Definition 2.2.1 (Squeeze Theorem). If there exists \(N >0\) such that \(a_n \leq b_n \leq c_n\) for any \(n\geq N\), and \begin {align*} \lim \limits _{n \to \infty } a_n = \lim \limits _{n \to \infty } c_n = L \end {align*}
Where \(L \in \mathbb {R}\) (i.e. finite), then \(\{b_n\}\) also has a limit and \(\lim \limits _{n \to \infty }b_n =L\)
Proof. Given that \(\lim \limits _{n\to \infty } a_n = \lim \limits _{n\to \infty } c_n = L\) and there exists some \(N > 0\) such that for all \(n>N\), \(a_n \leq b_n \leq c_n\).
Because \(a_n \to L\), for any \(\epsilon >0\), \(\exists N_1 > 0\) such for \(n>N_1\) \begin {align*} |a_n-L| < \epsilon \\ \implies -\epsilon <a_n-L< \epsilon \\ \implies L - \epsilon < a_n < L +\epsilon \end {align*}
Similarly, because, \(c_n \to L\), for the same \(\epsilon \) as above, \(\exists N_2 \in \mathbb {N}\), so that for \(n>N_2\) \begin {align*} |c_n-L| < \epsilon \\ \implies L-\epsilon < c_n < L+ \epsilon \end {align*}
Because \(\exists N > 0\) such that \(a_n\leq b_n \leq c_n\) for any \(n \geq N\), choose \(N' = \max \{N_1,N_2,N\}\), then for \(n> N'\), all the three conditions hold, and choose some arbitrary \(\epsilon \) \begin {gather*} a_n\leq b_n \leq c_n \\ L-\epsilon < a_n \leq b_n \leq c_n < L + \epsilon \\ \implies -\epsilon <b_n -L < \epsilon \\ \implies |b_n -L| < \epsilon \end {gather*} Hence for any arbitrary \(\epsilon \), when the above conditions hold, \(b_n \to L\) as well □
We need to show that \begin {align*} c_n \leq \dfrac {p(n)}{a^n} \leq b_n \\ \text {where} \lim \limits _{n \to \infty }c_n =0 \text { and }\lim \limits _{n \to \infty }b_n =0 \end {align*}
consider \(p(n)\) of some degree \(k \in \mathbb {N}\): \begin {align*} p(n) = b_kn^k + b_{k-1}n^{k-1}+ ... + b_1n+b_0 \end {align*}
For a polynomial \(p(n)\) of some degree \(k \in \mathbb {N}\) and some \(a >1\): \begin {align*} \lim \limits _{n \to \infty }\dfrac {p(n)}{a^n} &= \lim \limits _{n \to \infty }\dfrac {b_kn^k + b_{k-1}n^{k-1}+ ... + b_1n+b_0}{a^n}\\ &= \lim \limits _{n \to \infty } \left [ \dfrac {b_k}{a^n}n^k +\dfrac {b_{k-1}}{a^n}n^{k-1} +...+\dfrac {b_1}{a^n}n+\dfrac {b_0}{a^n} \right ] \end {align*}
consider any index \(j \in [0,k]\) \begin {align*} b_j\dfrac {1}{a^n}n^j = b_j\left (\dfrac {1}{a}\right )^nn^j \qquad \because a>1 \implies \frac {1}{a} <1 \text { and } \frac {1}{a} \in (-1,1)\\ \implies \text {from the lemma} \end {align*}
from the lemma: \begin {gather*} \lim \limits _{n \to \infty } b_j\left (\dfrac {1}{a}\right )^nn^j = \lim \limits _{n \to \infty }b_j \lim \limits _{n \to \infty } \left (\dfrac {1}{a}\right )^nn^j = \lim \limits _{n \to \infty }b_j \times 0 =0\\ \lim \limits _{n \to \infty } \left [ \dfrac {b_k}{a^n}n^k +\dfrac {b_{k-1}}{a^n}n^{k-1} +...+\dfrac {b_1}{a^n}n+\dfrac {b_0}{a^n} \right ] =0+0+...+0+0\\ \lim \limits _{n \to \infty }\dfrac {p(n)}{a^n} =0 \end {gather*}
The goal is to prove that: \begin {align*} b_n \leq p(n)^{\frac {1}{n}} \leq c_n \qquad \text {for } n > N, \text { for some }N\in \mathbb {N} \end {align*}
where \(b_n, c_n \to 1\) as \(n \to \infty \)
Let: \begin {align*} p(n)^{\frac {1}{n}} -1 = x_n\\ x_n\geq 0 \end {align*}
We need to bound \(x_n\). Note that \begin {align*} (1+x_n)^n = p(n)\\ \implies 1+nx_n+C_2^nx_n^2+...+nx_n^{n-1} +x_n^n = p(n) \end {align*}
WLOG, assume \(p(n)\) has a degree of \(k\) and that \(n>k\), using the general form of polynomials: \begin {align*} p(n) = a_kn^k + a_{k-1}n^{k-1}+...+a_1n+a_0\\ \end {align*}
For convenience, set \(M = \max \{|a_k|,|a_{k-1}|,...,|a_0|\}\), hence, for large \(n\) \begin {align*} p(n) \leq Mn^{k+1} \end {align*}
Now, \begin {align*} 1+nx_n+C_2^nx_n^2+...+nx_n^{n-1} +x_n^n \leq Mn^{k+1} \\ C^n_{k+2}x_n \leq 1+nx_n+C_2^nx_n^2+...+nx_n^{n-1} +x_n^n \leq Mn^{k+1} \end {align*}
Hence, \begin {align*} 0\leq x_n \leq \frac {Mn^{k+1}}{C^n_{k+2}}\\ \frac {Mn^{k+1}}{C^n_{k+2}} = \frac {M(k+2)!}{n\left ( 1-\frac {1}{n}\right )...\left ( 1-\frac {k+1}{n}\right )} \end {align*}
Clearly \begin {align*} \lim _{n\to \infty }\frac {M(k+2)!}{n\left ( 1-\frac {1}{n}\right )...\left ( 1-\frac {k+1}{n}\right )} = 0 \end {align*}
Hence, by the squeeze theorem, \(x_n\to 0\), hence \(p(n)^{\frac {1}{n}}\to 1\)