Prove that a n= ( n 2 + 2 n – 1)/(2 n 2 – n – 2) converges. Since 1/ n → 0 by Example 1, we have 2/ n → 0. Now we can prove Example 2 using Example 1. For the second statement, write a n– b n = a n + (-1) b n and apply the first statement. Hence, the result follows by multiplying ( a n) and ( c). Then:įor the first statement note that the constant sequence ( c, c, … ) converges to c. Let be sequences converging to K, L respectively. Having proven these basics, some consequences are immediate.Ĭorollary. Then the set of all | a n| is bounded by max, whenever n> N, we get: Now there are only finitely many a n for n≤ N, so we can let B be the maximum of these | a n|. Since, we pick ε=1, and by definition of convergence, there exists N such that when n > N, we have find M such that when n > M, | a n – K| M’, | b n – L| N, we have:Ģ.On the other hand, if L 0, since ε/2>0, we can And we’re done since there’re infinitely many odd n. If L ≥ 0, then for any odd n, we have | a n– L| = |-1- L| ≥ 1. there exists an ε, such that for any N, we can find n>N satisfying |a n-L| ≥ ε. įor our problem, suppose ( a n) → L.The sequence a n does not converge to L if and only if: ![]() “It is not true that there exists x, P( x) holds.” → “For any x, P( x) doesn’t hold.”Īpplying this rule-of-thumb recursively, we see that:.“It is not true that for every x, P( x) holds.” → “There exists x, P( x) doesn’t hold.”.what does it mean to say that a n does not converge to L? There’s a nice trick to take the negation of such statements: We need to take the negation of the definition, i.e. Prove that the sequence a n= (-1) n is not convergent. Which shows that the sequence converges to 3. ♦Įxample 3. For each ε>0, we need to establish an N such that when n > N, we have | a n – 0| 0, we set N = 6/ε. Prove that if a n= 1/ n, then ( a n) → 0. for every positive real ε, there exists N, such that whenever n>N, we have |a n – L| N, we have | a n– K| M, we have | a n– L| max( M, N) such thatĮxample 1.We say that (a n) converges to a real number L (written a n → L) if: One can think of it as a function N → R, where N is the set of positive integers. Our focus here is to provide a rigourous foundation for the statement “sequence ( a n) → L as n → ∞”.ĭefinition (Convergence). Definition of ConvergenceĪ sequence in R is given by ( a 1, a 2, a 3, …), where each a i is in R. Thus, L = sup(S) iff (i) L is an upper bound, (ii) for each ε>0, there exists x in S, x > L-ε. Since L-ε is not an upper bound, there exists x in S, x > L-ε. the set of real (or rational) numbers 0 0). A set may be upper-bounded without possessing a maximum, e.g.If min( S) exists, then so does inf( S), and min( S) = inf( S).If max( S) exists, then so does sup( S), and max( S) = sup( S).The maximum of T is called the infimum of S, denoted inf(S). Let S be lower-bounded, and T the set of lower bounds of S.The minimum of T is called the supremum of S, denoted sup(S). ![]() ![]()
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