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Math and science::Analysis::Tao::06. Limits of sequences

# Squeeze test

Let $$(a_n)_{n=m}^{\infty}$$, $$(b_n)_{n=m}^{\infty}$$ and $$(c_n)_{n=m}^{\infty}$$ be sequences of real numbers such that:
$a_n \le b_n \le c_n$
for all $$n \ge m$$. Suppose that $$(a_n)_{n=m}^{\infty}$$ and $$(c_n)_{n=m}^{\infty}$$ both converge to the same limit $$L$$. Then $$(b_n)_{n=m}^{\infty}$$ also converges to $$L$$.

This is not an iff relationship.

This proof is broken down using the comparison principle, then using the fact that if a sequence converges, then its limit superior and limit inferior both equal the limit of the sequence.

The comparison principle
Suppose that $$(a_n)_{n=m}^{\infty}$$ and $$(b_n)_{n=m}^{\infty}$$ are two sequences of real numbers such that $$a_n \le b_n$$ for all $$n \ge m$$. Then we have the inequalities:
\begin{aligned} \sup(a_n)_{n=m}^{\infty} &\le \sup(b_n)_{n=m}^{\infty} \\ \inf(a_n)_{n=m}^{\infty} &\le \inf(b_n)_{n=m}^{\infty} \\ \limsup_{n\rightarrow \infty} a_n &\le \limsup_{n\rightarrow \infty} b_n \\ \liminf_{n\rightarrow \infty} a_n &\le \liminf_{n\rightarrow \infty} b_n \end{aligned}

Tao, Analysis I
Chapter 6, p145