# Monotonic Sequence Definition

Sequence x_n is called increasing if x_1<x_2<...<x_n<x_(n+1)<..., i.e. if n'>n then x_(n')>x_n.

For example, {1,2,3,6,7,9,...} is increasing sequence, while {3,5,8,1,5,6,7,...} is not.

Sequence x_n is called non-decreasing if x_1<=x_2<=...<=x_n<=x_(n+1)<=..., i.e. if n'>n then x_(n')>=x_n.

For example, {1,2,3,3,7,7,7,9,...} is non-decreasing sequence, while {3,5,8,1,1,1,5,5,5,6,7,...} is not.

Sequence x_n is called decreasing if x_1>x_2>...>x_n>x_(n+1)>..., i.e. if n'>n then x_(n')<x_n.

For example, {15,12,10,8,6,...} is decreasing sequence, while {8,5,3,1,5,6,7,...} is not.

Sequence x_n is called non-increasing if x_1>=x_2>=...>=x_n>=x_(n+1)>=..., i.e. if n'>n then x_(n')<=x_n.

For example, {9,9,9,7,5,5,...} is non-increasing sequence, while {8,8,5,5,8,1,5,6,7,...} is not.

Increasing, decreasing, non-increasing and non-decreasing sequences have common name - they are called monotonic sequences.

Fact 1. Suppose that we are given monotonically increasing sequence x_n. If it is bounded from above: x_n<=M, where M is constant and n=1,2,3,..., then it has finite limit, otherwise x_n->+oo.

Fact 2. Suppose that we are given monotonically decreasing sequence x_n. If it is bounded from below: x_n>=m, where m is constant and n=1,2,3,..., then it has finite limit, otherwise x_n->-oo.

Note that above two facts are true even if sequence is monotonic starting from some number. For example, sequence {3,5,6,1,2,3,4,5,6,7,8,9,...} becomes increasing from number 4: 1,2,3,4,5,6,7,8,9,....

Example 1. Consider sequence x_n=(c^n)/(n!) where c>0 and n! =1*2*3*...*n.

When c>1 sequence has indeterminate form of type (oo)/(oo).

We have that x_(n+1)=(c^(n+1))/((n+1)!)=(c*c^n)/((n+1)n!)=x_n *c/(n+1).

This means that when c/(n+1)<1, i.e. when n>c-1 sequence becomes decreasing. In the same time it is bounded from below by 0. Therefore, using Fact 2 we state that sequence x_n has finite limit which denote by a.

To find this limit we take limit of both sides of the equation x_(n+1)=x_n*c/(n+1): lim x_(n+1)=lim x_n*c/(n+1) or a=a*0. From this we have that a=0.

Example 2. Find limit of the sequence that is given recursively: x_1=sqrt(c), x_(n+1)=sqrt(c+x_n) where c>0.

Let's write out a couple of members: x_1=sqrt(c),x_2=sqrt(c+sqrt(c)),x_3=sqrt(c+sqrt(c+sqrt(c))). It is clear that x_(n+1)>x_n so sequence is increasing. At the same time it is bounded from above by the number sqrt(c)+1. Thus, according to Fact 1 sequence has finite limit a.

 To find a rewrite recurrent equation x_(n+1)=sqrt(c+x_n) as x_(n+1)^2=c+x_n. Taking limits of both sides gives lim x_(n+1)^2=lim (c+x_n) or a^2=c+a. This is quadratic equation with respect to a, it has roots a_1=(1+sqrt(4c+1))/2 and a_2=(1-sqrt(4c+1))/2. But in this case limit can't be negative, because first member of sequence is positive and sequence is increasing. Thus, second root can't be limit of sequence. Therefore, a=(1+sqrt(4c+1))/2.