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A sequence \(\{a_n\}\) of real numbers is a function \(f:\mathbb N\to \mathbb R\) such that \(f(n)=a_n\). Sometimes it is denoted by \(\{a_n\}_{n=1}^{\infty}\) or by a few terms with a certain pattern \(\{a_1,a_2,a_3,\ldots\}\). A sequence can also be defined recursively.


  1. First four terms of the sequence \(\left\lbrace \frac{(-1)^n}{n}\right\rbrace\) are \(-1,\frac{1}{2},-\frac{1}{3},\frac{1}{4}\).

  2. \(\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4},\ldots\}\) can be written as \(\left\lbrace \frac{1}{n}\right\rbrace\).

  3. The Fibonacci sequence \(\{F_n\}_{n=0}^{\infty}\) is defined by \[F_n=F_{n-1}+F_{n-2},\; n\geq 2,\; F_0=0,\; F_1=1.\]

Definition. A sequence \(\{a_n\}\) converges to a real number \(L\) if \(x_n\) can be made arbitrarily close to \(L\) by taking sufficiently large \(n\) (for details, see the \(\varepsilon-N\) definition). It is denoted by \(\displaystyle\lim_{n\to \infty} a_n=L\) or \(a_n\to L\) as \(n\to \infty\). The sequence \(\{a_n\}\) diverges if it is not convergent.


  1. The sequence \(\left\lbrace \frac{1}{n}\right\rbrace\) converges to \(0\).

  2. \(\{ n^2 \}\) diverges to \(\infty\).

  3. \(\{(-1)^n\}\) diverges as its terms oscillate between \(1\) and \(-1\).

Note that if a sequence \(\{x_n\}\) diverges, its terms oscillate indefinitely like that of \(\{(-1)^n\}\) or it diverges to \(\pm \infty\).

Theorem. If \(a_n=f(n)\) for integers \(n\) and \(\displaystyle\lim_{x\to \infty} f(x)=L\), then \(\displaystyle\lim_{n\to \infty} a_n=L\).

Example. Show that \(\displaystyle\lim_{n\to \infty} \frac{n+\tan^{-1}n}{n}=1\).

Solution. Let \(f(x)=\displaystyle\frac{x+\tan^{-1}x}{x}\). Note that \(f(n)\) is the \(n\)th term of the sequence \(\left\lbrace \displaystyle\frac{n+\tan^{-1}n}{n} \right\rbrace\). \[ \displaystyle\lim_{x\to \infty} f(x)= \displaystyle\lim_{x\to \infty} \frac{x+\tan^{-1}x}{x} = \displaystyle\lim_{x\to \infty} \frac{1+\frac{1}{1+x^2}}{1} =1 \;\;\left(\text{by l'Hospitals's Rule} \right). \] Thus \(\displaystyle\lim_{n\to \infty} \frac{n+\tan^{-1}n}{n}=1\), i.e., the sequence \(\left\lbrace \displaystyle \frac{n+\tan^{-1}n}{n} \right\rbrace\) converges to \(1\).

Algebra of limits for sequences:
Theorem. The following are true for any convergent sequences \(\{a_n\}\) and \(\{b_n\}\):

  1. \(\displaystyle\lim_{n\to \infty}(a_n\pm b_n)=\left(\displaystyle\lim_{n\to \infty} a_n \right)\pm \left(\displaystyle\lim_{n\to \infty} b_n \right)\).

  2. \(\displaystyle\lim_{n\to \infty}(ca_n)=c\displaystyle\lim_{n\to \infty} a_n\) for all real numbers \(c\).

  3. \(\displaystyle\lim_{n\to \infty} a_nb_n=\left(\displaystyle\lim_{n\to \infty} a_n \right) \left(\displaystyle\lim_{n\to \infty} b_n \right)\).

  4. \(\displaystyle\lim_{n\to \infty} \frac{a_n}{b_n}=\frac{\displaystyle\lim_{n\to \infty} a_n}{\displaystyle\lim_{n\to \infty} b_n}\) whenever \(\displaystyle\lim_{n\to \infty} b_n\neq 0\).

  5. \(\displaystyle\lim_{n\to \infty} a_n^p=\left(\displaystyle\lim_{n\to \infty} a_n \right)^p\) for all real numbers \(p>0\) whenever \(a_n>0\).

Note that the convergence in the assumption is crucial. For \(\{a_n\}=\{\frac{1}{n}\}\) and \(\{b_n\}=\{n\}\), we have \(\{a_nb_n\}=\{1\}\). But \[\displaystyle\lim_{n\to \infty} a_nb_n=1\neq 0\cdot \infty\ =\left(\displaystyle\lim_{n\to \infty} a_n \right) \left(\displaystyle\lim_{n\to \infty} b_n\right).\]

Example. Determine if the sequence \(\{a_n\}\) converges or diverges.
(a) \(a_n=\displaystyle\frac{n^3}{n^3+1}\), (b) \(a_n=\displaystyle\frac{n^3}{n+1}\), (c) \(a_n=\displaystyle\sqrt{\frac{n+1}{9n+1}}\).

Solution. (a) \(\displaystyle\lim_{n\to \infty} a_n=\displaystyle\lim_{n\to \infty}\frac{n^3}{n^3+1} =\displaystyle\lim_{n\to \infty}\frac{1}{1+\frac{1}{n^3}} =\displaystyle\frac{1}{1+\displaystyle\lim_{n\to \infty}\frac{1}{n^3}} =\displaystyle\frac{1}{1+0} =1.\) Thus \(\left\lbrace \displaystyle\frac{n^3}{n^3+1} \right\rbrace\) converges to \(1\).

(b) \(\displaystyle\lim_{n\to \infty} a_n=\displaystyle\lim_{n\to \infty}\frac{n^3}{n+1} =\displaystyle\lim_{n\to \infty}\frac{n^2}{1+\frac{1}{n}} =\infty.\) Thus \(\left\lbrace \displaystyle\frac{n^3}{n^3+1} \right\rbrace\) diverges to \(\infty\).

(c) \(\displaystyle\lim_{n\to \infty} a_n =\displaystyle\lim_{n\to \infty}\sqrt{\frac{n+1}{9n+1}} =\displaystyle\sqrt{ \displaystyle\lim_{n\to \infty}\frac{n+1}{9n+1}} =\displaystyle\sqrt{ \displaystyle\lim_{n\to \infty}\frac{1+\frac{1}{n}}{9+\frac{1}{n}}} =\displaystyle\sqrt{ \frac{1+0}{9+0}} =\frac{1}{3}.\) Thus \(\left\lbrace \displaystyle\sqrt{\frac{n+1}{9n+1}} \right\rbrace\) converges to \(\frac{1}{3}\).

Theorem. If \(\{a_n\}\) is convergent and \(f\) is a continuous function, then \[\displaystyle\lim_{n\to \infty} f(a_n)=f\left( \displaystyle\lim_{n\to \infty} a_n \right).\]

Example. Since \(f(x)=\cos x\) is a continuous function, \[\displaystyle\lim_{n\to \infty} \cos\left( \frac{2}{n} \right)=\cos\left( \displaystyle\lim_{n\to \infty} \frac{2}{n} \right)=\cos 0=1.\] Thus \(\left\lbrace \cos\left( \frac{2}{n} \right) \right\rbrace\) converges to \(1\).

Theorem.(Sandwich Theorem) Suppose \(a_n\leq b_n\leq c_n\) for all natural numbers \(n\). If both \(\{a_n\}\) and \(\{c_n\}\) converge to \(L\), then \(\{b_n\}\) also converges to \(L\).

Example. Show that \(\displaystyle\lim_{n\to \infty} \frac{\sin(3n)}{n}=0\).

Solution. First note that \(-1\leq \sin(3n) \leq 1\) for all real numbers \(n\). Then for all natural numbers \(n\), \[-\frac{1}{n}\leq \frac{\sin(3n)}{n} \leq \frac{1}{n}.\] Since \(\displaystyle\lim_{n\to \infty} \frac{-1}{n}=\displaystyle\lim_{n\to \infty} \frac{1}{n}=0\), we have \(\displaystyle\lim_{n\to \infty} \frac{\sin(3n)}{n}=0\) by the Sandwich Theorem.

As consequences of the Sandwich Theorem, we have the following results:

  1. If \(\displaystyle\lim_{n\to \infty} |a_n|=0\), then \(\displaystyle\lim_{n\to \infty} a_n=0\).

  2. \(\left\lbrace r^n \right\rbrace\) is convergent if \(-1< r\leq 1\) and divergent otherwise. In particular, \[ \lim_{n\to \infty} r^n= \begin{cases} 0 & \text{if } -1< r <1 \\ 1 & \text{if } r=1. \end{cases} \]

(a) \(-|a_n|\leq a_n \leq |a_n|\) for all natural numbers \(n\). Now apply the Sandwich Theorem.

(b) If \(r=1\), then \(\left\lbrace r^n \right\rbrace=\{1\}\) is convergent. Suppose \(-1< r< 1\). Then \(0\leq |r|< 1\). Since \(\displaystyle\lim_{x\to \infty} |r|^x=0\), \(\displaystyle\lim_{n\to \infty} |r|^n=0\). Since \(|r|^n=|r^n|\), \(\displaystyle\lim_{n\to \infty} r^n=0\) by (a).
For \(r>1\), \(\displaystyle\lim_{x\to \infty} r^x=\infty\) and hence \(\displaystyle\lim_{n\to \infty} r^n=\infty\). For \(r=-1\), note that \(\{(-1)^n\}\) is divergent as its terms oscillate between \(1\) and \(-1\). Similar arguments can be applied for \(r<-1\).

Example. (a) Since \(\displaystyle\lim_{n\to \infty} \left| \frac{(-1)^n}{n} \right|=\displaystyle\lim_{n\to \infty} \frac{1}{n}=0\), \(\displaystyle\lim_{n\to \infty} \frac{(-1)^n}{n}=0\).

(b) \(\displaystyle\lim_{n\to \infty} \frac{2^n(-1)^n}{3^n}=\displaystyle\lim_{n\to \infty} \left( -\frac{2}{3} \right)^n=0\) as \(-1<-\frac{2}{3}\leq 1\). \(\displaystyle\lim_{n\to \infty} \left(\frac{3}{2} \right)^n\) does not exist as \(\frac{3}{2}>1\).

Definition. A sequence \(\{a_n\}\) is bounded if there exists a real number \(M\) such that \(|a_n|\leq M\) for all natural numbers \(n\).

Example. \(\{\frac{1}{n}\}\) is bounded since \(|\frac{1}{n}|\leq 1\) for all natural numbers \(n\).

Theorem. A convergent sequence is bounded.

Definition. A sequence \(\{a_n\}\) is increasing if \(a_{n+1} > a_n\) for all natural numbers \(n\) and decreasing if \(a_{n+1}< a_n\) for all natural numbers \(n\).

Theorem.(Monotone Convergence Theorem)

  1. An increasing sequence that is bounded above is convergent.

  2. A decreasing sequence that is bounded below is convergent.

Note that the Monotone Convergence Theorem is also true for non-decreasing and non-increasing sequences respectively.


  1. \(\{\frac{1}{n}\}\) is a decreasing sequence as \(a_{n+1}=\frac{1}{n+1}<\frac{1}{n}=a_n\) for all natural numbers \(n\). Also \(\{\frac{1}{n}\}\) is bounded below by \(0\) as \(0<\frac{1}{n}=a_n\) for all natural numbers \(n\). Since \(\{\frac{1}{n}\}\) is decreasing and bounded below, it is convergent by the Monotone Convergence Theorem.

  2. \(\{x_n\}\) where \(x_1=\sqrt{2}\) and \(x_n=\sqrt{2+x_{n-1}},\;n\geq 2\).
    We show \(\{x_n\}\) is increasing by induction. First note that \[x_1=\sqrt{2}< \sqrt{2+\sqrt{2}}=\sqrt{2+x_1}=x_2.\] Assume \(x_{n-1}< x_n\) for some \(n\in \mathbb N\). We show \(x_{n} < x_{n+1}\). Note that \[\begin{align*} & x_{n+1}^2-x_n^2=(2+x_n)-(2+x_{n-1})=x_n-x_{n-1}\\ \implies & x_{n+1}-x_n=(x_n-x_{n-1})/(x_{n+1}+x_n) \end{align*}\] Since \(x_{n-1} < x_n\) and \(x_n,x_{n+1} > 0\), we have \(x_{n+1}-x_n=(x_n-x_{n-1})/(x_{n+1}+x_n) > 0\), i.e., \(x_{n} < x_{n+1}\).
    We show \(\{x_n\}\) is bounded above also by induction. Note that \(x_1=\sqrt{2}<3\) and \(x_2=\sqrt{2+\sqrt{2}}<3\). Assume \( x_n < 3 \) for some \(n\in \mathbb N\). We show \( x_{n+1} < 3 \). Note that \[x_{n+1}=\sqrt{2+x_n}<\sqrt{2+3} < 3.\] By mathematical induction \(x_n < 3\) for all natural numbers \(n\).
    Since \(\{x_n\}\) is an increasing sequence bounded above by \(3\), it is convergent by the MCT. Suppose \(\displaystyle\lim_{n\to \infty} x_n=L\). To find \(L\), note that \(x_n^2=2+x_{n-1},\;n\geq 2.\) Taking limit of both sides we get \[\displaystyle\lim_{n\to \infty} x_n^2=\displaystyle\lim_{n\to \infty} (2+x_{n-1}) \implies (\displaystyle\lim_{n\to \infty} x_n)^2=2+ \displaystyle\lim_{n\to \infty} x_{n-1} \implies L^2=2+L \implies L=-1,2.\] Since \(x_n>0\) for all natural numbers \(n\), \(L\geq 0\). Thus \(L=2\), i.e., \(\displaystyle\lim_{n\to \infty} x_n=2\).

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