2.2: Limit Theorems

We now prove several theorems that facilitate the computation of limits of some sequences in terms of those of other simpler sequences.

Theorem \(\PageIndex{1}\)

Let \(\left\{a_{n}\right\}\) and \(\left\{b_{n}\right\}\) be sequences of real numbesr and let \(k\) be a real number. Suppose \(\left\{a_{n}\right\}\) converges to \(a\) and \(\left\{b_{n}\right\}\) converges to \(b\). Then the sequences \(\left\{a_{n}+b_{n}\right\}\), \(\left\{k a_{n}\right\}\), and \(\left\{a_{n} b_{n}\right\}\) converge and

1. \(\lim _{n \rightarrow \infty}\left(a_{n}+b_{n}\right)=a+b\);
2. \(\lim _{n \rightarrow \infty}\left(k a_{n}\right)=k a\);
3. \(\lim _{n \rightarrow \infty}\left(a_{n} b_{n}\right)=a b\);
4. If in addition \(b \neq 0\) and \(b_{n} \neq 0\) for \(n \in \mathbb{N}\)

Proof

(a) Fix any \(\varepsilon>0\). Since \(\left\{a_{n}\right\}\) converges to \(a\), there exists \(N_{1} \in \mathbb{N}\) such that

\[\left|a_{n}-a\right|<\frac{\varepsilon}{2} \text { for all } n \geq N_{1}.\]

Similarly, there exists \(N_{2} \in \mathbb{N}\) such that

\[\left|b_{n}-b\right|<\frac{\varepsilon}{2} \text { for all } n \geq N_{2}.\]

Let \(N=\max \left\{N_{1}, N_{2}\right\}\). For any \(n \geq N\), one has

\[\left|\left(a_{n}+b_{n}\right)-(a+b)\right| \leq\left|a_{n}-a\right|+\left|b_{n}-b\right|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon.\]

Therefore, \(\lim _{n \rightarrow \infty}\left(a_{n}+b_{n}\right)=a+b\). This proves (a).

(b) If \(k=0\), then \(ka=0\) and \(k a_{n}=0\) for all \(n\). The conclusion follows immediately. Suppose next that \(k \neq 0\). Given \(\varepsilon>0\), let \(N \in \mathbb{N}\) be such that \(\left|a_{n}-a\right|<\frac{\varepsilon}{|k|}\) for \(n \geq N\). Then for \(n \geq N\), \(\left|k a_{n}-k a\right|=|k|\left|a_{n}-a\right|<\varepsilon\). It follows that \(\lim _{n \rightarrow \infty}\left(k a_{n}\right)=k a\) as desired. This proves (b).

(c) Since \(\left\{a_{n}\right\}\) is convergent, it follows from Theorem 2.1.7 that is bounded. Thus, there exists \(M>0\) such that

\[\left|a_{n}\right| \leq M \text { for all } n \in \mathbb{N}.\]

For every \(n \in \mathbb{N}\), we have the following estimate:

\[\left|a_{n} b_{n}-a b\right|=\left|a_{n} b_{n}-a_{n} b+a_{n} b-a b\right| \leq\left|a_{n}\right|\left|b_{n}-b\right|+|b|\left|a_{n}-a\right|.\]

Let \(\varepsilon>0\). Since \(\left\{a_{n}\right\}\) converges to \(a\), we may choose \(N_{1} \in \mathbb{N}\) such that

\[\left|a_{n}-a\right|<\frac{\varepsilon}{2(|b|+1)} \text { for all } n \geq N_{1}.\]

Similarly, since \(\left\{b_{n}\right\}\) converges to \(b\), we may choose \(N_{2} \in \mathbb{N}\) such that

\[\left|b_{n}-b\right|<\frac{\varepsilon}{2 M} \text { for all } n \geq N_{2}.\]

Let \(N=\max \left\{N_{1}, N_{2}\right\}\). Then, for \(n \geq N\), it follows from (2.1) that

\[\left|a_{n} b_{n}-a b\right|<M \frac{\varepsilon}{2 M}+|b| \frac{\varepsilon}{2(|b|+1)}<\varepsilon \text { for all } n \geq N.\]

Therefore, \(\lim _{n \rightarrow \infty} a_{n} b_{n}=a b\). This proves (c).

(d) Let us first show that

\[\lim _{n \rightarrow \infty} \frac{1}{b_{n}}=\frac{1}{b}.\]

Since \(\left\{b_{n}\right\}\) converges to \(b\), there is \(N_{1} \in \mathbb{N}\) that

\[\left|b_{n}-b\right|<\frac{|b|}{2} \text { for } n \geq N_{1}.\]

It follows (using a triangle inequality) that, for such \(n\), \(-\frac{|b|}{2}<\left|b_{n}\right|-|b|<\frac{|b|}{2}\) and, hence, \(\frac{|b|}{2}<\left|b_{n}\right|\). For each \(n \geq N_{1}\), we have the following estimate

\[\left|\frac{1}{b_{n}}-\frac{1}{b}\right|=\frac{\left|b_{n}-b\right|}{\left|b_{n}\right||b|} \leq \frac{2\left|b_{n}-b\right|}{b^{2}}.\]

Now let \(\varepsilon>0\). Since \(\lim _{n \rightarrow \infty} b_{n}=b\), there exists \(N_{2} \in \mathbb{N}\) such that

\[\left|b_{n}-b\right|<\frac{b^{2} \varepsilon}{2} \text { for all } n \geq N_{2}.\]

Let \(N=\max \left\{N_{1}, N_{2}\right\}\). By (2.2), one has

\[\left|\frac{1}{b_{n}}-\frac{1}{b}\right| \leq \frac{2\left|b_{n}-b\right|}{b^{2}}<\varepsilon \text { for all } n \geq N.\]

It follows that \(\lim _{n \rightarrow \infty} \frac{1}{b_{n}}=\frac{1}{b}\).

Finally, we can apply part (c) and have

\[\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}=\lim _{n \rightarrow \infty} a_{n} \frac{1}{b_{n}}=\frac{a}{b}.\]

The proof is now complete. \(\square\)

Example \(\PageIndex{1}\)

Consider the sequence \(\left\{a_{n}\right\}\) given by

\[a_{n}=\frac{3 n^{2}-2 n+5}{1-4 n+7 n^{2}}.\]

Solution

Dividing numerator and denominator by \(n^{2}\), we can write

\[a_{n}=\frac{3-2 / n+5 / n^{2}}{1 / n^{2}-4 / n+7}\]

Therefore, by the limit theorems above,

\[\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} \frac{3-2 / n+5 / n^{2}}{1 / n^{2}-4 / n+7}=\frac{\lim _{n \rightarrow \infty} 3-\lim _{n \rightarrow \infty} 2 / n+\lim _{n \rightarrow \infty} 5 / n^{2}}{\lim _{n \rightarrow \infty} 1 / n^{2}-\lim _{n \rightarrow \infty} 4 / n+\lim _{n \rightarrow \infty} 7}=\frac{3}{7}.\]

Example \(\PageIndex{2}\)

Let \(a_{n}=\sqrt[n]{b}\), where \(b>0\). Consider the case where \(b>1\).

Solution

In this case, \(a_{n}>1\) for every \(n\). By the binomial theorem,

\[b=a_{n}^{n}=\left(a_{n}-1+1\right)^{n} \geq 1+n\left(a_{n}-1\right).\]

This implies

\[0<a_{n}-1 \leq \frac{b-1}{n}.\]

For each \(\varepsilon>0\), choose \(N>\frac{b-1}{\varepsilon}\). It follows that for \(n \geq N\),

\[\left|a_{n}-1\right|=a_{n}-1<\frac{b-1}{n} \leq \frac{b-1}{N}<\varepsilon.\]

Thus, \(\lim _{n \rightarrow \infty} a_{n}=1\).

In the case where \(b=1\), it is obvious that \(a_{n}=1\) for all \(n\) and, hence, \(\lim _{n \rightarrow \infty} a_{n}=1\).

If \(0<b<1\), let \(c=\frac{1}{b}\) and define

\[x_{n}=\sqrt[n]{c}=\frac{1}{a_{n}}.\]

Since \(c>1\), it has been shown that \(\lim _{n \rightarrow \infty} x_{n}=1\). This implies

\[\lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty} \frac{1}{x_{n}}=1.\]

Exercise \(\PageIndex{1}\)

Find the following limits:

1. \(\lim _{n \rightarrow \infty} \frac{3 n^{2}-6 n+7}{4 n^{2}-3}\),
2. \(\lim _{n \rightarrow \infty} \frac{1+3 n-n^{3}}{3 n^{3}-2 n^{2}+1}\).

Answer

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Exercise \(\PageIndex{2}\)

Find the following limts:

1. \(\lim _{n \rightarrow \infty} \frac{\sqrt{3 n}+1}{\sqrt{n}+\sqrt{3}}\),
2. \(\lim _{n \rightarrow \infty} \sqrt[n]{\frac{2 n+1}{n}}\).

Answer

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Exercise \(\PageIndex{3}\)

Find the following limits if they exist:

1. \(\lim _{n \rightarrow \infty}\left(\sqrt{n^{2}+n}-n\right)\).
2. \(\lim _{n \rightarrow \infty}\left(\sqrt[3]{n^{3}+3 n^{2}}-n\right)\).
3. \(\lim _{n \rightarrow \infty}\left(\sqrt[3]{n^{3}+3 n^{2}}-\sqrt{n^{2}+n}\right)\)
4. \(\lim _{n \rightarrow \infty}(\sqrt{n+1}-\sqrt{n})\).
5. \(\lim _{n \rightarrow \infty}(\sqrt{n+1}-\sqrt{n}) / n\).

Answer

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Exercise \(\PageIndex{4}\)

Find the following limts.

1. For \(|r|<1\) and \(b \i \mathbb{R}\), \(\lim _{n \rightarrow \infty}\left(b+b r+b r^{2}+\cdots+b r^{n}\right)\).
2. \(\lim _{n \rightarrow \infty}\left(\frac{2}{10}+\frac{2}{10^{2}}+\cdots+\frac{2}{10^{n}}\right)\).

Answer

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Exercise \(\PageIndex{5}\)

Prove or disprove the following statements:

1. If \(\left\{a_{n}\right\}\) and \(\left\{b_{n}\right\}\) are convergent sequences, then \(\left\{a_{n}+b_{n}\right\}\) is a convergent sequence.
2. If \(\left\{a_{n}\right\}\) and \(\left\{b_{n}\right\}\) are divergent sequences, then \(\left\{a_{n}+b_{n}\right\}\) is a divergent sequence.
3. If \(\left\{a_{n}\right\}\) and \(\left\{b_{n}\right\}\) are convergent sequences, then \(\left\{a_{n} b_{n}\right\}\) is a convergent sequence.
4. If \(\left\{a_{n}\right\}\) and \(\left\{b_{n}\right\}\) are divergent sequences, then \(\left\{a_{n} b_{n}\right\}\) is a divergent sequence.
5. If \(\left\{a_{n}\right\}\) and \(\left\{a_{n}+b_{n}\right\}\) are convergent sequences, then \(\left\{b_{n}\right\}\) is a convergent sequence.
6. If \(\left\{a_{n}\right\}\) and \(\left\{a_{n}+b_{n}\right\}\) are divergent sequences, then \(\left\{b_{n}\right\}\) is a divergent sequence.

Answer

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