\) converges to \(a\) if and only if for any \(\varepsilon>0\), there exists a real number \(N\) such that for any \(n \in \mathbb\) with \(n>N\), one has \[\left|a_-a\right|<\varepsilon. \nonumber\]
Example \(\PageIndex\)
Let \(a_=\frac\) for \(n \in \mathbb\). We claim that \(\lim _ a_=0\). 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 \[\left|a_-0\right|=\left|\frac\right|=\frac \leq \frac<\frac=\varepsilon. \nonumber\]
Example \(\PageIndex\)
We now generalize the previous example as follows. Let \(\alpha>0\) and consider the sequence given by \[a_=\frac> \text < for >n \in \mathbb. \nonumber\] Solution Let \(\varepsilon>0\). Choose an integer \(N\) such that \(N>\left(\frac\right)^\). For every \(n \geq N\), one has \(n>\left(\frac\right)^\) and, hence, \(n^>\frac\). This implies \[\left|\frac>-0\right|=\frac><\frac=\varepsilon.\] We conclude that \(\lim _ a_=0\).
Example \(\PageIndex\)
Consider the sequence \(\left\\) where \[a_=\frac+4>+n+5> \nonumber\] We will prove directly from the definition that this sequence converges to \(a= \frac\). Solution Let \(\varepsilon>0\). We first search for a suitable \(N\). To that end, we simplify and estimate the expression \(\left|a_-a\right|\). Notice that \[\begin
\left|a_-\frac\right| &=\left|\frac+4>+n+5>-\frac\right|=\left|\frac<2\left(3 n^+4\right)-3\left(2 n^+n+5\right)><2\left(2 n^+n+5\right)>\right|=\left|\frac<2\left(2 n^+n+5\right)>\right| \\
&=\frac<2\left(2 n^+n+5\right)><4 n^>=\frac
\end\] To guarantee that the last expression is less than \(\varepsilon\), it will suffice to choose \(N>\frac\). Indeed, if \(n \geq N\), we get \[\left|a_-a\right| \leq \frac \leq \frac<\frac<4 \frac>=\varepsilon.\]
Example \(\PageIndex\)
Let \(\left\\) be given by \[a_=\frac-1>-n>. \nonumber\] We claim \(\lim _ a_=\frac\). Solution Let \(\varepsilon>0\). We search for a suitable \(N\). First notice that \[\left|\frac-1>-n>-\frac\right|=\left|\frac-3-12 n^+4 n><3\left(3 n^-n\right)>\right|=\left|\frac<3\left(3 n^-n\right)>\right|\] Since \(n \geq 1\), we have \(n^ \geq n\) and \(4n>3\). Therefore we have \[\left|\frac-1>-n>-\frac\right|=\frac<3\left(3 n^-n\right)> \leq \frac<3\left(3 n^-n^\right)><6 n^>=\frac.\] Thus, if \(N>\frac\), we have, for \(n \geq N\) \[\left|\frac-1>-n>-\frac\right| \leq \frac \leq \frac<\varepsilon.\]
Example \(\PageIndex\)
Consider the sequence given by \[a_=\frac+5>+n>.\] We prove directly from the definition that \(\left\\right\>\) converges to \(\frac\). Solution Let \(\varepsilon>0\). Now, \[\left|\frac+5>+n>-\frac\right|=\left|\frac+20-4 n^-n><4\left(4 n^+n\right)>\right|=\frac<|20-n|><4\left(4 n^+n\right)>.\] If \(n \geq 20\), then \(|20-n|=n-20\). Therfore, for such \(n\) we have \[\left|\frac+5>+n>-\frac\right|=\frac<4\left(4 n^+n\right)> \leq \frac<16 n^>=\frac.\] Choose \(N>\max \left\<\frac, 20\right\>\). Then, for \(n \geq N\) we get \[\left|\frac+5>+n>-\frac\right| \leq \frac \leq \frac
The following result is quite useful in proving certain inequalities between numbers.
Lemma \(\PageIndex\)
Let \(\ell \geq 0\), If \(\ell0\), then \(\ell =0\). Proof This is easily proved by contraposition. If \(\ell >0\), then there is a positive number, for example \(\varepsilon=\ell / 2\), such that \(\varepsilon
Theorem \(\PageIndex\)
A convergent sequence \(\left\\) has at most one limit Proof Suppose \(\left\\) converges to \(a\) and \(b\). Then given \(\varepsilon>0\), there exist positive integers \(N_\) and \(N_\) such that \[\left|a_-a\right| n \geq N_\] and \[\left|a_-b\right| n \geq N_.\] Let \(N=\max \left\\). Then \[|a-b| \leq\left|a-a_\right|+\left|a_-b\right|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).
Lemma \(\PageIndex\)
Given real numbers \(a,b\), then \(a \leq b\) if and only if \(a0\). Proof Suppose \(a0\). And suppose, by way of contradiction, that \(a>b\). then set \(\varepsilon_=a-b\). Then \(\varepsilon_>0\). By assumption, we should have \(a
The following comparison theorem shows that (non-strict) inequalities are preserved "in the limit".
Theorem \(\PageIndex\) - Comparison Theorem.
Suppose \(\left\\) and \(\left\\right\>\) converge to \(a\) and \(b\), respectively, and \(a_ \leq b_\) for all \(n \in \mathbb\). Then \(a \leq b\). Proof For any \(\varepsilon>0\), there exist \(N_, N_ \in \mathbb\) such that \[a-\fracn \geq N_,\] \[b-\fracn \geq N_.\] Choose \(N>\max \left\\). Then \[a-\frac
Theorem \(\PageIndex\) - The Squeeze Theorem.
Definition \(\PageIndex\)
A sequence \(\left\\) is bounded above if the set \(\left\\) is bounded above. Similarly, the sequence \left\ is bounded below if the set \(\left\\) is bounded below. We say that the sequence \(\left\\) is bounded if the set \(\left\\) 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\\) is bounded if and only if there is \(M \in \mathbb\) such that \(\left|a_\right| \leq M\) for all \(n \in \mathbb\).
Theorem \(\PageIndex\)
A convergent sequence is bounded. Proof Suppose the sequence \(\left\\right\>\) converges to \(a\). Then, for \(\varepsilon=1\), there exists \(N \in \mathbb\) such that \[\left|a_-a\right| n \geq N.\] Since \(\left|a_\right|-|a| \leq \| a_|-| a|| \leq\left|a_-a\right|\), this implies \(\left|a_\right|<1+|a|\) for all \(n \geq N\). Set \[M=\max \left\<\left|a_<1>\right|, \ldots,\left|a_\right|,|a|+1\right\>\] Then \(\left|a_\right| \leq M\) for all \(n \in \mathbb\). Therefore, \(\left\\right\>\) is bounded. \(\square\)
Definition \(\PageIndex\)
Let \(\left\_^<\infty>\) be a sequence of real numbers. The sequence \(\left\\right\>_^<\infty>\) is called a subsequence of \(\left\_^<\infty>\) if there exists a sequence of increasing positive integers \[n_=a_>\) for each \(k \in \mathbb\).
Example \(\PageIndex\)
Consider the sequence \(a_=(-1)^\) for \(n \in \mathbb\). Solution Then \(\left\\right\>\) is a subsequence of \(\left\\right\>\) and \(a_=1\) for all \(k\) (here \(n_=2 k\) for all \(k\)). Similarly, \(\left\\right\>\) is also a subsequence of \(\left\\right\>\) and \(a_=-1\) for all \(k\) (here \(n_=2 k +1\) for all \(k\)).
Lemma \(\PageIndex\)
Let \(\left\_\) be a sequence of positive integers with \[n_ <\cdots\] Then \(n_\geq k\) for all \(k \in \mathbb\). Proof We use mathematical induction. When \(k=1\), it is clear that \(n_ \geq 1\) since \(n_\) is a positive integer. Assume \(n_ \geq k\) for some \(k\). Now \(n_>n_\) and, since \(n_\) and \(n_\) are integers, this implies, \(n_ \geq n_+1\). Therefore, \(n_ \geq k+1\) by the inductive hyposthesis. The conclusion now follows by the principle of mathematical induction. \(\square\)
Theorem \(\PageIndex\)
If a sequence \(\left\\) converges to \(a\), then any subsequence \(\left\>\right\>\) of \(\left\\) also converges to \(a\). Proof Suppose \(\left\\) converges to \(a\) and let \(\varepsilon>0\) be given. Then there exists \(N\) such that \[\left|a_-a\right| n \geq N.\] For any \(k \geq N\), since \(n_ \geq k\), we also have \[\left|a_-a\right|<\varepsilon.\] Thus, \(\left\>\right\>\) converges to \(a\) as \(k \rightarrow \infty\). \(\square\)
Example \(\PageIndex\)
Let \(a_=(-1)^\) for \(n \in \mathbb\). Solution Then the sequence \(\left\\right\>\) is divergent. Indeed suppose by contradiction that \(\lim _ a_=\ell\). Then every subsequence of \(\left\\right\>\) converges to a number \(\ell \in \mathbb\). From the previous theorem, it follows, in particular, that \(\ell=\lim _ a_=1 \text < and >\ell=\lim _ a_=-1\). This contradiction shows that the sequence is divergent.
Since the sequence \(\left\\) is bounded but not convergent, this example illustrates the fact that the converse of theorem 2.1.7 is not true.
Remark \(\PageIndex\)
Given a positive integer \(k_<0>\), it will be convenient to also talk about the sequence \(\left\\right\>_>\), 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 \(\left\\right\>\) whenever the integer \(k_<0>\) is clear form the context. For instance, we talk of the sequence \(\left\\right\>\) given by \[a_=\frac.\] although \(a_\) and \(a_\) are not defined. In all cases, the sequence must be defined from some integer onwards.
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Prove that if \(\left\\right\>\) is a convergent sequence, then \(\left\<|a_|\right\>\) is a convergent sequence. Is the converse true?
Answer
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Let \(\left\\right\>\) be a sequence. Prove that if the \(\left\<|a_|\right\>\) converges to 0, then \(\left\\right\>\) also converges to 0.
Answer
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Let \(\left\\right\>\) be a bounded sequence and let \(\left\\right\>\) be a sequence that converges to 0. Prove that the sequence \(\left\ y_\right\>\) converges to 0.
Answer
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Prove that the following limits are 0. (Hint: use Theorem 2.1.6.)
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Prove that if \(\lim _ a_=\ell>0\), then there exists \(N \in \mathbb\) such that \(a_>0\) for all \(n \geq N\).
Answer
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Prove that if \(\lim _ a_=\ell \neq 0\), then \(\lim _ \frac>=1\). Is the conclusion still true if \(\ell =0\)?
Answer
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Let \(\left\\right\>\) be a sequence of real numbers such that \(\lim _ a_=3\). Use Definition 2.1.1 to prove the following
- \(\lim _ 3 a_-7=2\);
- \(\lim _ \frac=\frac\); (Hint: prove first that there is \(N\) such that \(a_>1\) for \(n \geq N\).)
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Answer
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Prove that the sequence \(\left\\right\>\) with \(a_=\sin (n \pi / 2)\) is divergent.
Answer
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Consider a sequence \(\left\\right\>\).
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Given a sequence \(\left\\right\>\), define a new sequence \(\left\\right\>\) by
- Prove that if \(\lim _ a_=\ell\), then \(\lim _ b_=\ell\)
- Find a counterexample to show that the converse does not hold in general.
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