How To Determine if a Sequence is Monotonic[Examples and Solutions]

Example of Monotone Sequences

In my last post, i explained  the monotone sequences and their criteria of convergency,now i will be a giving solutions to problems of monotone sequences and how to identify if they are decreasing,increasing or constant.

Examples
Evaluate the following sequences and check if they are monotone converging.
(1) $a_n=\{1-\frac{1}{n}\}$, for $n\geq{1}$
(2) $b_n=\{n^3\}$ for $n\geq{1}$
(3) $c_n=\{(-1)^n\}$ for $n\geq{1}$
(4) $d_n=\{\frac{1}{n^2}\}$ for $n\geq{1}$

Solution:

(1) To evaluate all of the given sequences above we use the monotone formula which is $nth=a_{n+1}-a_n$.
Now let $a_n=\{1-\frac{1}{n}\}$ and $a_{n+1}=\{1-\frac{1}{n+1}\}$.
\begin{align*}
&nth=[1-\frac{1}{n+1}]-[1-\frac{1}{n}],\\
&\text{take the L.C.M of both sides}\\
&=\frac{n+1-1}{n+1}-[\frac{n-1}{n}]\\
&=\frac{n}{n+1}-[\frac{n-1}{n}]\\
&\text{Take the L.C.M}\\
&=\frac{n^2-(n+1)(n-1)}{n(n+1)}\\
&=\frac{n^2-(n^2-1)}{n(n+1)}\\
&=\frac{1}{n(n+1)}\ \text{for all}\ n\geq{1}\\
&\text{choose n=1}\\
&\frac{1}{n(n+1)}=\frac{1}{2}\\
&n=2\\
&\frac{1}{n(n+1)}=\frac{1}{6}\\
&n=3\\
&\frac{1}{n(n+1)}=\frac{1}{12}\\
&\text{hence it is strictly monotone increasing,because}\\
&\frac{1}{n(n+1)}>0
\end{align*}

(2) $b_n=\{n^3\}$ for $n\geq{1}$
using $nth=b_{n+1}-b_n$, let $b_n=n^3$ and $b_{n+1}=(n+1)^2$
\begin{align*}
&={n^3+3n^2+3n+1}-n^3\\
&=3n^2+3n+3\ \text{for} n\geq{1}\\
&n=1\\
&=3(1)^2+3(1)+3=9\\
&n=2\\
&3n^2+3n+3=21\\
&n=3\\
&3n^2+3n+3=39\\
&\text{hence it is monotone strictly increasing because}\\
&3n^2+3n+3>0
\end{align*}

(3) $c_n=\{(-1)^n\}$ for all $n\geq{1}$
using $nth=c_{n+1}-c_n$
\begin{align*}
&=(-1)^{n+1}-(-1)^n\\
&=(-1)^n.(-1)-(-1)^n\\
&\text{for n=1}\\
&(-1)^n.(-1)-(-1)^n=2\\
&\text{n=2}\\
&(-1)^n.(-1)-(-1)^n=-2\\
&\text{n=3}\\
&(-1)^n.(-1)-(-1)^n=2\\
\end{align*}
from here we can see that the nth values of $(-1)^n.(-1)-(-1)^n$ are not unique in the sense that some are positive and others are negative.hence $c_n=\{(-1)^n\}$ is not a monotone sequence.

(4) $d_n=\{\frac{1}{n^2}\}$ for $n\geq{1}$
using $nth=d_{n+1}-d_n$
\begin{align*}
&\frac{1}{(n+1)^2}-\frac{1}{n^2}\\
&=\frac{n^2-(n+1)^2}{n^2(n+1)^2}\\
&=\frac{n^2-(n^2+2n+1)}{n^2(n+1)^2}\\
&=\frac{-2n-1}{n^2(n+1)^2}\\
&n=1\\
&\frac{-2n-1}{n^2(n+1)^2}=\frac{-3}{4}\\
&n=2\\
&\frac{-2n-1}{n^2(n+1)^2}=\frac{-5}{36}\\
&n=3\\
&\frac{-2n-1}{n^2(n+1)^2}=\frac{-7}{144}\\
\end{align*}
Hence we can conclude that this monotone is strictly decreasing because $\frac{-2n-1}{n^2(n+1)^2}<0$ for all $n\geq{1}$.

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