Lower and Upper Bounds for Sequences

Theorem

Let $\left \langle {x_n} \right \rangle$ be a sequence in $\R$.

Let $x_n \to l$ as $n \to \infty$.

Then:

• $\forall n \in \N: x_n \ge a \implies l \ge a$;
• $\forall n \in \N: x_n \le b \implies l \le b$.

Corollary

Let $\left\langle{y_n}\right\rangle$ be another sequence in $\R$.

Let $y_n \to m$ as $n \to \infty$.

Suppose that for all $n \in \N$, $x_n \le y_n$.

Then $l \le m$, that is, $\displaystyle \lim_{n\to\infty} x_n \le \lim_{n\to\infty} y_n$.

This is often phrased as: limits preserve inequalities.

Proof

• $\forall n \in \N: x_n \ge a \implies l \ge a$:

Let $\epsilon > 0$.

Then $\exists N \in \N: n > N \implies \left|{x_n - l}\right| < \epsilon$.

So from Negative of Absolute Value: $l - \epsilon < x_n < l + \epsilon$.

But $x_n \ge a$, so $a \le x_n < l + \epsilon$.

Thus, for any $\epsilon > 0$, $a < l + \epsilon$.

From Real Plus Epsilon it follows that $a \le l$.

$\blacksquare$

• $\forall n \in \N: x_n \le b \implies l \le b$:

If $x_n \le b$ it follows that $-x_n \ge -b$ and the above result can be used.

$\blacksquare$

Warning

Let $\left \langle {x_n} \right \rangle$ be a sequence in $\R$.

Let $x_n \to l$ as $n \to \infty$.

Then it is not the case that:

• $\forall n \in \N: x_n > a \implies l > a$;
• $\forall n \in \N: x_n < b \implies l < b$.

Take the examples:

• $\displaystyle \left \langle {x_n} \right \rangle = \frac 1 n$
• $\displaystyle \left \langle {y_n} \right \rangle = -\frac 1 n$

Then :

$\displaystyle \forall n \in \N^*: \frac 1 n > 0, -\frac 1 n < 0$.

From Power of Reciprocal: Corollary, we have

• $x_n \to 0$
• $y_n \to 0$

as $n \to \infty$.

However, it is clearly false that $0 > 0$ and $0 < 0$.

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 4.23, \ \S 4.24$