Sequences: Limit Theorems

Transcription

1 Sequences: Limit Theorems Limit Theorems Philippe B. Laval KSU Today Philippe B. Laval (KSU) Limit Theorems Today 1 / 20

2 Introduction These limit theorems fall in two categories. 1 The first category deals with ways to combine sequences. Like numbers, sequences can be added, multiplied, divided,... Theorems from this category deal with the ways sequences can be combined and how the limit of the result can be obtained. If a sequence can be written as the combination of several "simpler" sequences, the idea is that it should be easier to find the limit of the "simpler" sequences. These theorems allow us to write a limit in terms of easier limits. however, we still have limits to evaluate. 2 The second category of theorems deal with specific sequences and techniques applied to them. Usually, computing the limit of a sequence involves using theorems from both categories. Philippe B. Laval (KSU) Limit Theorems Today 2 / 20

3 Triangle Inequality In many proofs or problems, different versions of the triangle inequality are often used. As a reminder, here are the different versions of the triangle inequality students must remember. a b a b a b a + b and a b a b a + b a + b Philippe B. Laval (KSU) Limit Theorems Today 3 / 20

4 Limit Properties Theorem Let x be a number such that ɛ > 0, x < ɛ, then x = 0. Theorem If a sequence converges, then its limit is unique. Theorem If a sequence converges, then it is bounded, that is there exists a number M > 0 such that a n M for all n. Philippe B. Laval (KSU) Limit Theorems Today 4 / 20

5 Limit Properties Theorem If a sequence converges, then ɛ > 0, N : m, n > N = a n a m < ɛ The above theorem simply says that if a sequence converges, then the difference between any two terms gets smaller and smaller. It should also be clear to the reader that if a n L, then so does a n+k where k is any natural number. Find lim cos nπ Philippe B. Laval (KSU) Limit Theorems Today 5 / 20

6 Limit Properties Theorem Suppose that (a n ) converges. Then, any subsequence (a nk ) also converges and has the same limit. This theorem is often used to show that a given sequence diverges. To do so, it is enough to find two subsequences which do not converge to the same limit. Alternatively, once can find a subsequence which diverges. Study the convergence of cos nπ Philippe B. Laval (KSU) Limit Theorems Today 6 / 20

7 Limit Laws The theorems below are useful when finding the limit of a sequence. Finding the limit using the definition is a long process which we will try to avoid whenever possible. Theorem Let (a n ) and (b n ) be two sequences such that a n a and b n b with a and b real numbers. Then, the following results hold: 1 lim (a n ± b n ) = (lim a n ) ± (lim b n ) = a ± b 2 lim (a n b n ) = (lim a n ) (lim b n ) = ab ( ) an 3 if lim b n = b 0 then lim = lim a n = a b n lim b n b 4 lim a n = lim a n = a 5 if a n 0 then lim a n 0 6 if a n b n then lim a n lim b n 7 if lim a n = a 0 then lim a n = lim a n = a Philippe B. Laval (KSU) Limit Theorems Today 7 / 20

8 Limit Laws: Remarks Parts 1, 2 and 3 of the above theorem hold even when a and b are ± as long as the right hand side in each part is defined. You will recall the following rules when working with : 1 + = = ( ) ( ) = 2 = ( ) = ( ) = Philippe B. Laval (KSU) Limit Theorems Today 8 / 20

9 Limit Laws: Remarks If x is any real number, then 1 + x = x + = 2 + x = x = x 3 = x = 0 { x 4 if x > 0 0 = if x < 0 { 5 if x > 0 x = x = if x < 0 { 6 if x > 0 ( ) x = x ( ) = if x < 0 Philippe B. Laval (KSU) Limit Theorems Today 9 / 20

10 Limit Laws: Remarks However, the following are still indeterminate forms. Their behavior is unpredictable. Finding what they are equal to requires more advanced techniques such as l Hôpital s rule. 1 + and 2 0 and 0 3 and , 0, 0, and 1 Philippe B. Laval (KSU) Limit Theorems Today 10 / 20

11 Limit Laws: s Use the limit laws above to compute the following limits. s lim 1 n 2 s lim n2 +2n+5 2n 2 1 s lim ( n + 1 n ) Philippe B. Laval (KSU) Limit Theorems Today 11 / 20

12 More Theorems on Limits When finding n such that x n L < ɛ, we usually simplify x n L to some expression involving n. Let E 1 (n) denote this expression. This gives us the inequality E 1 (n) < ɛ which we have to solve for n. If it is too hard, we then try to find a second expression we will call E 2 (n) such that E 1 (n) < E 2 (n) < ɛ. E 2 (n) should be such that solving the inequality E 2 (n) < ɛ is feasible and easier. In order to achieve this, several tricks are used. We recall some useful results, as well as some theorems below. Philippe B. Laval (KSU) Limit Theorems Today 12 / 20

13 More Theorems on Limits We begin with a well-known theorem. Theorem (squeeze theorem) If a n L, c n L and a n b n c n, then b n L. Theorem If 0 < a < 1 then a n 0 We look at a few examples, to see how all these results come into play. Philippe B. Laval (KSU) Limit Theorems Today 13 / 20

14 More Theorems on Limits: s lim cos n n 2 Find lim 4n n! Find lim 3n 4 n+2 Philippe B. Laval (KSU) Limit Theorems Today 14 / 20

15 More Theorems on Limits Sometimes, before we can apply the squeeze theorem, we need to apply other results to simplify the given sequence. We list a few theorems which will help us do this. Theorem (Bernoulli s inequality) If x 1, and n is a natural number, then (1 + x) n 1 + nx. Theorem (binomial theorem) (a + b) n = a n +na n 1 b+ Corollary (1 + x) n = 1 + nx + n (n 1) a n 2 b 2 n (n 1) (n 2) + a n 3 b nab n 1 +b n. 2 3! n (n 1) x n (n 1) (n 2) x nx n 1 + x n. 3! Philippe B. Laval (KSU) Limit Theorems Today 15 / 20

16 More Theorems on Limits Corollary (1 + x) n = 1 + nx + n (n 1) x n (n 1) (n 2) x nx n 1 + x n. 3! In particular, when x 0, then (1 + x) n is greater than any part of the right hand side. For example, we obtain Bernoulli s inequality: (1 + x) n 1 + nx. We could also write (1 + x) n n (n 1) x 2. And so 2 on. This is useful to get approximations on quantities like 3 n. We rewrite it as 3 n = (1 + 2) n. Philippe B. Laval (KSU) Limit Theorems Today 16 / 20

17 More Theorems on Limits: s Find lim n 4 n Find lim n2 4 n Find lim 4n 2n Philippe B. Laval (KSU) Limit Theorems Today 17 / 20

18 More Theorems on Limits Finally, we give a theorem which generalizes some of the examples we did above. Theorem All limits are taken as n. 1 If p > 0 then lim 1 n p = 0. 2 If p > 0 then lim n p = 1. 3 lim n n = 1. 4 If p > 1 and α R then lim nα p n = 0. 5 If p < 1 then lim p n = 0 6 p R, lim pn n! = 0. Philippe B. Laval (KSU) Limit Theorems Today 18 / 20

19 Sequences - Limit More s ( ) 1 n Find lim n 2 Find lim n n 3n 3 n Find lim n 5 n+2 Philippe B. Laval (KSU) Limit Theorems Today 19 / 20

20 Exercises See the problems at the end of section 4.3 from my notes on sequences. Philippe B. Laval (KSU) Limit Theorems Today 20 / 20