Limit Superior/Examples/(-1)^n (1 + n^-1)
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Example of Limit Superior
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 superior of $\sequence {a_n}$ is given by:
- $\ds \map {\limsup_{n \mathop \to \infty} } {a_n} = 1$
This is not the same as:
- $\ds \sup_{n \mathop \ge 1} \paren {-1}^n \paren {1 + \dfrac 1 n}$
Proof
Let $L$ be the set of all real numbers which are the limit of some subsequence of $\sequence {a_n}$.
We have that:
\(\ds \paren {-1}^{2 n} \paren {1 + \dfrac 1 {2 n} }\) | \(=\) | \(\ds 1 + \dfrac 1 {2 n}\) | ||||||||||||
\(\ds \) | \(\to\) | \(\ds 1\) | \(\ds \text {as $n \to \infty$}\) |
and:
\(\ds \paren {-1}^{2 n - 1} \paren {1 + \dfrac 1 {2 n - 1} }\) | \(=\) | \(\ds -\paren {1 + \dfrac 1 {2 n - 1} }\) | ||||||||||||
\(\ds \) | \(\to\) | \(\ds -1\) | \(\ds \text {as $n \to \infty$}\) |
- $\sequence {a_n}$ contains no subsequence which converges to a limit different from $-1$ or $1$.
Hence:
- $L = \set {-1, 1}$
and it follows that:
- $\ds \map {\limsup_{n \mathop \to \infty} } {\paren {-1}^n \paren {1 + \dfrac 1 n} } = 1$
But note that:
- $\ds \sup_{n \mathop \ge 1} \paren {-1}^n \paren {1 + \dfrac 1 n}$
occurs when $n = 2$, at which point:
- $\paren {-1}^n \paren {1 + \dfrac 1 n} = \paren {-1}^2 \paren {1 + \dfrac 1 2} = \dfrac 3 2$
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 5$: Subsequences: Lim sup and lim inf: Exercise $\S 5.15 \ (1) \ \text {(i)}$