Convergence of Sequences

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1 James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University September 5, 2018

2 Outline 1 2 Homework

3 Definition Let (a n ) n k be a sequence of real numbers. Let a be a real number. We say the sequence a n converges to a if ɛ > 0, N n > N = a n a < ɛ We usually just write a n a as n to indicate this convergence. We call a the limit of the sequence (a n ) n k. The sequence (a n ) n k does not converge to a number a if we can find a positive number ɛ so that no matter what N we choose, there is always at least one n > N with a n a > ɛ. We write this using mathematical language as ɛ > 0, N n > N with a n a ɛ If a sequence does not converge, we say it diverges.

4 The sequence ( 1) n n 1 does not converge. The range of this sequence is the set { 1, 1}. We will show there is no number a so that ( 1) n a for this sequence. Case 1: Let a = 1. Note the difference between 1 and +1 is 2 which suggests we pick a tolerance ɛ < 2 to show the sequence does not converge to 1 as we want to isolate which value we are trying to get close to. Let ɛ = 1. Then ( 1) n 1 = { 1 1 = 0, if n is even. 1 1 = 2, if n is odd. Now pick an N. Then there is an odd integer larger than N, say 2N + 1, for which ( 1) 2N+1 1 = 2 > ɛ. Since we can do this for all N, we see the sequence can not converge to 1.

5 Case -1: We can repeat this argument for this case. Let ɛ = 1. Then ( 1) n ( 1) = { = 2, if n is even = 0, if n is odd. Now pick an N. Then there is an even integer larger than N, say 2N, for which ( 1) 2N ( 1) = 2 > ɛ. Since we can do this for all N, we see the sequence can not converge to 1. Case a 1, 1: If a is not 1 or 1, let d 1 be the distance for a to 1 which is a 1 and let d 2 be the distance to 1 which is a ( 1). Let d be the minimum of these two distances, d = min(d 1, d 2 ). Then, we have ( 1) n a = { 1 a = d1, if n is even. 1 a = d 2, if n is odd.

6 If we pick ɛ = (1/2)d, we see if both cases ( 1) n a > d for any value of n. Thus, ( 1) n can not converge to this a either. Since all choices of a are now exhausted, we can say this sequence does not converge. Comment This argument works for any sequence with a finite range. For example, for a n = sin(nπ/4) for n 1, the range is { 1, 2/2, 0, 2/2, 1}. We can argue that the limit can not be any of the values in the range by choosing an ɛ less that 1/2 of the distances between the values in the range. If we let the range values be y 1,..., y 5, the minimum distance to a number a other than these values is is min 1 i 5 a y i. We use then use the last argument above choosing ɛ as before. Thus, no a can be the limit of this sequence. Comment No sequence with a finite range can have a limit unless there is only one value in its range. We do another example next.

7 Now pick an N. Then any n > N with cos(nπ/4) 1 satisfies cos(nπ/4) ( 1) > ɛ =.1. Since we can do this for all N, we see the sequence can not converge to 1. The sequence cos(nπ/4) n 1 does not converge. The block for this sequence is B = { 2/2, 0, 2/2, 1, 2/2, 0, 2/2, 1} and the range of this sequence is the set { 1, 2/2, 0, 2/2, 1}. We show there is no number a so that cos(nπ/4) a for this sequence. Case -1: Let a = 1. Note the minimum distance between range values is 1 2/2.3 which suggests we pick ɛ <.3 to show the sequence does not converge to 1. Let ɛ =.1. Then = 0, if n for range value 1 2/2 + 1 =.303, for range value 2/2 cos(nπ/4) ( 1) = = 1, for range value 0 2/2 + 1 = 1.707, for range value 2/ = 2.0, for range value 1

8 Case 2/2: Let a = 2/2. Let ɛ =.1 for the same reason as before. Then cos(nπ/4) ( 2/2) = 1 + 2/2 =.303, if n for range value 1 2/2 + 2/2 = 0, for range value 2/ /2 =.707, for range value 0 2/2 + 2/2 = 1.414, for range value 2/ /2 = 1.707, for range value 1 Now pick an N. Then any n > N whose sequence value cos(nπ/4) 2/2 satisfies cos(nπ/4) ( 2/2) > ɛ =.1. Since we can do this for all N, we see the sequence can not converge to 2/2. We will leave it to you to mimic these arguments for the remaining cases. Next, we will do the case where a is not equal to any of the range values.

9 Case a is not a range value: If a is not equal to a range value, let d 1 be the distance for a to 1, a ( 1), d 2 the distance for a to 2/2, a ( 2/2), d 3 the distance for a to 0, a 0, d 4 the distance for a to 2/2, a 2/2 and d 5 the distance for a to 1, a 1, and let d 2 be the distance to 1 which is a ( 1). Let d be the minimum of these five distances, d = min{d 1,..., d 5 } and choose ɛ = d/2. Then, we have 1 a = d 1, if n for range value 1 2/2 a = d 2, for range value 2/2 cos(nπ/4) a = 0 a = d 3, for range value 0 2/2 a = d 4, for range value 2/2 1 a = d 5, for range value 1 For ɛ = d/2, in all cases cos(nπ/4) a > d/2 for any value of n. Hence, for any N, any n > N gives cos(nπ/4) a > ɛ. Thus cos(nπ/4) can not converge to an a that is not a range value. Since all choices of a are now exhausted, we can say this sequence does not converge.

10 The sequence (1 + (4/n)) n 1 converges to 1. The range of this sequence is not finite and indeed does not repeat any values. However, we can guess that as n gets large, the values in the sequence are closer and closer to 1. So let s assume a = 1 as our guess for the limit. Pick an arbitrary ɛ > 0 and consider a n a = ( (1 + (4/n) ) 1 = 4/n = 4 1/n. Pick any N so that 4/N < ɛ or N > 4/ɛ. Then for any n > N, n > 4/ɛ or 4/n < ɛ. So we have shown n > N > 4/ɛ ( (1 + (4/n) ) 1 < ɛ Since ɛ was chosen arbitrarily, we know the sequence converges to 1.

11 Here is our template: Proposition: a n a. Step 1: Identitfy what a is. Step 2: Choose ɛ > 0 arbitrarily. Step 3: Now follow this argument: original sequence - proposed limit = simplify using algebra etc to get a new expression use triangle inequality etc to get a new expression again = call this last step the overestimate. We have now = overestimate < ɛ Step 4: Solve for n in terms of ɛ to give a simple equation. Step 5: Choose N to satisfy the inequality you get from Step 4. Step 6: Then for any n > N, overestimate < ɛ and we have (original) (proposed limit) < ɛ proving a n a.

12 ( 1+4n 5+6n ) n 1 converges. We can guess the value of the limit is a = 2/3 so pick ɛ > 0 arbitrarily. Consider 1 + 4n 5 + 6n 2 3(1 + 4n) 2(5 + 6n) = 3 (3) (5 + 6n) = 7 (3) (5 + 6n) The denominator can be underestimated: n 18n which implies 1/(3(5 + 6n) < 1/(18n). Thus, our original calculation gives 1 + 4n 5 + 6n 2 < (7/18) (1/n) 3 Now set this overstimate less than ɛ and solve for n.

13 Now 7/(18n) < ɛ n > 7/(18ɛ) and so choosing N > 7/(18ɛ) implies the following chain: n N n > 7/(18ɛ) = original limit < ɛ This shows 1+4n 5+6n 2 3.

14 ( 1+4n 5 6n ) n 1 converges. We can guess the value of the limit is a = 2/3 so pick ɛ > 0 arbitrarily. Consider 1 + 4n 5 6n 2 3(1 + 4n) + 2(5 6n) = 3 (3) (5 6n) = 13 (3) (5 6n) We want to underestimate 15 18n = 18n 15. Now 18n 15 is positive for n 1 so we don t need the absolute values here. Pick a number smaller than 18. We usually go for half or 9. Set 18n 15 > 9n and figure out when this is true. A little thought tells us this works when n 2. So 18n 15 > 9n when n 2 implying 13/(15 18n) < 13/(9n) when n 2. Thus, 1 + 4n 5 6n + 2 < (13/(9n), n 2 3

15 Next, set this overestimate less than ɛ to get 13/(9n) < ɛ or n > 13/(9ɛ). But we must also make sure N 2 too, so if we set N > max(13/(9ɛ), 2) implies the following chain: n N n > 13/(9ɛ) = original limit < ɛ This shows 1+4n 5 6n 2 3.

16 11+4n+8n 2 6+7n 2 converges. We guess the limit is a = 8/7. Consider n + 8n n 2 8 7(11 + 4n + 8n 2 ) 8(6 + 7n 2 ) = 7 7(6 + 7n 2 ) n = 7(6 + 7n 2 ) Now, for the numerator, n 29n + 28n = 57n and for the denominator, 7(6 + 7n 2 ) > 49n 2 and so n + 8n n 2 8 < 57n 7 49n 2 = 57 49n

17 Set 57/(49n) < ɛ to get n > 57/(49ɛ). Now pick any N > 57/(49ɛ) and we see n > N implies original limit < ɛ. We conclude 8/7 is the limit of the sequence.

18 11+4n 8n 2 6 n 2n 2 converges. We guess the limit is a = 4. Consider n 8n n 8n 2 4(6 n 2n 2 ) 6 n 2n 2 4 = 6 n 2n n = 6 n 2n 2 Now, for the numerator, n n 13n + 8n = 21n by the triangle inequality and for the denominator, 6 n 2n 2 = 2n 2 + n 6. A good choice for underestimating here is n 2 ; i.e. use half of the coefficient of the n 2. For n > 2, the absolute values are not needed and so we see when 2n 2 + n 6 > n 2 implying n 2 + n 6 > 0. This is true for all n 3 so there are restrictions here on n. We have 6 n 2n 2 n 2 if n 3.

19 Thus, if n n 8n 2 6 n 2n 2 4 < 21n n 2 = 21 (1/n) Set 21/n < ɛ to get n > 3/ɛ. Now pick any N > max(21/ɛ, 3) and we see n > N implies original limit < ɛ. We conclude 4 is the limit of the sequence.

20 (3 + sin(2n + 5)/n) n 1 converges We suspect the limit is 3. Hence, pick ɛ > 0 arbitrary and consider sin(2n + 5) sin(2n + 5) = n n 1/n because sin 1 no matter its argument. Set this overestimate less than ɛ to conclude if N > 1/ɛ, then n > N implies original limit < ɛ. Thus, we see this sequence converges to 3.

21 Homework Homework Prove the sequence (sin(2nπ/5)) n 1 does not converge. 6.2 Prove the sequence ( 3+5n 6+8n ) n 1 converges. 6.3 Prove the sequence ( 4+5n 6 7n ) n 1 converges. 6.4 Prove the sequence ( 4+5n 2 6 3n 4n 2 ) n 1 converges.

Convergence of Sequences

Convergence of Sequences James K. Peterson Department of Biological Sciences and Department of Mathematical Sciences Clemson University February 12, 2018 Outline Convergence of Sequences Definition Let