Product of Limits of Real Sequences (1 + x over n)^n and (1 - x over n)^n equals 1
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Theorem
Let $\sequence {a_n}$ be the sequence defined as:
- $a_n = \paren {1 + \dfrac x n}^n$
Let $\sequence {b_n}$ be the sequence defined as:
- $b_n = \paren {1 - \dfrac x n}^n$
Then the product of the limits of $\sequence {a_n}$ and $\sequence {b_n}$ equals $1$
Proof
From Real Sequence $\paren {1 + \dfrac x n}^n$ is Convergent, $\sequence {a_n}$ is convergent.
Setting $x \to -x$, it follows that $\sequence {\paren {1 + \dfrac {\paren {-x} } n}^n} = \sequence {b_n}$ is also convergent.
Then:
\(\ds \paren {1 - \dfrac x n}^n \paren {1 - \dfrac x n}^n\) | \(=\) | \(\ds \paren {\paren {1 - \dfrac x n} \paren {1 - \dfrac x n} }^n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {1 - \dfrac {x^2} {n^2} }^n\) | Difference of Two Squares | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {\paren {1 - \dfrac {x^2} {n^2} }^{n^2} }^{1/n}\) |
From Real Sequence $\paren {1 + \dfrac x n}^n$ is Convergent, $\sequence {\paren {1 - \dfrac {x^2} {n^2} }^{n^2} }$ is convergent to a positive limit.
It follows that:
- $\exists a, b \in \R_{>0}: a < \paren {1 - \dfrac {x^2} {n^2} }^{n^2} < b$
for sufficiently large $n$.
Hence:
- $a^{1/n} < \paren {1 - \dfrac x n}^n \paren {1 - \dfrac x n}^n < b^{1/n}$
The result follows from Limit of Root of Positive Real Number and the Squeeze Theorem for Real Sequences.
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 5$: Subsequences: Exercise $\S 5.7 \ (6)$