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Sequences

In simple words sequence is a list of numbers written in definite order: $$${a}_{{1}},{a}_{{2}},\ldots,{a}_{{n}}$$$. $$${a}_{{1}}$$$ is first term, $$${a}_{{2}}$$$ is second term, and, in general, $$${a}_{{n}}$$$ is n-th term. We deal with infinite sequences, so each term $$${a}_{{n}}$$$ will have a successor $$${a}_{{{n}+{1}}}$$$.

Notice that for every positive integer n there is a corresponding number $$${a}_{{n}}$$$ and so a sequence can be defined as a function whose domain is the set of positive integers. But we usually write $$${a}_{{n}}$$$ instead of the function notation f(n) for the value of the function at the number n.

The sequence $$${\left\{{a}_{{1}},{a}_{{2}},\ldots,{a}_{{n}}\right\}}$$$ is also denoted by $$${\left\{{a}_{{n}}\right\}}$$$ or $$${{\left\{{a}_{{n}}\right\}}_{{{n}={1}}}^{{\infty}}}$$$.

If sequence is given by formula $$${a}_{{n}}={f{{\left({n}\right)}}}$$$ then we calculate $$${a}_{{n}}$$$ by calculating $$${f{{\left({n}\right)}}}$$$. For example, if $$${a}_{{n}}=\frac{{1}}{{{n}+{1}}}$$$ then $$${a}_{{3}}=\frac{{1}}{{{3}+{1}}}=\frac{{1}}{{4}}$$$.

Example 1. Some sequences can be defined by giving a formula for the n-th term. In the following examples we give three descriptions of the sequence: one by using the preceding notation, another by using the defining formula, and a third by writing out the terms of the sequence. Notice that n doesn't have to start at 1.

  1. $$${{\left\{\frac{{n}}{{{{n}}^{{2}}+{1}}}\right\}}_{{{n}={1}}}^{{\infty}}}$$$, $$${a}_{{n}}=\frac{{n}}{{{{n}}^{{2}}+{1}}}$$$, $$${\left\{\frac{{1}}{{2}},\frac{{2}}{{5}},\frac{{3}}{{10}},\frac{{4}}{{17}},\ldots,\frac{{n}}{{{{n}}^{{2}}+{1}}},\ldots\right\}}$$$
  2. $$${\left\{{{\left(-{1}\right)}}^{{n}}{{2}}^{{n}}\right\}}$$$, $$${a}_{{n}}={{\left(-{1}\right)}}^{{n}}{{2}}^{{n}}$$$, $$${\left\{-{2},{4},-{8},{16},\ldots,{{\left(-{1}\right)}}^{{n}}{{2}}^{{n}},\ldots\right\}}$$$
  3. $$${{\left\{\frac{{1}}{{\sqrt{{{n}-{2}}}}}\right\}}_{{{n}={3}}}^{{\infty}}}$$$, $$${a}_{{n}}=\frac{{1}}{\sqrt{{{n}-{2}}}},{n}>{2}$$$, $$${\left\{{1},\frac{{1}}{\sqrt{{{2}}}},\frac{{1}}{\sqrt{{{3}}}},\frac{{1}}{{2}},\ldots\frac{{1}}{\sqrt{{{n}-{2}}}},\ldots\right\}}$$$
  4. $$${{\left\{{\sin{{\left(\frac{{\pi{n}}}{{4}}\right)}}}\right\}}_{{{n}={0}}}^{{\infty}}}$$$, $$${a}_{{n}}={\sin{{\left(\frac{{\pi{n}}}{{4}}\right)}}},{n}\ge{0}$$$, $$${\left\{{0},\frac{{1}}{\sqrt{{{2}}}},{1},\frac{{1}}{\sqrt{{{2}}}},{0},\ldots,{\sin{{\left(\frac{{\pi{n}}}{{4}}\right)}}},\ldots\right\}}$$$

Note, that there are sequences that don't have a simple defining equation.

Example 2.

  1. Sequence $$${\left\{{p}_{{n}}\right\}}$$$ where $$${p}_{{n}}$$$ is population of USA as of January 1 in the year n.
  2. If $$${a}_{{n}}$$$ is n-th decimal digit of number $$$\pi$$$ then $$${\left\{{a}_{{n}}\right\}}={\left\{{1},{4},{1},{5},{9},{2},{6},{5},\ldots\right\}}$$$.
  3. The Fibonacci sequence $$${\left\{{f}_{{n}}\right\}}$$$ is defined recursively as $$${f}_{{1}}={1},{f}_{{2}}={1}$$$,$$${f}_{{n}}={f}_{{{n}-{1}}}+{f}_{{{n}-{2}}}$$$, $$${n}\ge{3}$$$. In other words, n-th member is sum of two previous. So, $$${\left\{{f}_{{n}}\right\}}={\left\{{1},{1},{2},{3},{5},{8},{13},{21},{34},{55},\ldots\right\}}$$$.

Now, consider the sequence $$${a}_{{n}}=\frac{{n}}{{{n}+{1}}}$$$. Since $$$\frac{{n}}{{{n}+{1}}}={1}-\frac{{1}}{{{n}+{1}}}$$$ then $$${1}-\frac{{n}}{{{n}+{1}}}=\frac{{1}}{{{n}+{1}}}$$$. We see that this difference can be made as small as we like by taking n sufficiently large. We indicate it by writing $$$\lim_{{{n}\to\infty}}\frac{{n}}{{{n}+{1}}}={1}$$$.

Definition. A sequence $$${\left\{{a}_{{n}}\right\}}$$$ has the limit L and we write $$$\lim_{{{n}\to\infty}}{a}_{{n}}={L}$$$ or $$${a}_{{n}}\to{L}$$$ as $$${L}\to\infty$$$ if we can make the terms $$${a}_{{n}}$$$ as close to L as we like by taking n sufficiently large. If $$$\lim_{{{n}\to\infty}}{a}_{{n}}$$$ exists, we say the sequence converges (or is convergent). Otherwise, we say the sequence diverges (or is divergent).

Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity.

If you compare Definition with definition of a limit of a function at infinity you will see that the only difference between $$$\lim_{{{x}\to\infty}}{f{{\left({x}\right)}}}={L}$$$ and $$$\lim_{{{n}\to\infty}}{a}_{{n}}={L}$$$ is that n is required to be an integer.
Thus, we have the following theorem.

Theorem 1. If $$$\lim_{{{x}\to\infty}}{f{{\left({x}\right)}}}={L}$$$ and $$${f{{\left({n}\right)}}}={a}_{{n}}$$$ when n is an integer, then $$$\lim_{{{n}\to\infty}}{a}_{{n}}={L}$$$.

In particular, since $$$\lim_{{{x}\to\infty}}\frac{{1}}{{{x}}^{{r}}}={0}$$$ when $$${r}>{0}$$$, we have that $$$\lim_{{{n}\to\infty}}\frac{{1}}{{{n}}^{{r}}}={0}$$$ when $$${r}>{0}$$$.

If $$${a}_{{n}}$$$ becomes large as $$${n}$$$ grows we use notation $$$\lim_{{{n}\to\infty}}{a}_{{n}}=\infty$$$. In this case sequence $$${\left\{{a}_{{n}}\right\}}$$$ is divergent, but in a special way. We say that $$${\left\{{a}_{{n}}\right\}}$$$ diverges to $$$\infty$$$.

The Limits Laws also hold for the limits of sequences and their proof are similar.

Limit Laws for Convergent Sequences

  1. If $$${\left\{{a}_{{n}}\right\}}$$$ and $$${\left\{{b}_{{n}}\right\}}$$$ are convergent sequences and c is arbitrary constant then;
  2. $$$\lim_{{{n}\to\infty}}{\left({a}_{{n}}+{b}_{{n}}\right)}=\lim_{{{n}\to\infty}}{a}_{{n}}+\lim_{{{n}\to\infty}}{b}_{{n}}$$$
  3. $$$\lim_{{{n}\to\infty}}{\left({a}_{{n}}-{b}_{{n}}\right)}=\lim_{{{n}\to\infty}}{a}_{{n}}-\lim_{{{n}\to\infty}}{b}_{{n}}$$$
  4. $$$\lim_{{{n}\to\infty}}{c}={c}$$$
  5. $$$\lim_{{{n}\to\infty}}{c}{a}_{{n}}={c}\lim_{{{n}\to\infty}}{a}_{{n}}$$$
  6. $$$\lim_{{{n}\to\infty}}{\left({a}_{{n}}{b}_{{n}}\right)}=\lim_{{{n}\to\infty}}{a}_{{n}}\cdot\lim_{{{n}\to\infty}}{b}_{{n}}$$$
  7. $$$\lim_{{{n}\to\infty}}\frac{{{a}_{{n}}}}{{{b}_{{n}}}}=\frac{{\lim_{{{n}\to\infty}}{a}_{{n}}}}{{\lim_{{{n}\to\infty}}{b}_{{n}}}}$$$ if $$$\lim_{{{n}\to\infty}}{b}_{{n}}\ne{0}$$$
  8. $$$\lim_{{{n}\to\infty}}{{a}_{{n}}^{{p}}}={{\left(\lim_{{{n}\to\infty}}{a}_{{n}}\right)}}^{{p}}$$$ if $$${p}>{0}$$$ and $$${a}_{{n}}>{0}$$$

The Squeeze Theorem can also be adapted for sequences as follows:

Squeeze Theorem for Sequences. If $$${a}_{{n}}\le{b}_{{n}}\le{c}_{{n}}$$$ for $$${n}\ge{n}_{{0}}$$$ and $$$\lim_{{{n}\to\infty}}{a}_{{n}}=\lim_{{{n}\to\infty}}{c}_{{n}}={L}$$$ then $$$\lim_{{{n}\to\infty}}{b}_{{n}}={L}$$$.

Another Useful theorem is consequence of Squeeze Theorem:

Theorem 2. If $$$\lim_{{{n}\to\infty}}{\left|{a}_{{n}}\right|}={0}$$$ then $$$\lim_{{{n}\to\infty}}{a}_{{n}}={0}$$$.

Proof. Since $$$-{\left|{a}_{{n}}\right|}\le{a}_{{n}}\le{\left|{a}_{{n}}\right|}$$$ and $$$\lim_{{{n}\to\infty}}-{\left|{a}_{{n}}\right|}=\lim_{{{n}\to\infty}}{\left|{a}_{{n}}\right|}={0}$$$ then by Squeeze Theorem $$$\lim_{{{n}\to\infty}}{a}_{{n}}={0}$$$.

Example 3. Find $$$\lim_{{{n}\to\infty}}\frac{{n}}{{{n}-{1}}}$$$.

As with limits of functions we divide numerator and denominator by the highest power of n that occurs in the denominator and then use the Limit Laws:

$$$\lim_{{{n}\to\infty}}\frac{{1}}{{{1}-\frac{{1}}{{n}}}}=\frac{{\lim_{{{n}\to\infty}}{1}}}{{\lim_{{{n}\to\infty}}{1}-\lim_{{{n}\to\infty}}\frac{{1}}{{n}}}}=\frac{{1}}{{{1}-{0}}}={1}$$$.

Example 4. Calculate $$$\lim_{{{n}\to\infty}}\frac{{{\ln{{\left({n}\right)}}}}}{{n}}$$$.

Notice that both numerator and denominator approach infinity as $$${n}\to\infty$$$. We can't apply l'Hospital's Rule directly because it applies not to sequences but to functions of a real variable. However, we can apply l'Hospital's Rule to the related function $$${f{{\left({x}\right)}}}=\frac{{{\ln{{\left({x}\right)}}}}}{{x}}$$$ and obtain $$$\lim_{{{x}\to\infty}}\frac{{{\ln{{\left({x}\right)}}}}}{{x}}=\lim_{{{x}\to\infty}}\frac{{{\left({\ln{{\left({x}\right)}}}\right)}'}}{{{x}'}}=\lim_{{{x}\to\infty}}\frac{{\frac{{1}}{{x}}}}{{1}}=\lim_{{{x}\to\infty}}\frac{{1}}{{x}}={0}$$$.

Therefore, by Theorem 1, $$$\lim_{{{n}\to\infty}}\frac{{{\ln{{\left({n}\right)}}}}}{{n}}={0}$$$.

Example 5. Determine whether the sequence $$${a}_{{n}}={{\left(-{1}\right)}}^{{n}}$$$ is convergent or divergent.

Let's write a couple of terms in sequence: $$${\left\{-{1},{1},-{1},{1},-{1},{1},\ldots\right\}}$$$.

Since the terms oscillate between 1 and -1 infinitely often, $$${a}_{{n}}$$$ does not approach any number. Thus, $$$\lim_{{{n}\to\infty}}{{\left(-{1}\right)}}^{{n}}$$$ does not exist; that is, the sequence is divergent.

Example 6. Evaluate $$$\lim_{{{n}\to\infty}}\frac{{{{\left(-{1}\right)}}^{{n}}}}{{n}}$$$ if it is exists.

Since $$$\lim_{{{n}\to\infty}}{\left|\frac{{{{\left(-{1}\right)}}^{{n}}}}{{n}}\right|}=\lim_{{{n}\to\infty}}\frac{{1}}{{n}}={0}$$$ then, by Theorem 2, $$$\lim_{{{n}\to\infty}}\frac{{{{\left(-{1}\right)}}^{{n}}}}{{n}}={0}$$$.

Example 7. Discuss convergence of sequence $$${a}_{{n}}=\frac{{{n}!}}{{{n}}^{{n}}}$$$ where $$${n}!={1}\cdot{2}\cdot{3}\cdot\ldots\cdot{n}$$$.

Both numerator and denominator approach infinity as $$${n}\to\infty$$$ but here we have no corresponding function for use with l'Hospital's Rule (x! is not defined when x is not an integer). However, we can rewrite n-th term as $$${a}_{{n}}=\frac{{{1}\cdot{2}\cdot{3}\cdot\ldots\cdot{n}}}{{{n}\cdot{n}\cdot{n}\cdot\ldots\cdot{n}}}$$$.

Now, $$${a}_{{n}}=\frac{{1}}{{n}}{\left(\frac{{{2}\cdot{3}\cdot\ldots\cdot{n}}}{{{n}\cdot{n}\cdot\ldots\cdot{n}}}\right)}\le\frac{{1}}{{n}}{\left(\frac{{{n}\cdot{n}\cdot\ldots\cdot{n}}}{{{n}\cdot{n}\cdot\ldots\cdot{n}}}\right)}=\frac{{1}}{{n}}$$$.

So, $$${0}<{a}_{{n}}\le\frac{{1}}{{n}}$$$ and since $$$\lim_{{{n}\to\infty}}\frac{{1}}{{n}}={0}$$$ then by Squeeze Theorem $$$\lim_{{{n}\to\infty}}{a}_{{n}}={0}$$$.

Example 8. For what values of r is the sequence $$${\left\{{{r}}^{{n}}\right\}}$$$ convergent?

We know that $$$\lim_{{{x}\to\infty}}{{a}}^{{x}}=\infty$$$ if a>1 and $$$\lim_{{{x}\to\infty}}{{a}}^{{x}}={0}$$$ for 0<a<1. Therefore, putting a=r and using Theorem 1 we obtain that $$$\lim_{{{n}\to\infty}}{{r}}^{{n}}={\left\{\begin{array}{c}{0}{\quad\text{if}\quad}{0}<{a}<{1}\\\infty{\quad\text{if}\quad}{a}>{1}\\ \end{array}\right.}$$$.

If r=0 then $$$\lim_{{{n}\to\infty}}{{0}}^{{n}}={0}$$$.

If r=1 then $$$\lim_{{{n}\to\infty}}{{1}}^{{n}}={1}$$$.

If $$$-{1}<{r}<{0}$$$ then $$${0}<{\left|{r}\right|}<{1}$$$, so $$$\lim_{{{n}\to\infty}}{\left|{{r}}^{{n}}\right|}=\lim_{{{n}\to\infty}}{{\left|{r}\right|}}^{{n}}={0}$$$, so ,by Theorem 2, $$$\lim_{{{n}\to\infty}}{{r}}^{{n}}={0}$$$.

If $$${r}\le-{1}$$$ then sequence $$${\left\{{r}_{{n}}\right\}}$$$ diverges as in Example 5.

Fact. The sequence $$${\left\{{r}_{{n}}\right\}}$$$ is convergent if $$$-{1}<{r}\le{1}$$$ and divergent for all other values of r: $$$\lim_{{{n}\to\infty}}{{r}}^{{n}}={\left\{\begin{array}{c}{0}{\quad\text{if}\quad}-{1}<{r}<{1}\\{1}{\quad\text{if}\quad}{r}={1}\\ \end{array}\right.}$$$.