Limit Superior

Definition (Limit superior)

Let (an)n∈N⊆Rˉ(a_{n})_{n\in\mathbb{N}}\subseteq \bar{\mathbb{R}}. Define a new sequence (bn)n∈N(b_{n})_{n\in\mathbb{N}} as bn=sup⁡k≥nakb_{n}=\sup_{k\ge n}a_{k}where we observe b1≥b2≥…b_{1}\ge b_{2}\ge\dots (i.e. monotonically decreasing). Then we define the limit superior of our sequence as lim sup⁡n≥1an:=inf⁡n≥1bn=inf⁡n≥1sup⁡k≥nak\limsup_{n\ge 1}a_{n}:=\inf_{n\ge 1}b_{n}=\inf_{n\ge 1}\sup_{k\ge n}a_{k}Let T={t∈Rˉ:∃(ank)k∈N⊂(an)n∈N:lim⁡k→∞ank=t}T=\{ t\in \bar{\mathbb{R}}: \exists(a_{n_{k}})_{k\in\mathbb{N}}\subset(a_{n})_{n\in\mathbb{N}}:\lim_{ k \to \infty }a_{n_{k}}=t \}. Then lim sup⁡n≥1an=sup⁡t∈Tt\limsup_{ n \ge 1 }a_{n}=\sup_{t\in T}t

Remark

To provide intuition for this you can think of the lim sup⁡\limsup as giving “the largest value that the sequence approaches infinitely often”. While the supremum gives you the static LUB of a sequence the lim sup⁡\limsup focuses on the “tail behaviour” of a sequence and what values the sequence “settles down to” as n approaches infinity. This is important for understanding the long-term behaviour of a sequence.

Definition (Limit Inferior)

Let (an)n∈N⊆Rˉ(a_{n})_{n\in\mathbb{N}}\subseteq \bar{\mathbb{R}}. Define a new sequence (cn)n∈N(c_{n})_{n\in\mathbb{N}} as cn=inf⁡k≥nakc_{n}=\inf_{k\ge n}a_{k}where we observe c1≤c2≤…c_{1}\le c_{2}\le\dots (i.e. monotonically increasing). Then we define the limit inferior of our sequence as lim inf⁡n≥1an:=sup⁡n≥1cn=sup⁡n≥1inf⁡k≥nak\liminf_{n\ge 1}a_{n}:=\sup_{n\ge 1}c_{n}=\sup_{n\ge 1}\inf_{k\ge n}a_{k}Let T={t∈Rˉ:∃(ank)k∈N⊂(an)n∈N:lim⁡k→∞ank=t}T=\{ t\in \bar{\mathbb{R}}: \exists(a_{n_{k}})_{k\in\mathbb{N}}\subset(a_{n})_{n\in\mathbb{N}}:\lim_{ k \to \infty }a_{n_{k}}=t \}. Then lim sup⁡n≥1an=sup⁡t∈Tt\limsup_{ n \ge 1 }a_{n}=\sup_{t\in T}t

Proposition (Properties of Limit Inferior & Limit Superior)

Let (an)n∈N⊆Rˉ(a_{n})_{n\in\mathbb{N}}\subseteq \bar{\mathbb{R}}. The liminf and limsup of (an)n∈N(a_{n})_{n\in\mathbb{N}} have the following properties:

  1. lim inf⁡n≥1an≤lim sup⁡n≥1an\liminf_{n\ge 1}a_{n}\le\limsup_{n\ge 1}a_{n}
  2. lim sup⁡n≥1(−an)=−lim inf⁡n≥1an\limsup_{n\ge 1}(-a_{n})=-\liminf_{n\ge 1}a_{n}
  3. lim sup⁡n≥1an=lim inf⁡n≥1an  ⟹  lim⁡n→∞an exists and equals both\limsup_{n\ge 1}a_{n}=\liminf_{n\ge 1}a_{n}\implies \lim_{ n \to \infty } a_{n}\text{ exists and equals both}
  4. lim sup⁡n≥1(an+αn)≤lim sup⁡n≥1an+lim sup⁡n≥1αn\limsup_{n\ge 1}(a_{n}+\alpha_{n})\le \limsup_{n\ge 1}a_{n}+\limsup_{n\ge 1}\alpha_{n}iff we don’t encounter ∞−∞\infty-\infty or −∞+∞-\infty+\infty
  5. If an≤αn,∀n≥1a_{n}\le \alpha_{n},\forall n\ge 1 then lim inf⁡n≥1an≤lim inf⁡n≥1αnlim sup⁡n≥1an≤lim sup⁡n≥1αn\begin{align*} &\liminf_{n\ge 1}a_{n}\le\liminf_{n\ge 1}\alpha_{n}\\ &\limsup_{n\ge 1}a_{n}\le\limsup_{n\ge 1}\alpha_{n} \end{align*}

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