5.3: Real Analysis - Convergent Sequences

Notation \(5.3.1\).

For \(x \in \mathbb{R}\), \(|x|\) denotes the absolute value of \(x\): \[|x|=\left\{\begin{aligned}
x & \text { if } x \geq 0 , \\
-x & \text { if } x<0 .
\end{aligned}\right.\]

You may assume the following basic properties of absolute value (without proof):

Lemma \(5.3.2\).

For \(x, y, z \in \mathbb{R}\), we have:

1. \(|x| \geq 0\) (and \(|x|=0 \Leftrightarrow x=0\)).
2. \(|x| = |−x|\).
3. \(|x+y| \leq |x|+|y|\). (“triangle inequality”)
4. \(|x y|=|x| \cdot|y|\).
5. \(−|x| \leq x \leq |x|\).
6. \(\exists N \in \mathbb{N}, N>|x|\).
7. If \(|x| < |y|\) and \(z \neq 0\), then \(|xz| < |yz|\).
8. If \(|x| > |y| \neq 0\), then \(1/|x| < 1/|y|\).

Definition \(5.3.3\).

Assume \(a_{1}, a_{2}, a_{3}, \ldots\) is an infinite sequence of real numbers, and \(L \in \mathbb{R}\). We say that the sequence converges to \(L\) (and write \(a_{n} \rightarrow L\)) iff \[\forall \epsilon>0, \exists N \in \mathbb{N}, \forall n>N,\left|a_{n}-L\right|<\epsilon .\]

Other Terminology.

When \(a_{n} \rightarrow L\), we can also say that the limit of the sequence is \(L\).

Example \(5.3.4\).

Let \(t \in \mathbb{R}\). If \(a_{n} = t\) for all \(n\), then \(a_{n} \rightarrow t\).

Solution

PROOF.

Given\(\epsilon>0\), let \(N = 0\). Given \(n > N\), we have \(\left|a_{n}-t\right|=|t-t|=|0|=0<\epsilon\).

Example \(5.3.5\).

If \(a_{n} = 1/n\) for all \(n\), then \(a_{n} \rightarrow 0\).

Scratchwork. To prove \(a_{n} \rightarrow 0\), we want: \[\left|a_{n}-0\right| \stackrel{?}{<} \epsilon \quad 1 / n \stackrel{?}{<} \epsilon \quad 1 / \epsilon \stackrel{?}{<} n\]
Since \(n > N\), it suffices to choose \(N > 1 / \epsilon\).

Solution

PROOF.

Given \(\epsilon > 0\), Lemma \(5.3.2(6)\) tells us there exists \(N \in \mathbb{N}\), such that \(N > 1 / \epsilon\). Given \(n > N\), we have \[\begin{aligned}
\left|a_{n}-0\right| &=1 / n & &\left(a_{n}=1 / n>0\right) \\
&<1 / N & &(n>N \text { and Lemma 5.3.2(8)) }\\
&<\epsilon & &(N>1 / \epsilon \text { and Lemma 5.3.2(8)) }
\end{aligned}\]

Exercise \(5.3.6\).

Show that if \(a_{n} = n/(n + 1)\) for all \(n\), then \(a_{n} \rightarrow 1\).

Proposition \(5.3.7\).

If \(a_{n} \rightarrow L\) and \(b_{n} \rightarrow M\), then \(a_{n} + b_{n} \rightarrow L + M\).

Scratchwork. To prove \(a_{n} + b_{n} \rightarrow L + M\), \[\text { we want to make }\left|\left(a_{n}+b_{n}\right)-(L+M)\right| \text { small (less than } \epsilon \text { ). }\]
What we know is that we can make \(\left|a_{n}-L\right|\) and \(\left|b_{n}-M\right|\) as small as we like. By the triangle inequality, we have \[\left|\left(a_{n}-L\right)+\left(b_{n}-M\right)\right|<\left|a_{n}-L\right|+\left|b_{n}-M\right|\]
By simple algebra, the left-hand side is equal to \(\left|\left(a_{n}+b_{n}\right)-(L+M)\right|\), so we just need to make the right-hand side less than \(\epsilon\). This will be true if \(\left|a_{n}-L\right|\) and \(\left|b_{n}-M\right|\) are both less than \(\epsilon / 2\).

Since \(a_{n} \rightarrow L\), there is some large \(N_{a}\), such that \(\left|a_{n}-L\right|<\epsilon / 2\) for all \(n > N_{a}\). Similarly, since \(b_{n} \rightarrow M\), there is some large \(N_{b}\), such that \(\left|b_{n}-M\right|<\epsilon / 2\) for all \(n > N_{b}\). Now, we just need know that \(n\) will be larger than both \(N_{a}\) and \(N_{b}\) whenever \(n > N\). So we should choose \(N\) to be whichever of \(N_{a}\) and \(N_{b}\) is larger. That is, we let \(N\) be the maximum of \(N_{a}\) and \(N_{b}\), which is denoted max(Na, Nb).

Proof

Given \(\epsilon > 0\), we know that \(\epsilon / 2 > 0\). Hence:

• Since \(a_{n} \rightarrow L\), we know \(\exists N_{a} \in \mathbb{N}, \forall n>N_{a},\left|a_{n}-L\right|<\epsilon / 2\).
• Since \(b_{n} \rightarrow M\), we know \(\exists N_{b} \in \mathbb{N}, \forall n>N_{b},\left|b_{n}-M\right|<\epsilon / 2\).

Let \(N=\max \left(N_{a}, N_{b}\right) \in \mathbb{N}\), so \(N \geq N_{a}\) and \(N \geq N_{b}\).

Given \(n > N\):

1. We have \(n > N \geq N_{a}\), so \(\left|a_{n}-L\right|<\epsilon / 2\).
2. We have \(n > N \geq N_{b}\), so \(\left|b_{n}-M\right|<\epsilon / 2\).

Therefore \[\begin{aligned}
\left|\left(a_{n}+b_{n}\right)-(L+M)\right| &=\left|\left(a_{n}-L\right)+\left(b_{n}-M\right)\right| & & \text { (high-school algebra) } \\
& \leq\left|a_{n}-L\right|+\left|b_{n}-M\right| & & \text { (triangle inequality) } \\
&<\epsilon / 2+\epsilon / 2 & &((*) \text { and }(* *)) \\
&=\epsilon . & &
\end{aligned}\]

Exercise \(5.3.8\).

Assume \(a_{n} \rightarrow L\), and \(c \in \mathbb{R}\). Do these proofs directly from the definition of convergence.

1. Show \(−a_{n} \rightarrow −L\).
2. Show \(a_{n} + c \rightarrow L + c\).
3. Show \(2a_{n} \rightarrow 2L\).
4. Show \(ca_{n} \rightarrow cL\) if \(c \neq 0\).
5. Show that if \(L > 0\), then \(\exists N \in \mathbb{N}\), such that \(a_{n} > 0\) for all \(n > N\).
6. (harder) Show that if \(L = 1\), then \(1/a_{n} \rightarrow 1\).

Exercise \(5.3.9\).

Assume \(a_{n} \rightarrow L\) and \(b_{n} \rightarrow M\).

1. Show that if \(M = 0\), and \(|a_{n}| \leq 2\) for all \(n\), then \(a_{n}b_{n} \rightarrow 0\).
2. (harder) Show \(a_{n}b_{n} \rightarrow LM\).