Monotonic Sequence Definition
Sequence $$${x}_{{n}}$$$ is called increasing if $$${x}_{{1}}<{x}_{{2}}<\ldots<{x}_{{n}}<{x}_{{{n}+{1}}}<\ldots$$$, i.e. if $$${n}'>{n}$$$ then $$${x}_{{{n}'}}>{x}_{{n}}$$$.
For example, $$${\left\{{1},{2},{3},{6},{7},{9},\ldots\right\}}$$$ is increasing sequence, while $$${\left\{{3},{5},{8},{1},{5},{6},{7},\ldots\right\}}$$$ is not.
Sequence $$${x}_{{n}}$$$ is called non-decreasing if $$${x}_{{1}}\le{x}_{{2}}\le\ldots\le{x}_{{n}}\le{x}_{{{n}+{1}}}\le\ldots$$$, i.e. if $$${n}'>{n}$$$ then $$${x}_{{{n}'}}\ge{x}_{{n}}$$$.
For example, $$${\left\{{1},{2},{3},{3},{7},{7},{7},{9},\ldots\right\}}$$$ is non-decreasing sequence, while $$${\left\{{3},{5},{8},{1},{1},{1},{5},{5},{5},{6},{7},\ldots\right\}}$$$ is not.
Sequence $$${x}_{{n}}$$$ is called decreasing if $$${x}_{{1}}>{x}_{{2}}>\ldots>{x}_{{n}}>{x}_{{{n}+{1}}}>\ldots$$$, i.e. if $$${n}'>{n}$$$ then $$${x}_{{{n}'}}<{x}_{{n}}.$$$
For example, $$${\left\{{15},{12},{10},{8},{6},\ldots\right\}}$$$ is decreasing sequence, while $$${\left\{{8},{5},{3},{1},{5},{6},{7},\ldots\right\}}$$$ is not.
Sequence $$${x}_{{n}}$$$ is called non-increasing if $$${x}_{{1}}\ge{x}_{{2}}\ge\ldots\ge{x}_{{n}}\ge{x}_{{{n}+{1}}}\ge\ldots$$$, i.e. if $$${n}'>{n}$$$ then $$${x}_{{{n}'}}\le{x}_{{n}}$$$.
For example, $$${\left\{{9},{9},{9},{7},{5},{5},\ldots\right\}}$$$ is non-increasing sequence, while $$${\left\{{8},{8},{5},{5},{8},{1},{5},{6},{7},\ldots\right\}}$$$ 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: $$${x}_{{n}}\le{M}$$$, where $$${M}$$$ is constant and $$${n}={1},{2},{3},\ldots$$$, then it has finite limit, otherwise $$${x}_{{n}}\to+\infty$$$.
Fact 2. Suppose that we are given monotonically decreasing sequence $$${x}_{{n}}$$$. If it is bounded from below: $$${x}_{{n}}\ge{m}$$$, where $$${m}$$$ is constant and $$${n}={1},{2},{3},\ldots$$$, then it has finite limit, otherwise $$${x}_{{n}}\to-\infty$$$.
Note that above two facts are true even if sequence is monotonic starting from some number. For example, sequence $$${\left\{{3},{5},{6},{1},{2},{3},{4},{5},{6},{7},{8},{9},\ldots\right\}}$$$ becomes increasing from number 4: $$${1},{2},{3},{4},{5},{6},{7},{8},{9},\ldots$$$.
Example 1. Consider sequence $$${x}_{{n}}=\frac{{{{c}}^{{n}}}}{{{n}!}}$$$ where $$${c}>{0}$$$ and $$${n}!={1}\cdot{2}\cdot{3}\cdot\ldots\cdot{n}$$$.
When $$${c}>{1}$$$ sequence has indeterminate form of type $$$\frac{{\infty}}{{\infty}}$$$.
We have that $$${x}_{{{n}+{1}}}=\frac{{{{c}}^{{{n}+{1}}}}}{{{\left({n}+{1}\right)}!}}=\frac{{{c}\cdot{{c}}^{{n}}}}{{{\left({n}+{1}\right)}{n}!}}={x}_{{n}}\cdot\frac{{c}}{{{n}+{1}}}$$$.
This means that when $$$\frac{{c}}{{{n}+{1}}}<{1}$$$, 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 $$${x}_{{{n}+{1}}}={x}_{{n}}\cdot\frac{{c}}{{{n}+{1}}}$$$: $$$\lim{x}_{{{n}+{1}}}=\lim{x}_{{n}}\cdot\frac{{c}}{{{n}+{1}}}$$$ or $$${a}={a}\cdot{0}$$$. 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 $$$\lim{{x}_{{{n}+{1}}}^{{2}}}=\lim{\left({c}+{x}_{{n}}\right)}$$$ or $$${{a}}^{{2}}={c}+{a}$$$. This is quadratic equation with respect to $$${a}$$$, it has roots $$${a}_{{1}}=\frac{{{1}+\sqrt{{{4}{c}+{1}}}}}{{2}}$$$ and $$${a}_{{2}}=\frac{{{1}-\sqrt{{{4}{c}+{1}}}}}{{2}}$$$. 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, $$${a}=\frac{{{1}+\sqrt{{{4}{c}+{1}}}}}{{2}}$$$.