# 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: $\lim_{{{n}\to\infty}}{{r}}^{{n}}={\left\{\begin{array}{c}{0}{\quad\text{if}\quad}-{1}<{r}<{1}\\{1}{\quad\text{if}\quad}{r}={1}\\ \end{array}\right.}$.