A sequence \((x_n)_{n=1}^{\infty}\subseteq\mathbb{R}\) converges to \(\infty\) if \(\forall\, M\in \mathbb{R}\text{,}\)\(\exists N\in\mathbb{Z}_{\geq 1}\) such that for all \(n\in\mathbb{Z}, n\geq N, x_n\geq M\text{.}\) In this case we'll write \(\lim_{n\rightarrow\infty}x_n=\infty\text{.}\) Similarly \(\lim_{n\rightarrow\infty}x_n=-\infty\) iff \(\exists N\in\mathbb{Z}{\geq 1}\) such that for all \(n\in\mathbb{Z},n\geq N, x_n<M\text{.}\)
Remark2.28.
If \(\lim_{n\rightarrow\infty}x_n=\pm\infty\text{,}\) we'll still call \((x_n)_{n=1}^{\infty}\) as a divergent sequence. (might say diverges to infinity)
Every monotone sequence in \(\mathbb{R}\) has a limit in \(\overline{\mathbb{R}}=\mathbb{R}\cup\{\infty,-\infty\}\text{.}\)
Definition2.30.
Let \((x_n)_{n=1}^{\infty}\subseteq\mathbb{R}\) be given.
The limit superior of \((x_n)_{n=1}^{\infty}\) is\(\limsup_{n\rightarrow\infty}x_n=\lim_{n\rightarrow\infty}(\sup_{k\geq n}x_k)=\inf_{n\geq 1}(\sup_{k\geq n}x_k)\)
Similarly the limit inferior of \((x_n)_{n=1}^{\infty}\) is\(\liminf_{n\rightarrow\infty}x_n=\lim_{n\rightarrow\infty}(\inf_{k\geq n}x_k)=\sup_{n\geq 1}(\inf_{k\geq n}x_k)\)
\(\exists\) integer \(N\geq 1\) such that \(\forall\) int \(n\geq N\text{,}\)\(x_n<C\text{.}\)
Theorem2.33.
For any sequence \((x_n)_{n=1}^{\infty}\subseteq \mathbb{R}\text{,}\)\(\liminf_{n\rightarrow\infty}x_n\leq \limsup_{n\rightarrow\infty}x_n\) and \(\liminf_{n\rightarrow\infty}x_n= \limsup_{n\rightarrow\infty}x_n\) iff \(\lim_{n\rightarrow\infty}x_n\) exists and in this case \(\lim_{n\rightarrow\infty}x_n=\liminf_{n\rightarrow\infty}x_n= \limsup_{n\rightarrow\infty}x_n\text{.}\)
For each integer \(n\geq 1\text{,}\)\(\inf_{k\geq n}x_k\leq \sup_{k\geq n}x_k\text{.}\) Now take limit as \(n\rightarrow\infty\text{.}\) Now suppose \(x=\lim_{n\rightarrow\infty}x_n\) exists. We'll assume \(x\in\mathbb{R}\) (Exercise: modify the proof for \(x=\pm\infty\)). We'll show \(\liminf_{n\rightarrow\infty}x_n\geq x-\varepsilon\) and \(\limsup_{n\rightarrow\infty}x_n\leq x+\varepsilon\text{.}\) Let \(N\in\mathbb{Z}_{\geq 1}\) be st for all \(n\geq N, |x_n-x|<\varepsilon\text{.}\) This implies \(x-\varepsilon<x_n<x+\varepsilon\) for all \(n\geq N\text{.}\) Now, \(\inf_{n\geq N}x_n\geq x-\varepsilon\) and \(\sup_{n\geq N}x_n\leq x+\varepsilon\text{.}\) Therefore, \(\liminf_{n\rightarrow\infty}x_n=\sup_{m\geq 1}\inf_{n\geq m}x_n\geq \inf_{n\geq N}x_n\geq x-\varepsilon\text{.}\) Similarly, \(\limsup_{n\rightarrow\infty}x_n\leq x+\varepsilon\) since (1) and (2) had for all \(\varepsilon>0, \liminf_{n\rightarrow\infty}x_n\geq_{(1)} x\geq_{(2)} \limsup_{n\rightarrow\infty}x_n\geq \liminf_{n\rightarrow\infty}x_n\text{.}\) So \(x=\liminf_{n\rightarrow\infty}x_n=\limsup_{n\rightarrow\infty}x_n\text{.}\) Conversely, suppose \(\liminf_{n\rightarrow\infty}x_n=\limsup_{n\rightarrow\infty}x_n:=x\text{.}\) We'll assume again that \(x\in\mathbb{R}\text{.}\) Fix \(\varepsilon>0\text{.}\) Since \(\limsup_{n\rightarrow\infty}x_n<x+\varepsilon\) and \(\liminf_{n\rightarrow\infty}x_n>x-\varepsilon\text{,}\) there is an \(N\in\mathbb{Z}, N\geq 1\) such that for all \(n\geq N\text{,}\)\(x_n<x+\varepsilon\) and \(x_n>x-\varepsilon\text{.}\) But this means \(|x-x_n|<\varepsilon\) for all \(n\geq N\text{,}\) so \(\lim_{n\rightarrow\infty}x_n=x\text{.}\)
Remark2.34.
Often to show a sequence \((x_n)_{n=1}^{\infty}\) converges to \(x\) one shows \(\limsup_{n\rightarrow\infty}x_n\leq x\) and \(\liminf_{n\rightarrow\infty}x_n\geq x\text{.}\)
Corollary2.35.
If \((x_n)_{n=1}^{\infty}\subseteq \mathbb{R}\) is a sequence and \((x_{n_k})_{k=1}^{\infty}\subseteq \mathbb{R}\text{,}\) then
Second inequality is known (what we ended weds with). We'll prove the first – the third is similar.Note that \(n_1\geq 1\) and \(n_{k+1}>n_k\text{,}\) and so \(n_k\geq k\,\forall\, k\geq 1\text{.}\) This implies that \(\forall\,k\geq 1\text{,}\)\(\{m\in\mathbb{Z}\,|\,m\geq k\}\supseteq\{n_m\,|\,m\in\mathbb{Z},m\geq k\}\text{,}\) and so \(\inf_{m\geq k}x_m\leq \inf_{m\geq k}x_{n_m}\)Taking the limit as \(k\rightarrow\infty\text{,}\) we have \(\lim_{k\rightarrow\infty}\inf_{m\geq k}x_m\leq \lim_{k\rightarrow\infty}\inf_{m\geq k}x_{n_m}\) and \(\liminf_{k\rightarrow\infty}x_k\leq \liminf_{k\rightarrow\infty}x_{n_k}\text{.}\)
Theorem2.36.
If \((x_n)_{n=1}^{\infty}\subseteq \mathbb{R}\) is a sequence, then \((x_n)_{n=1}^{\infty}\) converges iff every subsequence of \((x_n)_{n=1}^{\infty}\) converges. Moreover, in this case, for every subsequence \((x_{n_k})_{k=1}^{\infty}\)
(\(\impliedby\)) is trivial. \((x_n)_{n=1}^{\infty}\) is a subsequence of itself.(\(\implies\)) and “moreover”: assume \((x_n)_{n=1}^{\infty}\) converges and write \(x=\lim_{n\rightarrow\infty}x_n\text{.}\) Let \((x_{n_k})_{k=1}^{\infty}\) be a subsequence. We'll show \(\lim_{k\rightarrow\infty}x_{n_k}=x\text{.}\)By the previous result,
Indeed \((\impliedby)\) still trivial. \((\implies)\) Suppose \(x_n\rightarrow x\in\mathbb{C}\) and \((x_{n_k})_{k=1}^{\infty}\subseteq \mathbb{C}\) is a subsequence. We know \((\text{Re}(x_n))_{n=1}^{\infty}\subseteq \mathbb{R}\) converges to \(\text{Re}(x)\text{,}\) and so the subsequence \((\text{Re}(x_{n_k}))_{k=1}^{\infty}\subseteq \mathbb{R}\) converges to \(\text{Re}(x)\text{.}\) Similarly \((\text{Im}(x_{n_k})_{k=1}^{\infty}\) converges to \(\text{Im}(x)\text{.}\) Now \(x_{n_k}=\text{Re}(x_{n_k})+i\text{Im}(x_{n_k})\rightarrow\text{Re}(x)+i\text{Im}(x)=x\)
Lemma2.38.
For every sequence \((x_n)_{n=1}^{\infty}\subseteq \mathbb{R}\) there are subsequences \((x_{m_k})_{k=1}^{\infty}\) and \((x_{n_k})_{k=1}^{\infty}\) of \((x_{n})_{n=1}^{\infty}\) so that
Suppose \(\limsup_{n\rightarrow\infty}x_n>-\infty\text{.}\) Fix a strictly increasing sequence \((C_k)_{k=1}^{\infty}\subseteq\mathbb{R}\) with \(\sup_{k\geq 1}C_k=\limsup_{n\rightarrow\infty}x_n\)(eg when \(\limsup_{n\rightarrow\infty}x_n<\infty\text{,}\)\(C_k=\limsup_{n\rightarrow\infty}x_n-\frac{1}{k}\) and when \(\limsup_{n\rightarrow\infty}x_n=\infty, C_k=k\)).Claim: There is a subsequence \((x_{n_k})_{k=1}^{\infty}\) of \((x_n)_{n=1}^{\infty}\) such that \(x_{n_k}>C_k\) for all \(k\geq 1\text{.}\)Assume the claim for now (we'll come back to it). Then \(\liminf_{k\rightarrow\infty}x_{n_k}\geq \liminf_{k\rightarrow\infty}C_k=\limsup_{n\rightarrow\infty}x_n\geq \limsup_{k\rightarrow\infty}x_{n_k}\text{.}\) So \(\lim_{k\rightarrow\infty}x_{n_k}=\limsup_{n\rightarrow\infty}x_n\) as required.Proof of claim: (constructed inductively) First note that \(\inf_{n\geq 1}\sup_{l\geq n}x_l=\limsup_{n\rightarrow\infty}x_n>C_1\text{.}\) We have \(\sup_{l\geq n}x_l>C_1\) for all \(n\geq 1\text{.}\) In particular \(\sup_{l\geq 1}x_l>C_1\text{.}\) There is an \(n_1\geq 1\) such that \(x_{n_1}>C_1\text{.}\)Assume now we've constructed \(n_1<n_2<\dots<n_k\) with \(x_{n_i} > C_i\) for all \(i=1,\dots,k\text{.}\) Since \(\limsup_{n\rightarrow\infty}x_n>C_{k+1}\text{,}\) we have \(\sup_{n\geq l}x_n>C_{k+1}\) for all \(l\geq 1\text{.}\) In particular, taking \(l=n_k+1\text{,}\)\(\sup_{n\geq n_k+1}x_n>C_{k+1}\text{.}\) There is an \(n_{k+1}\geq n_k+1>n_k\) with \(x_{n_{k+1}}>C_{k+1}\text{.}\) This proves the claim.
If \((x_n)_{n=1}^{\infty}\subseteq \mathbb{R}\text{,}\) is bounded, then \(\limsup_{n\rightarrow\infty}x_n\in\mathbb{R}\text{.}\) Now apply the previous result.Now if \((x_n)_{n=1}^{\infty}\subseteq\mathbb{C}\) is bounded, then \((\text{Re}(x_n))_{n=1}^{\infty}\subseteq \mathbb{R}\) is also bounded (as \(\left|{\text{Re}(z)}\right|\leq \left|{z}\right|\) for all \(z\in\mathbb{C}\text{.}\) So there is a subsequence \((x_{n_k})_{k=1}^{\infty}\) of \((x_n)_{n=1}^{\infty}\) such that \((\text{Re}(x_{n_k}))_{k=1}^{\infty}\) converges. Note that \((x_{n_k})_{k=1}^{\infty}\subseteq\mathbb{C}\) is bounded, and hence \((\Im(x_{n_k}))_{k=1}^{\infty}\subseteq\mathbb{R}\) is bounded. So there is a subsequence \((x_{n_{k_l}})_{l=1}^{\infty}\) of \((x_{n_k})_{k=1}^{\infty}\) so that \((\Im(x_{n_{k_l}}))_{l=1}^{\infty}\) converges. As \((\text{Re}(x_{n_{k_l}}))_{l=1}^{\infty}\) of \((\text{Re}(x_{n_k}))_{k=1}^{\infty}\) and the latter sequence converges, so does the former. But now \(x_{n_{k_l}}=\text{Re}(x_{n_{k_l}})+i\Im((x_{n_{k_l}}))\text{,}\) so \((x_{n_{k_l}})_{l=1}^{\infty}\) converges.
