Question

How to determine whether the sequences are increasing, decreasing or not monotonic?

Answer 1

Hint: We first try to describe the relation between the slope of the curve and the characteristics of it being increasing, decreasing or not monotonic. We find the differentiation of the curve by taking its ${{t}_{n}}$, the ${{n}^{th}}$ term of the sequence. Depending on the value of slope we get the characteristics of the function.

Complete step-by-step solution:
Let us assume that for a given sequence ${{a}_{n}},n=1\left( 1 \right)n$. We need to find if the sequence is increasing, decreasing or not monotonic.
We first try to find the general term of the sequence. We express the terms as ${{t}_{n}}$, the ${{n}^{th}}$ term of the series. We also take the $f\left( n \right)={{t}_{n}}$.
We take differentiation of the function and find the slope of the function.
So, $\dfrac{df}{dn}={{f}^{'}}\left( n \right)$ is the slope of the function.
Now, if the slope at any fixed point is negative which means $\dfrac{df}{dn}<0$ then the function is decreasing and if $\dfrac{df}{dn}>0$ then the function is increasing.
If the changes for the whole curve happens very rapidly then the function is not monotone.
Let's take as an example where $f\left( x \right)=\dfrac{1}{6x+3}$.
We find the slope of the function by taking $\dfrac{df}{dx}={{f}^{'}}\left( x \right)$.
So, $\dfrac{df}{dx}={{f}^{'}}\left( x \right)=\dfrac{-6}{{{\left( 6x+3 \right)}^{2}}}$.
Now for any value of $x$, the value of $\dfrac{df}{dx}={{f}^{'}}\left( x \right)=\dfrac{-6}{{{\left( 6x+3 \right)}^{2}}}<0$ as ${{\left( 6x+3 \right)}^{2}}\ge 0$.
The function is monotonically decreasing the whole function.

Note: We can also find the value of $x$ for which if we get ${{x}_{1}}>{{x}_{2}}$ and $f\left( {{x}_{1}} \right)>f\left( {{x}_{2}} \right)$, the curve is increasing. If we find ${{x}_{1}}<{{x}_{2}}$ and $f\left( {{x}_{1}} \right)>f\left( {{x}_{2}} \right)$, the curve is decreasing. The change of values is equal to the slope.