(8.2 b,e) For each of the following sequences, determine its limit (usual Calculus reasoning will be helpful here), then prove that the sequence does indeed converge to this limit.
Let . Consider . Then for ,
So .
Let . Consider and . Then
So .
(8.6) Let be a sequence in .
Prove that if and only if . (Hint: This is an “if and only if” statement, so you will need to prove both directions of the statement.)
() Suppose . Consider . Then there exists so that implies that .
Observe that which is less than by above. So .
() The reverse direction is very similar to the forward direction.
Prove that if that exists but diverges.
We have that diverges by Example 4 in Section 8 (pp41-42). Also, which converges to as it is constant.
(7.4) Give examples of
A sequence of irrational numbers converging to a rational number
For example, let .
A sequence of rational numbers converging to an irrational number
For example, let which converges to .
For each of the following false statements, give a counterexample.
If and are two divergent sequences, their sum also diverges.
For example, let diverge and let .
If a sequence converges, so too does the sequence .
For example, let .
If the sequence converges and the sequence diverges, the product also diverges.
For example, let be any divergent sequence and let .
Another example, let be any divergent, nonzero sequence and let .
(9.12) Suppose for all and that the limit exists.
Show that if then .
Proof. As there exists a number by the Completeness Axiom. Consider which is positive as .
Claim: There exists such that for all , . Let be that for which
for all , which exists as . (Notice: For simplicity’s sake, we’re taking here so that even satisfies the convergence inequality. We may do this, as once we know with the strict inequality exists, we can just choose the next natural number larger by the Archimedean Property.) So, for such we have that
Notice that the right part of this inequality implies that , as claimed.
Claim: For all natural numbers , . We prove this using induction on . The base case of we have already shown, by taking in the inequality which is true by choice of .
Suppose now that . We have that , so we know that . By our inductive hypothesis, we have that
so by induction the claim is true.
Claim: . Observe that we have now that . Consider ; then consider , which is a natural number as and . So
Notice we’re actually considering the sequence here. But think about why if we can prove the limit for this sequence, we’ve also proved the limit for the full sequence as well.
The LHS has limit , and the RHS has limit (why?). So by the squeeze theorem, the sequence has limit 0 and so as well. By question (2a), this means .
Show that if then .
Consider the sequence . So
So and by question 2a, . As is a positive sequence and we have that .