Limit Superior

Definition (Limit superior)

Let (an)nNRˉ(a_{n})_{n\in\mathbb{N}}\subseteq \bar{\mathbb{R}}. Define a new sequence (bn)nN(b_{n})_{n\in\mathbb{N}} as bn=supknakb_{n}=\sup_{k\ge n}a_{k}where we observe b1b2b_{1}\ge b_{2}\ge\dots (i.e. monotonically decreasing). Then we define the limit superior of our sequence as lim supn1an:=infn1bn=infn1supknak\limsup_{n\ge 1}a_{n}:=\inf_{n\ge 1}b_{n}=\inf_{n\ge 1}\sup_{k\ge n}a_{k}Let T={tRˉ:(ank)kN(an)nN:limkank=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 supn1an=suptTt\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)nNRˉ(a_{n})_{n\in\mathbb{N}}\subseteq \bar{\mathbb{R}}. Define a new sequence (cn)nN(c_{n})_{n\in\mathbb{N}} as cn=infknakc_{n}=\inf_{k\ge n}a_{k}where we observe c1c2c_{1}\le c_{2}\le\dots (i.e. monotonically increasing). Then we define the limit inferior of our sequence as lim infn1an:=supn1cn=supn1infknak\liminf_{n\ge 1}a_{n}:=\sup_{n\ge 1}c_{n}=\sup_{n\ge 1}\inf_{k\ge n}a_{k}Let T={tRˉ:(ank)kN(an)nN:limkank=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 supn1an=suptTt\limsup_{ n \ge 1 }a_{n}=\sup_{t\in T}t

Proposition (Properties of Limit Inferior & Limit Superior)

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

  1. lim infn1anlim supn1an\liminf_{n\ge 1}a_{n}\le\limsup_{n\ge 1}a_{n}
  2. lim supn1(an)=lim infn1an\limsup_{n\ge 1}(-a_{n})=-\liminf_{n\ge 1}a_{n}
  3. lim supn1an=lim infn1an    limnan 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 supn1(an+αn)lim supn1an+lim supn1αn\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 anαn,n1a_{n}\le \alpha_{n},\forall n\ge 1 then lim infn1anlim infn1αnlim supn1anlim supn1αn\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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