# Sequences

In simple words sequence is a list of numbers written in definite order: a_1,a_2,...,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 {a_1,a_2,...,a_n} is also denoted by {a_n} or {a_n}_(n=1)^(oo).

If sequence is given by formula a_n=f(n) then we calculate a_n by calculating f(n). For example, if a_n=1/(n+1) then a_3=1/(3+1)=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. {n/(n^2+1)}_(n=1)^(oo), a_n=n/(n^2+1), {1/2,2/5,3/10,4/17,...,n/(n^2+1),...}
2. {(-1)^n2^n}, a_n=(-1)^n 2^n, {-2,4,-8,16,...,(-1)^n 2^n,...}
3. {1/(sqrt(n-2))}_(n=3)^(oo), a_n=1/sqrt(n-2),n>2, {1,1/sqrt(2),1/sqrt(3),1/2,...1/sqrt(n-2),...}
4. {sin((pin)/4)}_(n=0)^(oo), a_n=sin((pin)/4),n>=0, {0,1/sqrt(2),1,1/sqrt(2),0,...,sin((pin)/4),...}

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

Example 2.

1. Sequence {p_n} 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 {a_n}={1,4,1,5,9,2,6,5,...}.
3. The Fibonacci sequence {f_n} is defined recursively as f_1=1,f_2=1,f_n=f_(n-1)+f_(n-2), n>=3. In other words, n-th member is sum of two previous. So, {f_n}={1,1,2,3,5,8,13,21,34,55,...}.

Now, consider the sequence a_n=n/(n+1). Since n/(n+1)=1-1/(n+1) then 1-n/(n+1)=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->oo)n/(n+1)=1.

Definition. A sequence {a_n} has the limit L and we write lim_(n->oo)a_n=L or a_n->L as L->oo if we can make the terms a_n as close to L as we like by taking n sufficiently large. If lim_(n->oo)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->oo)f(x)=L and lim_(n->oo)a_n=L is that n is required to be an integer.
Thus, we have the following theorem.

Theorem 1. If lim_(x->oo)f(x)=L and f(n)=a_n when n is an integer, then lim_(n->oo)a_n=L.

In particular, since lim_(x->oo)1/x^r=0 when r>0, we have that lim_(n->oo)1/n^r=0 when r>0.

If a_n becomes large as n grows we use notation lim_(n->oo)a_n=oo. In this case sequence {a_n} is divergent, but in a special way. We say that {a_n} diverges to oo.

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

Limit Laws for Convergent Sequences

1. If {a_n} and {b_n} are convergent sequences and c is arbitrary constant then;
2. lim_(n->oo)(a_n+b_n)=lim_(n->oo)a_n+lim_(n->oo)b_n
3. lim_(n->oo)(a_n-b_n)=lim_(n->oo)a_n-lim_(n->oo)b_n
4. lim_(n->oo)c=c
5. lim_(n->oo)ca_n=clim_(n->oo)a_n
6. lim_(n->oo)(a_nb_n)=lim_(n->oo)a_n*lim_(n->oo)b_n
7. lim_(n->oo)(a_n)/(b_n)=(lim_(n->oo)a_n)/(lim_(n->oo)b_n) if lim_(n->oo)b_n!=0
8. lim_(n->oo)a_n^p=(lim_(n->oo)a_n)^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<=b_n<=c_n for n>=n_0 and lim_(n->oo)a_n=lim_(n->oo)c_n=L then lim_(n->oo)b_n=L.

Another Useful theorem is consequence of Squeeze Theorem:

Theorem 2. If lim_(n->oo)|a_n|=0 then lim_(n->oo)a_n=0.

Proof. Since -|a_n|<=a_n<=|a_n| and lim_(n->oo)-|a_n|=lim_(n->oo)|a_n|=0 then by Squeeze Theorem lim_(n->oo)a_n=0.

Example 3. Find lim_(n->oo)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->oo)1/(1-1/n)=(lim_(n->oo)1)/(lim_(n->oo)1-lim_(n->oo)1/n)=1/(1-0)=1.

Example 4. Calculate lim_(n->oo)(ln(n))/n.

Notice that both numerator and denominator approach infinity as n->oo. 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(x)=(ln(x))/x and obtain lim_(x->oo)(ln(x))/x=lim_(x->oo)((ln(x))')/(x')=lim_(x->oo)(1/x)/1=lim_(x->oo)1/x=0.

Therefore, by Theorem 1, lim_(n->oo)(ln(n))/n=0.

Example 5. Determine whether the sequence a_n=(-1)^n is convergent or divergent.

Let's write a couple of terms in sequence: {-1,1,-1,1,-1,1,...}.

Since the terms oscillate between 1 and -1 infinitely often, a_n does not approach any number. Thus, lim_(n->oo)(-1)^n does not exist; that is, the sequence is divergent.

Example 6. Evaluate lim_(n->oo)((-1)^n)/n if it is exists.

Since lim_(n->oo)|((-1)^n)/n|=lim_(n->oo)1/n=0 then, by Theorem 2, lim_(n->oo)((-1)^n)/n=0.

Example 7. Discuss convergence of sequence a_n=(n!)/n^n where n! =1*2*3*...*n.

Both numerator and denominator approach infinity as n->oo 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=(1*2*3*...*n)/(n*n*n*...*n).

Now, a_n=1/n((2*3*...*n)/(n*n*...*n))<=1/n((n*n*...*n)/(n*n*...*n))=1/n.

So, 0<a_n<=1/n and since lim_(n->oo)1/n=0 then by Squeeze Theorem lim_(n->oo)a_n=0.

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

We know that lim_(x->oo)a^x=oo if a>1 and lim_(x->oo)a^x=0 for 0<a<1. Therefore, putting a=r and using Theorem 1 we obtain that lim_(n->oo)r^n={(0 if 0<a<1),(oo if a>1):}.

If r=0 then lim_(n->oo)0^n=0.

If r=1 then lim_(n->oo)1^n=1.

If -1<r<0 then 0<|r|<1, so lim_(n->oo)|r^n|=lim_(n->oo)|r|^n=0, so ,by Theorem 2, lim_(n->oo)r^n=0.

If r<=-1 then sequence {r_n} diverges as in Example 5.

Fact. The sequence {r_n} is convergent if -1<r<=1 and divergent for all other values of r: lim_(n->oo)r^n={(0 if -1<r<1),(1 if r=1):}.