Inequality Rule for Real Sequences
Theorem
Let $\sequence {x_n}$ and $\sequence {y_n}$ be sequences in $\R$.
Let $\sequence {x_n}$ and $\sequence {y_n}$ be convergent to the following limits:
\(\ds \lim_{n \mathop \to \infty} x_n\) | \(=\) | \(\ds l\) | ||||||||||||
\(\ds \lim_{n \mathop \to \infty} y_n\) | \(=\) | \(\ds m\) |
Let there exist $N \in \N$ such that:
- $\forall n \ge N: x_n \le y_n$
Then:
- $l \le m$
Proof 1
Suppose $l > m$.
Then:
- $m = \dfrac m 2 + \dfrac m 2 < \dfrac {l + m} 2 < \dfrac l 2 + \dfrac l 2 = l$
Let $\epsilon = \dfrac {l - m} 2$.
Then:
- $\epsilon > 0$
We are given that:
- $\ds \lim_{n \mathop \to \infty} x_n = l$
By definition of the limit of a real sequence, we can find $N_1$ such that:
- $\forall n \ge N_1: \size {x_n - l} < \epsilon$
where $\size {x_n - l}$ denotes the absolute value of $x_n - l$
Suppose $n \ge N_1$
If $x_n \ge l$ then $x_n > \dfrac {l + m} 2$
If $x_n < l$ then:
\(\ds \epsilon\) | \(>\) | \(\ds \size {x_n - l}\) | as $n > N_1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds l - x_n\) | as $x_n < l$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {l - m} 2\) | \(>\) | \(\ds l - x_n\) | as $\epsilon = \dfrac {l - m} 2$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x_n\) | \(>\) | \(\ds \dfrac {l + m} 2\) | rearranging terms |
In either case $x_n > \dfrac {l+m} 2$
We are also given that:
- $\ds \lim_{n \mathop \to \infty} y_n = m$
Similarly we can find $N_2$ such that:
- $\forall n > N_2: \size {y_n - m} < \epsilon$
Suppose $n \ge N_2$
If $y_n \le m$ then $y_n < \dfrac {l + m} 2$
If $y_n > m$ then:
\(\ds \epsilon\) | \(>\) | \(\ds \size {y_n - l}\) | as $n > N_2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds y_n - m\) | as $y_n > m$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {l - m} 2\) | \(>\) | \(\ds y_n - m\) | as $\epsilon = \dfrac {l - m} 2$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {l + m} 2\) | \(>\) | \(\ds y_n\) | rearranging terms |
In either case $y_n < \dfrac {l + m} 2$
Let $N = \max \set {N_1, N_2}$.
Then if $n > N$, both the above inequalities will be true:
- $n > N_1$
- $n > N_2$
Thus $\forall n > N$:
- $y_n < \dfrac {l + m} 2 < x_n$
It has been shown that:
- $l > m \implies \forall n \in \N: \exists m \ge n: x_n > y_n$
Taking the contrapositive:
- $\exists N \in \N: \forall n \ge N: x_n \le y_n \implies l \le m$
Hence the result.
$\blacksquare$
Proof 2
Consider the sequence $\sequence {z_n}$ defined by:
- $z_n := y_n - x_n$
From Sum Rule for Real Sequences:
- $z_n \to m - l$ as $n \to \infty$
Furthermore, the assumption that $x_n \le y_n$ for all $n \in \N$ means that:
- $\forall n \in \N: z_n \ge 0$
Applying the Lower and Upper Bounds for Sequences to the sequence $\sequence {z_n}$ leads to the conclusion that $m - l \ge 0$.
That is:
- $l \le m$
$\blacksquare$
Also known as
The Inequality Rule for Real Sequences is also presented on $\mathsf{Pr} \infty \mathsf{fWiki}$ as: