Theorem
- If {an} and {bn} are convergent
- Then, {anbn} are also convergent
Proof
- Suppose {an} converges to some a∈R
- Suppose {bn} converges to some b∈R
- We WTS: {anbn} converges (∃l∈R,∀ϵ>0,∃N>0 s.t ∀n∈N if n>N then, ∣an−l∣<ϵ)
- Choose l=ab
- Let ϵ>0 be arbitrary
- Invoke 1,2 with ϵ1=1+∣b∣12ϵ,ϵ2=∣a∣+112ϵ,e3=1
- Note that ∃N1>0 s.t ∀n∈N,n>N1 then, ∣an−a∣<ϵ1=1+∣b∣12ϵ as {an} converges
- Note that ∃N2>0 s.t ∀n∈N,n>N2 then, ∣bn−b∣<ϵ2=∣a∣+112ϵ as {bn} converges
- Note that ∃N3>0 s.t ∀n∈N,n>N3 then, ∣bn−b∣<e3=1 as {bn} converges)
- ∣bn∣−∣b∣≤∣bn−b∣<1 by Reverse Triangle Inequality
- ⟹∣bn∣−∣b∣<1
- ⟹∣bn∣<1+∣b∣
- Choose N=max{N1,N2,N3}>0
- Suppose n>N
- Then, ∣anbn−ab∣=∣anbn−abn+abn−ab∣
- =∣bn(an−a)+a(bn−b)∣≤∣bn(an−a)∣+∣a(bn−b)∣ by Triangle Inequality
- =∣bn∣∣an−a∣+∣a∣∣bn−b∣ by Absolute Function properties
- =(1+∣b∣)∣an−a∣+∣a∣∣bn−b∣
- <(1+∣b∣)1+∣b∣12ϵ+∣a∣∣a∣+112ϵ≤2ϵ+(1)2ϵ=ϵ as ∣a∣+11≤1
- □