Double Sequence Limits: UNIZOR.COM - Math4Teens - Calculus - Limit of Sequence - Double Limits
Notes to a video lecture on http://www.unizor.com Double Limits Sometimes we are interested in sequences that depend on two natural numbers like {am,n}. For example, am,n=arctan(m)+2−n. Now, if m→∞ and n→∞, how to calculate the limit of am,n? It might be limm→∞ [limn→∞ (am,n)] or limn→∞ [limm→∞ (am,n)] depending on which index, m or n, we will use to go to a limit first, and there is no guarantee that the answers will be the same. Let's check both methods. limn→∞ (am,n) = = limn→∞ (arctan(m) + 2−n) = = arctan(m) limm→∞ (arctan(m)) = π/2 If we change the order of limits, limm→∞ (am,n) = = limm→∞ (arctan(m) + 2−n) = = π/2 + 2−n limn→∞ (π/2 + 2−n) = π/2 As you see, the results are the same. In some way, the equality of these two limits might be considered similar to a standard accounting procedure of checking the calculations. Imagine an M⨯N matrix with numbers. If you summarize the numbers within each of M rows with row #i having a sum Ri and summarize all these row totals by i from 1 to M, you will get a total of all numbers in an original table. If you summarize the numbers within each of N columns with column #j having a sum Sj and summarize all these column totals by j from 1 to N, you should get exactly the same total of all numbers. If these two calculated totals are not equal, that is if Σi∈[1,M] Ri ≠ Σj∈[1,N] Sj then your calculations are wrong. The equality of two limits in the example above is not coincidental. We shall prove the following theorem. Theorem IF limn→∞ (am,n) = bm limm→∞ (bm) = B limm→∞ (am,n) = cn limn→∞ (cn) = C THEN B = C Proof Assume, B≠C and |B−C|=d. Since limm→∞(bm)=B, there exists some natural number M1 such that |B−bm| < d/4 for all m>M1 Since limn→∞(am,n)=bm, there exists some natural number N1 such that |am,n−bm| < d/4 for all n>N1 Then, for all pairs m,n such that m>M1 and n>N1 both following inequalities are true: |B − bm| < d/4 and |bm − am,n| < d/4 Therefore, |B−am,n| < d/2 That is, our double sequence am,n with both indices m and n sufficiently large (m>M1 and n>N1) is closer to number B than d/2. Absolutely similarly we can prove that this same sequence am,n with both indices sufficiently large (m>M2 and n>N2) is closer to number C than d/2. Here is a proof in mathematical symbolics. limn→∞ (cn) = C ∴ ∴ ∃M2: |C−cn| < d/4 ∀m>M2 limm→∞ (am,n) = cn ∴ ∴ ∃N2: |cn−am,n| < d/4 ∀n>N2 m>M2 ∧ n>N2 ∴ ∴ |C−cn| < d/4 ∧ |cn−am,n| < d/4 ∴ ∴ |C−am,n| < d/2 Choosing M=max(M1,M2) and N=max(N1,N2) we see that all am,n with m>M and n>N should be closer to B than d/2 and at the same time should be closer to C than d/2. With the distance between B and C equal to d it's impossible. We came to contradiction that signifies that our initial assumption that B≠C is wrong. Therefore, B=C. End of proof.