state and prove that sandwich theorem of infinite sequence of real number
Answers
Answer:
Step-by-step explanation:
Please bear with me here and please try to read it all and spot any mistakes or errors as I'm trying to prove this result but I'm unsure of whether I have done it or not. THANK YOU.
Suppose we have the following statement (an)→ℓ,(bn)→ℓ
a
n
ℓ
b
n
ℓ
and we have
an≤cn≤bn
a
n
c
n
b
n
then (cn)→ℓ
c
n
ℓ
. I think I have a proof which goes as follow ;
an≤cn≤bn≤ ⇒0≤cn−an≤bn−an
a
n
c
n
b
n
0
c
n
a
n
b
n
a
n
, as the terms are all larger than 0, taking the absolute value will not change any of the signs of the inequalities. So we have
0≤|cn−an|≤|bn−an|.
0
c
n
a
n
b
n
a
n
Now consider
|bn−an|=|(bn−ℓ)+(ℓ−an)|≤|bn−ℓ|+|an−ℓ| (by triangle inequality).
b
n
a
n
b
n
ℓ
ℓ
a
n
b
n
ℓ
a
n
ℓ
(by triangle inequality)
Using the definition of a sequence tending to a value, if (an)→ℓ
a
n
ℓ
then ∃N1∈N s.t ∀n>N,|an−ℓ|<ϵ ,∀ϵ>0.
∃
N
1
N
s.t
∀
n
N
a
n
ℓ
ϵ
∀
ϵ
0
We do the same for (bn)
b
n
but replacing N1
N
1
with N2
N
2
and using the same ϵ
ϵ
without loss of generality. So we can now say that
|bn−an|≤|bn−ℓ|+|an−ℓ|<2ϵ.
b
n
a
n
b
n
ℓ
a
n
ℓ
2
ϵ
So we have
0≤|cn−an|≤|bn−an|<ϵ0, where ϵ0=2ϵ.
0
c
n
a
n
b
n
a
n
ϵ
0
where
ϵ
0
2
ϵ
So we can conclude (using sandwich theorem for null sequences) that (cn−