Sandwich Rule for Sequences Proof Sandwich rule for sequences:
Let (an)n∈N, (bn)n∈N and (cn)n∈N be sequences with the following properties.
(a) Both (an)n∈N and (cn)n∈N converge to the same real number r.
(b) an ≤ bn ≤ cn for all n ∈ N.
Then also (bn)n∈N converges to r.
Proof. :
We show that (bn − an)n∈N is a null sequence.
Then by the sum rule,
limn→∞ bn exists and is equal to:
limn→∞
(an + bn − an) = limn→∞
an + limn→∞
(bn − an) = r + 0 = r,
as desired.
By (b) we have 0 ≤ bn − an ≤ cn − an and by the sandwich rule for null sequences
1.5.3(iii) it suffices to show that (cn − an)n∈N is a null sequence.
However, by the
scalar multiplication rule, (−an)n∈N converges with limn→∞(−an) = −r and so by
the sum rule again
limn→∞
(cn − an) = limn→∞
(cn + (−an)) = limn→∞
cn + limn→∞
(−an) = r − r = 0,
as desired.
I understand the majority of the proof but don't understand how to show that (bn − an)n∈N is a null sequence from the information given
 A: $\limsup b_n \le \lim c_n = r, \quad \liminf b_n \ge \lim a_n = r$, hence $\limsup b_n = \liminf b_n = r$, hence $\lim b_n = r$.
A: You want to prove that $b_n-a_n$ is a null sequence.
The key idea is to show instead that $c_n-a_n$ is a null sequence, and then since $0\leq b_n-a_n\leq c_n-a_n$, it follows from the sandwich rule for null sequences (1.5.3(iii) in your notes/book) that $b_n-a_n$ is also a null sequence.
To prove that $c_n-a_n$ is a null sequence, you calculate the limit $\lim_{n\to\infty} c_n-a_n$ as follows:
$$\lim_{n\to\infty} c_n - a_n = \lim_{n\to\infty}c_n + (-a_n) = \lim_{n\to\infty} c_n + \lim_{n\to\infty} (-a_n) = \lim_{n\to\infty} c_n - \lim_{n\to\infty} a_n = r-r = 0.$$
The conclusion is thus that $c_n-a_n$ is a null sequence, and that this implies by the above that $b_n-a_n$ is a null sequence as well.
To finish the proof, since $b_n-a_n$ is a null sequence, you have that $\lim\limits_{n\to\infty} b_n-a_n = 0$. Since the limit of $a_n$ exists, the limit of $b_n$ must also exist, and
$$\lim_{n\to\infty} b_n = \lim_{n\to\infty} (b_n-a_n) + \lim_{n\to\infty} a_n = 0+r = r.$$
Thus $b_n$ converges to $r$.
