2.1: Convergence
- Page ID
- 49099
Let \(\left\{a_{n}\right\}\) be a sequence of real numbers. We say that the sequence \(\left\{a_{n}\right\}\) converges to \(a \in \mathbb{R}\) if, for any \(\varepsilon>0\), there exists a positive integer \(N\) such that for any \(n \in \mathbb{N}\) with \(n \geq N\), one has
\[\left|a_{n}-a\right|<\varepsilon\left(\text { or equivalently }, a-\varepsilon<a_{n}<a+\varepsilon\right).\]
In this case, we call \(a\) the limit of the sequence (see Theorem 2.1.3 below) and write \(\lim _{n \rightarrow \infty} a_{n}=a\). If the sequence \(\left\{a_{n}\right\}\) does not converge, we call the sequence divergent.
It follows directly from the definition, using the Archimedean property, that a sequence \(\left\{a_{n}\right\}\) converges to \(a\) if and only if for any \(\varepsilon>0\), there exists a real number \(N\) such that for any \(n \in \mathbb{N}\) with \(n>N\), one has
\[\left|a_{n}-a\right|<\varepsilon. \nonumber\]
Let \(a_{n}=\frac{1}{n}\) for \(n \in \mathbb{N}\). We claim that \(\lim _{n \rightarrow \infty} a_{n}=0\). We verify it using the definition.
Solution
Let \(\varepsilon>0\). Choose an integer \(N>1 / \varepsilon\). (Note that such an integer \(N\) exists due to the Archimidean Property.) Then, if \(n \geq N\), we get
\[\left|a_{n}-0\right|=\left|\frac{1}{n}\right|=\frac{1}{n} \leq \frac{1}{N}<\frac{1}{1 / \varepsilon}=\varepsilon. \nonumber\]
We now generalize the previous example as follows. Let \(\alpha>0\) and consider the sequence given by
\[a_{n}=\frac{1}{n^{\alpha}} \text { for } n \in \mathbb{N}. \nonumber\]
Solution
Let \(\varepsilon>0\). Choose an integer \(N\) such that \(N>\left(\frac{1}{\varepsilon}\right)^{1 / \alpha}\). For every \(n \geq N\), one has \(n>\left(\frac{1}{\varepsilon}\right)^{1 / \alpha}\) and, hence, \(n^{\alpha}>\frac{1}{\varepsilon}\). This implies
\[\left|\frac{1}{n^{\alpha}}-0\right|=\frac{1}{n^{\alpha}}<\frac{1}{1 / \varepsilon}=\varepsilon.\]
We conclude that \(\lim _{n \rightarrow \infty} a_{n}=0\).
Consider the sequence \(\left\{a_{n}\right\}\) where
\[a_{n}=\frac{3 n^{2}+4}{2 n^{2}+n+5} \nonumber\]
We will prove directly from the definition that this sequence converges to \(a= \frac{3}{2}\).
Solution
Let \(\varepsilon>0\). We first search for a suitable \(N\). To that end, we simplify and estimate the expression \(\left|a_{n}-a\right|\). Notice that
\[\begin{aligned}
\left|a_{n}-\frac{3}{2}\right| &=\left|\frac{3 n^{2}+4}{2 n^{2}+n+5}-\frac{3}{2}\right|=\left|\frac{2\left(3 n^{2}+4\right)-3\left(2 n^{2}+n+5\right)}{2\left(2 n^{2}+n+5\right)}\right|=\left|\frac{-7-3 n}{2\left(2 n^{2}+n+5\right)}\right| \\
&=\frac{3 n+7}{2\left(2 n^{2}+n+5\right)}<\frac{10 n}{4 n^{2}}=\frac{10}{4 n}
\end{aligned}\]
To guarantee that the last expression is less than \(\varepsilon\), it will suffice to choose \(N>\frac{10}{4 \varepsilon}\). Indeed, if \(n \geq N\), we get
\[\left|a_{n}-a\right| \leq \frac{10}{4 n} \leq \frac{10}{4 N}<\frac{10}{4 \frac{10}{4 \varepsilon}}=\varepsilon.\]
Let \(\left\{a_{n}\right\}\) be given by
\[a_{n}=\frac{4 n^{2}-1}{3 n^{2}-n}. \nonumber\]
We claim \(\lim _{n \rightarrow \infty} a_{n}=\frac{4}{3}\).
Solution
Let \(\varepsilon>0\). We search for a suitable \(N\). First notice that
\[\left|\frac{4 n^{2}-1}{3 n^{2}-n}-\frac{4}{3}\right|=\left|\frac{12 n^{2}-3-12 n^{2}+4 n}{3\left(3 n^{2}-n\right)}\right|=\left|\frac{4 n-3}{3\left(3 n^{2}-n\right)}\right|\]
Since \(n \geq 1\), we have \(n^{2} \geq n\) and \(4n>3\). Therefore we have
\[\left|\frac{4 n^{2}-1}{3 n^{2}-n}-\frac{4}{3}\right|=\frac{4 n-3}{3\left(3 n^{2}-n\right)} \leq \frac{4 n-3}{3\left(3 n^{2}-n^{2}\right)}<\frac{4 n}{6 n^{2}}=\frac{4}{6 n}.\]
Thus, if \(N>\frac{4}{6 \varepsilon}\), we have, for \(n \geq N\)
\[\left|\frac{4 n^{2}-1}{3 n^{2}-n}-\frac{4}{3}\right| \leq \frac{4}{6 n} \leq \frac{4}{6 N}<\varepsilon.\]
Consider the sequence given by
\[a_{n}=\frac{n^{2}+5}{4 n^{2}+n}.\]
We prove directly from the definition that \(\left\{a_{n}\right\}\) converges to \(\frac{1}{4}\).
Solution
Let \(\varepsilon>0\). Now,
\[\left|\frac{n^{2}+5}{4 n^{2}+n}-\frac{1}{4}\right|=\left|\frac{4 n^{2}+20-4 n^{2}-n}{4\left(4 n^{2}+n\right)}\right|=\frac{|20-n|}{4\left(4 n^{2}+n\right)}.\]
If \(n \geq 20\), then \(|20-n|=n-20\). Therfore, for such \(n\) we have
\[\left|\frac{n^{2}+5}{4 n^{2}+n}-\frac{1}{4}\right|=\frac{n-20}{4\left(4 n^{2}+n\right)} \leq \frac{n}{16 n^{2}}=\frac{1}{16 n}.\]
Choose \(N>\max \left\{\frac{1}{16 \varepsilon}, 20\right\}\). Then, for \(n \geq N\) we get
\[\left|\frac{n^{2}+5}{4 n^{2}+n}-\frac{1}{4}\right| \leq \frac{1}{16 n} \leq \frac{1}{16 N}<\varepsilon.\]
The following result is quite useful in proving certain inequalities between numbers.
Let \(\ell \geq 0\), If \(\ell<\varepsilon\) for all \(\varepsilon>0\), then \(\ell =0\).
- Proof
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This is easily proved by contraposition. If \(\ell >0\), then there is a positive number, for example \(\varepsilon=\ell / 2\), such that \(\varepsilon<\ell\). \(\square\)
A convergent sequence \(\left\{a_{n}\right\}\) has at most one limit
- Proof
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Suppose \(\left\{a_{n}\right\}\) converges to \(a\) and \(b\). Then given \(\varepsilon>0\), there exist positive integers \(N_{1}\) and \(N_{2}\) such that
\[\left|a_{n}-a\right|<\varepsilon / 2 \text { for all } n \geq N_{1}\]
and
\[\left|a_{n}-b\right|<\varepsilon / 2 \text { for all } n \geq N_{2}.\]
Let \(N=\max \left\{N_{1}, N_{2}\right\}\). Then
\[|a-b| \leq\left|a-a_{N}\right|+\left|a_{N}-b\right|<\varepsilon / 2+\varepsilon / 2=\varepsilon.\]
Since \(\varepsilon>0\) is arbitrary, by Lemma 2.1.2, \(|a-b|=0\) and, hence, \(a=b\). \(\square\)
The following lemma is simple generalization of (2.1.2).
Given real numbers \(a,b\), then \(a \leq b\) if and only if \(a<b+\varepsilon\) for all \(\varepsilon>0\).
- Proof
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Suppose \(a<b+\varepsilon\) for all \(\varepsilon>0\). And suppose, by way of contradiction, that \(a>b\). then set \(\varepsilon_{0}=a-b\). Then \(\varepsilon_{0}>0\). By assumption, we should have \(a<b+\varepsilon_{0}=b+a-b=a\), which is a contradiction. It follows that \(a \leq b\).
The other direction follows immediately from the order axioms. \(\square\)
The following comparison theorem shows that (non-strict) inequalities are preserved "in the limit".
Suppose \(\left\{a_{n}\right\}\) and \(\left\{b_{n}\right\}\) converge to \(a\) and \(b\), respectively, and \(a_{n} \leq b_{n}\) for all \(n \in \mathbb{N}\). Then \(a \leq b\).
- Proof
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For any \(\varepsilon>0\), there exist \(N_{1}, N_{2} \in \mathbb{N}\) such that
\[a-\frac{\varepsilon}{2}<a_{n}<a+\frac{\varepsilon}{2}, \quad \text { for } n \geq N_{1},\]
\[b-\frac{\varepsilon}{2}<b_{n}<b+\frac{\varepsilon}{2}, \quad \text { for } n \geq N_{2}.\]
Choose \(N>\max \left\{N_{1}, N_{2}\right\}\). Then
\[a-\frac{\varepsilon}{2}<a_{N} \leq b_{N}<b+\frac{\varepsilon}{2}.\]
Thus, \(a<b+\varepsilon\) for any \(\varepsilon>0\). Using Lemma 2.1.4 we conclude \(a \leq b\). \(\square\)
Suppose the sequences \(\left\{a_{n}\right\}\), \(\left\{b_{n}\right\}\), and \(\left\{c_{n}\right\}\) satisfy
\[a_{n} \leq b_{n} \leq c_{n} \text { for all } n \in \mathbb{N},\]
and \(\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} c_{n}=\ell\). Then \(\lim _{n \rightarrow \infty} b_{n}=\ell\).
- Proof
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Fix any \(\varepsilon>0\). Since \(\lim _{n \rightarrow \infty} a_{n}=\ell\), there exists \(N_{1} \in \mathbb{N}\) such that
\[\ell-\varepsilon<a_{n}<\ell+\varepsilon \nonumber\]
for all \(n \geq N_{1}\). Similarly, since \(\displaystyle \lim _{n \rightarrow \infty} c_{n}=\ell\), there exists \(N_{2} \in \mathbb{N}\) such that
\[\ell-\varepsilon<c_{n}<\ell+\varepsilon \nonumber\]
for all \(n \geq N_{2}\). Let \(N=\max \left\{N_{1}, N_{2}\right\}\). Then, for \(n \geq N\), we have
\[\ell-\varepsilon<a_{n} \leq b_{n} \leq c_{n}<\ell+\varepsilon, \nonumber\]
which implies \(\left|b_{n}-\ell\right|<\varepsilon\). Therefore, \(\lim _{n \rightarrow \infty} b_{n}=\ell\). \(\square\)
A sequence \(\left\{a_{n}\right\}\) is bounded above if the set \(\left\{a_{n}: n \in \mathbb{N}\right\}\) is bounded above. Similarly, the sequence \left\{a_{n}: n \in \mathbb{N}\right\} is bounded below if the set \(\left\{a_{n}: n \in \mathbb{N}\right\}\) is bounded below. We say that the sequence \(\left\{a_{n}\right\}\) is bounded if the set \(\left\{a_{n}: n \in \mathbb{N}\right\}\) is bounded, that is, if it is both bounded above and bounded below.
It follows from the observation after Definition 1.5.1 that the sequence \(\left\{a_{n}\right\}\) is bounded if and only if there is \(M \in \mathbb{R}\) such that \(\left|a_{n}\right| \leq M\) for all \(n \in \mathbb{N}\).
A convergent sequence is bounded.
- Proof
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Suppose the sequence \(\left\{a_{n}\right\}\) converges to \(a\). Then, for \(\varepsilon=1\), there exists \(N \in \mathbb{N}\) such that
\[\left|a_{n}-a\right|<1 \text { for all } n \geq N.\]
Since \(\left|a_{n}\right|-|a| \leq \| a_{n}|-| a|| \leq\left|a_{n}-a\right|\), this implies \(\left|a_{n}\right|<1+|a|\) for all \(n \geq N\). Set
\[M=\max \left\{\left|a_{1}\right|, \ldots,\left|a_{N-1}\right|,|a|+1\right\}\]
Then \(\left|a_{n}\right| \leq M\) for all \(n \in \mathbb{N}\). Therefore, \(\left\{a_{n}\right\}\) is bounded. \(\square\)
Let \(\left\{a_{n}\right\}_{n=1}^{\infty}\) be a sequence of real numbers. The sequence \(\left\{b_{n}\right\}_{n=1}^{\infty}\) is called a subsequence of \(\left\{a_{n}\right\}_{n=1}^{\infty}\) if there exists a sequence of increasing positive integers
\[n_{1}<n_{2}<n_{3}<\cdots,\]
such that \(b_{k}=a_{n_{k}}\) for each \(k \in \mathbb{N}\).
Consider the sequence \(a_{n}=(-1)^{n}\) for \(n \in \mathbb{N}\).
Solution
Then \(\left\{a_{2 k}\right\}\) is a subsequence of \(\left\{a_{n}\right\}\) and \(a_{2 k}=1\) for all \(k\) (here \(n_{k}=2 k\) for all \(k\)). Similarly, \(\left\{a_{2 k+1}\right\}\) is also a subsequence of \(\left\{a_{n}\right\}\) and \(a_{2 k+1}=-1\) for all \(k\) (here \(n_{k}=2 k +1\) for all \(k\)).
Let \(\left\{n_{k}\right\}_{k}\) be a sequence of positive integers with
\[n_{1}<n_{2}<n_{3}<\cdots\]
Then \(n_{k} \geq k\) for all \(k \in \mathbb{N}\).
- Proof
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We use mathematical induction. When \(k=1\), it is clear that \(n_{1} \geq 1\) since \(n_{1}\) is a positive integer. Assume \(n_{k} \geq k\) for some \(k\). Now \(n_{k+1}>n_{k}\) and, since \(n_{k}\) and \(n_{k+1}\) are integers, this implies, \(n_{k+1} \geq n_{k}+1\). Therefore, \(n_{k+1} \geq k+1\) by the inductive hyposthesis. The conclusion now follows by the principle of mathematical induction. \(\square\)
If a sequence \(\left\{a_{n}\right\}\) converges to \(a\), then any subsequence \(\left\{a_{n_{k}}\right\}\) of \(\left\{a_{n}\right\}\) also converges to \(a\).
- Proof
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Suppose \(\left\{a_{n}\right\}\) converges to \(a\) and let \(\varepsilon>0\) be given. Then there exists \(N\) such that
\[\left|a_{n}-a\right|<\varepsilon \text { for all } n \geq N.\]
For any \(k \geq N\), since \(n_{k} \geq k\), we also have
\[\left|a_{n_{k}}-a\right|<\varepsilon.\]
Thus, \(\left\{a_{n_{k}}\right\}\) converges to \(a\) as \(k \rightarrow \infty\). \(\square\)
Let \(a_{n}=(-1)^{n}\) for \(n \in \mathbb{N}\).
Solution
Then the sequence \(\left\{a_{n}\right\}\) is divergent. Indeed suppose by contradiction that
\(\lim _{n \rightarrow \infty} a_{n}=\ell\).
Then every subsequence of \(\left\{a_{n}\right\}\) converges to a number \(\ell \in \mathbb{R}\). From the previous theorem, it follows, in particular, that
\(\ell=\lim _{k \rightarrow \infty} a_{2 k}=1 \text { and } \ell=\lim _{k \rightarrow \infty} a_{2 k+1}=-1\).
This contradiction shows that the sequence is divergent.
Since the sequence \(\left\{a_{n}\right\}\) is bounded but not convergent, this example illustrates the fact that the converse of theorem 2.1.7 is not true.
Given a positive integer \(k_{0}\), it will be convenient to also talk about the sequence \(\left\{a_{n}\right\}_{n \geq k_{0}}\), that is, a function defined only for the integers greater than or equal to \(k_{0}\). For simplicity of notation, we may also denote this sequence by \(\left\{a_{n}\right\}\) whenever the integer \(k_{0}\) is clear form the context. For instance, we talk of the sequence \(\left\{a_{n}\right\}\) given by
\[a_{n}=\frac{n+1}{(n-1)(n-2)}.\]
although \(a_{1}\) and \(a_{2}\) are not defined. In all cases, the sequence must be defined from some integer onwards.
Exercise \(\PageIndex{1}\)
Prove the following directly from the definition of limit.
- \(\lim _{n \rightarrow \infty} \frac{2 n^{2}+2}{3 n^{3}+1}=0\).
- \(\lim _{n \rightarrow \infty} \frac{n^{2}+1}{5 n^{2}+n+1}=\frac{1}{5}\).
- \(\lim _{n \rightarrow \infty} \frac{2 n^{3}+1}{4 n^{3}-n}=\frac{1}{2}\).
- \(\lim _{n \rightarrow \infty} \frac{3 n^{2}+5}{6 n^{2}+n}=\frac{1}{2}\).
- \(\lim _{n \rightarrow \infty} \frac{4 n^{2}-1}{n^{2}-n}=4\).
- Answer
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Exercise \(\PageIndex{2}\)
Prove that if \(\left\{a_{n}\right\}\) is a convergent sequence, then \(\left\{|a_{n}|\right\}\) is a convergent sequence. Is the converse true?
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Exercise \(\PageIndex{3}\)
Let \(\left\{a_{n}\right\}\) be a sequence. Prove that if the \(\left\{|a_{n}|\right\}\) converges to 0, then \(\left\{a_{n}\right\}\) also converges to 0.
- Answer
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Exercise \(\PageIndex{4}\)
Prove that \(\lim _{n \rightarrow \infty} \frac{\sin n}{n}=0\).
- Answer
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Exercise \(\PageIndex{5}\)
Let \(\left\{x_{n}\right\}\) be a bounded sequence and let \(\left\{y_{n}\right\}\) be a sequence that converges to 0. Prove that the sequence \(\left\{x_{n} y_{n}\right\}\) converges to 0.
- Answer
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Exercise \(\PageIndex{6}\)
Prove that the following limits are 0. (Hint: use Theorem 2.1.6.)
- \(\lim _{n \rightarrow \infty} \frac{n+\cos \left(n^{2}-3\right)}{2 n^{2}+1}\).
- \(\lim _{n \rightarrow \infty} \frac{3^{n}}{n !}\).
- \(\lim _{n \rightarrow \infty} \frac{n !}{n^{n}}\).
- \(\lim _{n \rightarrow \infty} \frac{n^{2}}{3^{n}}\). (Hint: see Exercise 1.3.4(c)).
- Answer
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Exercise \(\PageIndex{7}\)
Prove that if \(\lim _{n \rightarrow \infty} a_{n}=\ell>0\), then there exists \(N \in \mathbb{N}\) such that \(a_{n}>0\) for all \(n \geq N\).
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Exercise \(\PageIndex{8}\)
Prove that if \(\lim _{n \rightarrow \infty} a_{n}=\ell \neq 0\), then \(\lim _{n \rightarrow \infty} \frac{a_{n+1}}{a_{n}}=1\). Is the conclusion still true if \(\ell =0\)?
- Answer
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Exercise \(\PageIndex{9}\)
Let \(\left\{a_{n}\right\}\) be a sequence of real numbers such that \(\lim _{n \rightarrow \infty} a_{n}=3\). Use Definition 2.1.1 to prove the following
- \(\lim _{n \rightarrow \infty} 3 a_{n}-7=2\);
- \(\lim _{n \rightarrow \infty} \frac{a_{n}+1}{a_{n}}=\frac{4}{3}\); (Hint: prove first that there is \(N\) such that \(a_{n}>1\) for \(n \geq N\).)
- Answer
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Exercise \(\PageIndex{10}\)
Let \(a_{n} \geq 0\) for all \(n \in \mathbb{N}\). Prove that if \(\lim _{n \rightarrow \infty} a_{n}=\ell\), then \(\lim _{n \rightarrow \infty} \sqrt{a_{n}}=\sqrt{\ell}\).
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Exercise \(\PageIndex{11}\)
Prove that the sequence \(\left\{a_{n}\right\}\) with \(a_{n}=\sin (n \pi / 2)\) is divergent.
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Exercise \(\PageIndex{12}\)
Consider a sequence \(\left\{a_{n}\right\}\).
- Prove that \(\lim _{n \rightarrow \infty} a_{n}=\ell\) if and only if \(\lim _{k \rightarrow \infty} a_{2 k}=\ell\) and \(\lim _{k \rightarrow \infty} a_{2 k+1}=\ell\).
- Prove that \(\lim _{n \rightarrow \infty} a_{n}=\ell\) if and only if \(\lim _{k \rightarrow \infty} a_{3 k}=\ell\), \(\lim _{k \rightarrow \infty} a_{3 k+1}=\ell\), and \(\lim _{k \rightarrow \infty} a_{3 k+2}=\ell\).
- Answer
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Exercise \(\PageIndex{13}\)
Given a sequence \(\left\{a_{n}\right\}\), define a new sequence \(\left\{b_{n}\right\}\) by
\[b_{n}=\frac{a_{1}+a_{2}+\ldots+a_{n}}{n}.\]
- Prove that if \(\lim _{n \rightarrow \infty} a_{n}=\ell\), then \(\lim _{n \rightarrow \infty} b_{n}=\ell\)
- Find a counterexample to show that the converse does not hold in general.
- Answer
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