Inequality Rule for Real Sequences

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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:

$\ds \lim_{n \mathop \to \infty} x_n = l$
$\ds \lim_{n \mathop \to \infty} y_n = m$

Let there exist $N \in \N$ such that:

$\forall n \ge N: x_n \le y_n$


$l \le m$

Proof 1

Suppose $l > m$.


$m = \dfrac m 2 + \dfrac m 2 < \dfrac {l + m} 2 < \dfrac l 2 + \dfrac l 2 = l$

Let $\epsilon = \dfrac {l - m} 2$.


$\epsilon > 0$

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.


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:

$l \le m$


Also known as

The Inequality Rule for Real Sequences is also presented on $\mathsf{Pr} \infty \mathsf{fWiki}$ as: