Limit Inferior/Examples/(-1)^n (1 + n^-1)

From ProofWiki
Jump to navigation Jump to search

Example of Limit Inferior

Let $\sequence {a_n}$ be the sequence defined as:

$\forall n \in \N_{>0}: a_n = \paren {-1}^n \paren {1 + \dfrac 1 n}$


The limit inferior of $\sequence {a_n}$ is given by:

$\displaystyle \map {\liminf_{n \mathop \to \infty} } {a_n} = -1$


This is not the same as:

$\displaystyle \inf_{n \mathop \ge 1} {\paren {-1}^n \paren {1 + \dfrac 1 n} }$


Proof

Let $L$ be the set of all real numbers which are the limit of some subsequence of $\sequence {a_n}$.

We have that:

\(\displaystyle \paren {-1}^{2 n} \paren {1 + \dfrac 1 {2 n} }\) \(=\) \(\displaystyle 1 + \dfrac 1 {2 n}\)
\(\displaystyle \) \(\to\) \(\displaystyle 1\) \(\displaystyle \text {as $n \to \infty$}\)

and:

\(\displaystyle \paren {-1}^{2 n - 1} \paren {1 + \dfrac 1 {2 n - 1} }\) \(=\) \(\displaystyle -\paren {1 + \dfrac 1 {2 n - 1} }\)
\(\displaystyle \) \(\to\) \(\displaystyle -1\) \(\displaystyle \text {as $n \to \infty$}\)
$\sequence {a_n}$ contains no subsequence which converges to a limit different from $-1$ or $1$.

Hence:

$L = \set {-1, 1}$

and it follows that:

$\displaystyle \map {\liminf_{n \mathop \to \infty} } {\paren {-1}^n \paren {1 + \dfrac 1 n} } = 0$


But note that:

$\displaystyle \sup_{n \mathop \ge 1} \paren {-1}^n \paren {1 + \dfrac 1 n}$

occurs when $n = 1$, at which point:

$\paren {-1}^n \paren {1 + \dfrac 1 n} = \paren {-1}^1 \paren {1 + \dfrac 1 1} = -2$

$\blacksquare$


Sources