Suppose that {Bn} converges to B, that Bn is not equal to 0 for all n, and that b is not equal to 0 prove that

Question

a) Prove that the sequence {|Bn|} is bounded below by a positive number by first showing that |Bn|>|B|/2 for all sufficiently large n.

b) Prove that the sequence {1/Bn} is bounded.

c) Prove that {1/Bn} converges to 1/B by noting that there exists a positive constant M such that |1/Bn - 1/B|=
d) Finish the proof by writting An/Bn=An(1/Bn) and using the result for the product of two convergent sequences.

Answer 1

It looks like we're trying to prove that {An/Bn} converges to A/B, where {An} is a sequence converging to A.

So, proceeding as suggested:
(a) Let ε = |B|/2 > 0 since B ≠ 0. Since {Bn} converges to B, there exists an N ∈ Z+ such that for all n > N, |Bn - B| < ε = |B|/2. By the triangle inequality |B| <= |Bn| + |Bn-B|, so |Bn| >= |B| - |Bn-B| > |B| - |B/2| = |B|/2.
So for all n > N, |Bn| > |B|/2. Let k = min {|Bn|: 1<=n<=N}. Since none of the Bn are 0 and this is a finite set, k is well-defined and is positive. Let r = min {k, |B|/2} > 0. Then for all n, |Bn| > r. So {|Bn|} is bounded below by r.

(b) This follows directly from (a): Let R = 1/r > 0. Then since |Bn| > r for every n, 0 < |1/Bn| = 1/|Bn| < R. So -R < 1/Bn < R, and thus the sequence {1/Bn} is bounded.

(c) |1/Bn - 1/B| = |(B-Bn)/(Bn.B)| = |Bn-B| / [|Bn|.|B|]
Since |Bn| > r, 1/|Bn| < 1/r. So let M = 1/(r|B|) > 0. Then
|1/Bn - 1/B| = |Bn-B| / [|Bn|.|B|] < M|Bn-B| for all n.
This is sufficient to show that {1/Bn} converges to {1/B}. [To see this: let ε > 0 and let N > 0 be such that for all n>N, |Bn-B| < ε/M. Then for all n > N, |1/Bn-1/B| < M|Bn-B| < M(ε/M) = ε. So {1/Bn} converges to 1/B.]

d) As it says: An/Bn = An (1/Bn); {An} -> A, {1/Bn} -> 1/B, so {An/Bn} -> A(1/B) = A/B.

Answer 2

that's "rediculously" ordinary, in case you be attentive to the delta-epsilon definition of a shrink. permit epsilon be B/2 interior the definition. there is an N such that for all n > N, B - B/2 < bn < B + B/2 i.e. B/2 < bn < 3B/2 2/B > a million/bn > 2/3B considering the fact that B > 0, 2/B > |a million/bn| > 0