xn x and n n2 y n y . 2 2 Let now n =max n1, n2 . We have n n xn y n x y xn x y n y 2 2 . By the Triangle Inequality Index FAQ Limits of Products The same argument as for sums can be used to prove the following result. Assume that the limits lim xn x and lim y n y Theorem n are finite. Then n lim xn y n x y . n Remark Observe that the limits lim xn y n and lim xn y n may exist n n and be finite even if the limits lim xn and lim y n do not exist. n n 1 . Then lim y n 0 and 2 n n Examples the limit lim xn does not exist. However, lim xn y n 0. Let xn 1 n and y n n n Index n FAQ Squeeze Theorem for Sequences Theorem Assume that n : xn y n zn and that lim xn lim zn a. n n Then the limit lim y n exists and n lim y n lim xn lim zn . n Proof n n Let 0. Since lim xn lim zn a, nx nz such n n that n nx xn a and n nz zn a . Let ny max nx , nz . Then n ny a y n max a xn , a zn . This follows since xn y n znn. Index FAQ Using the Squeeze/Pinching Theorem Example Solution n! . n nn Compute lim This is difficult to compute using the standard methods because n! is defined only if n is a natural number. So the values of the sequence in question are not given by an elementary function to which we could apply tricks like L’Hospital’s Rule. n! Here each term k/n < 1. Observe that 0< n for all n 0. n Next observe that n ! 1 2 3 n 1 n 1 2 3 n n nnn nn n n n n 1 n 1 . n n n Hence 0 n! 1 . n n n 1 n! 0, also lim n 0 by the Squeeze Theorem. n n n n Since lim Index FAQ Using the Squeeze Theorem sin(n ) Does the sequence converge? n cos(n ) If it does, find its limit. Problem Solution 1 sin(n ) 1 We have Hence and 1 cos(n ) 1 for all n 2,3,4, . 1 sin(n ) 1 . n 1 n cos(n ) n 1 1 1 Since lim lim 0 we conclude that the sequence n n -1 n n -1 sin(n ) sin(n ) 0. converges and that nlim n cos(n ) n cos(n ) Index FAQ Monotonous Sequences A sequence (a1,a2,a3,…) is increasing if an ≤ an+1 for all n. The sequence (a1,a2,a3,…) is decreasing if an+1 ≤ an for all n. Definition The sequence (a1,a2,a3,…) is monotonous if it is either increasing or decreasing. The sequence (a1,a2,a3,…) is bounded if there are numbers M and m such that m ≤ an ≤ M for all n. Theorem A bounded monotonous sequence always has a finite limit. Observe that it suffices to show that the theorem for increasing sequences (an) since if (an) is decreasing, then consider the increasing sequence (-an). Index FAQ Monotonous Sequences A bounded monotonous sequence always has a finite limit. Theorem Let (a1,a2,a3,…) be an increasing bounded sequence. Proof Then the set {a1,a2,a3,…} is bounded from the above. By the fact that the set of real numbers is complete, s=sup {a1,a2,a3,…} is finite. Claim lim an s. n Index FAQ Monotonous Sequences A bounded monotonous sequence always has a finite limit. Theorem Let (a1,a2,a3,…) be an increasing bounded sequence. Proof Let s=sup {a1,a2,a3,…}. Claim lim an s. n Proof of the Claim Let 0. We have to find a number n with the property that n n an s . Since s sup an , there is an element an such that s an s. Since an is increasing n n s an an s. Hence n n an s . This means that lim an s. n Index FAQ SUMMARY 1. Notion of a sequence 2. Notion of a limit of a sequence 3. The limit of a convergent sequence is unique. 4. Every convergent sequence is bounded. 5. Any bounded increasing (or decreasing) sequence is convergent. Note that if the sequence is increasing (resp. decreasing), then the limit is the least-upper bound (resp. greatest-lower bound) of the numbers Index FAQ SUMMARY 6. If two sequences are convergent and we compose their +, -, *. /, 1/.. then the limit of this composed sequence exists and is the +, -, *. /, 1/..of the original limiting values. 7. Squeeze/Pinching theorem Index FAQ