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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:
\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


Tao, Analysis I
Chapter 6, p145