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`.