# 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, ${a}=\frac{{{1}+\sqrt{{{4}{c}+{1}}}}}{{2}}$.