(related to Proposition: How Convergence Preserves the Order Relation of Sequence Members)

By assumption, \((a_n)_{n\in\mathbb N}\) and \((b_n)_{n\in\mathbb N}\) are convergent real sequences with the limits \(\lim_{n\rightarrow\infty} a_n=a\) and \(\lim_{n\rightarrow\infty} b_n=b\). We have to show the following:

\( ( 1 ) \) If \(a_n \le b_n\) for all \(n\in\mathbb N\), then \(a \le b\).

Using the definition of the order relation for real numbers, the relations "\(\le\)" and "\( > \)" are mutually exclusive. Thus, let assume that \(a_n \le b_n\) for all \(n\in\mathbb N\), but \(a > b\). Set \(\epsilon:=(a-b)/2 > 0\). Since there are indices \(N_a(\epsilon),N_b(\epsilon)\in\mathbb N\) such that

\[|a_n-a| < \epsilon~\forall n\ge N_a(\epsilon)\quad\quad\text{and}\quad\quad|b_n-b| < \epsilon~\forall n\ge N_b(\epsilon),\]

then, for all \(n\ge\max(N_a(\epsilon),N_b(\epsilon))\), it follows

\[a_n > a - \epsilon\quad\quad\text{and}\quad\quad b_n < b + \epsilon .\]

By construction, we have \(a - \epsilon=b +\epsilon\), so it follows \(a_n > b_n\) for all \(n\ge \max(N_a(\epsilon),N_b(\epsilon))\). But \(a_n \le b_n\) by assumption, which is a contradiction.

\( ( 2 ) \) From \(a_n < b_n\) for all \(n\in\mathbb N\) it does not (!) follow that \(a < b\).

It is sufficient to find only one counterexample, for which \(a_n < b_n\) for all \(n\in\mathbb N\), but \(a \not < b\). This counterexample can be given for the sequences \(a_n = 0\) and \(b_n=1/n\) for all \(n\in\mathbb N\), for which we have

\[\lim_{n\rightarrow\infty} a_n=\lim_{n\rightarrow\infty} 0= 0 = \lim_{n\rightarrow\infty} \frac 1n=\lim_{n\rightarrow\infty} b_n\]

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  1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983