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

## 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:

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

$\blacksquare$