18. Sequences

c2. Monotonic Sequences

A sequence \(a_n\) is increasing if \(a_n \le a_{n+1}\) for all \(n\).
A sequence \(a_n\) is strictly increasing if \(a_n\lt a_{n+1}\) for all \(n\).

A sequence \(a_n\) is decreasing if \(a_n \ge a_{n+1}\) for all \(n\).
A sequence \(a_n\) is strictly decreasing if \(a_n\gt a_{n+1}\) for all \(n\).

A sequence is monotonic if it is increasing or decreasing.
A sequence is strictly monotonic if it is strictly increasing or strictly decreasing.

Determine if each sequence is increasing, decreasing or neither and further if it is strictly increasing or decreasing.

1. \(a_n=\dfrac{1}{n^2}\)

   Strictly decreasing

   Since \(1 \le n\lt n+1\), we have \(n^2 < (n+1)^2\) and \(\dfrac{1}{n^2}>\dfrac{1}{(n+1)^2}\).
   So \(\left\{ \dfrac{1}{n^2}\right\}\) is strictly decreasing.

2. \(a_n=2-\dfrac{1}{n}\)

   Strictly increasing

   Since \(1 \le n\lt n+1\), we have \(\dfrac{1}{n}>\dfrac{1}{n+1}\) and \(-\dfrac{1}{n} \lt -\dfrac{1}{n+1}\) and \(2-\dfrac{1}{n} \lt 2-\dfrac{1}{n+1}\).
   So \(\left\{ 2-\dfrac{1}{n}\right\}\) is strictly increasing.

3. \(a_n=n+(-1)^n\)

   Neither increasing nor decreasing

   We write out the first \(6\) terms of the sequence \(a_n=n+(-1)^n\): \[\begin{aligned} a_1&=1+(-1)^1=0 \\ a_2&=2+(-1)^2=3 \\ a_3&=3+(-1)^3=2 \\ a_4&=4+(-1)^4=5 \\ a_5&=5+(-1)^5=4 \\ a_6&=6+(-1)^6=7 \end{aligned}\] So the sequence is neither increasing nor decreasing.

4. \(a_n=2n+(-1)^n\)

   Increasing but not strictly increasing

   We write out the first \(6\) terms of the sequence \(a_n=2n+(-1)^n\): \[\begin{aligned} a_1&=2+(-1)^1=1 \\ a_2&=4+(-1)^2=5 \\ a_3&=6+(-1)^3=5 \\ a_4&=8+(-1)^4=9 \\ a_5&=10+(-1)^5=9 \\ a_6&=12+(-1)^6=13 \end{aligned}\] So we suspect the sequence is increasing but not strictly increasing.
   
   To prove this rigorously, we separately check \(a_n \le a_{n+1}\) when \(n\) is even or odd.
   
   If \(n\) is even, then \(a_n=2n+(-1)^n=2n+1\) and \(a_{n+1}=2(n+1)+(-1)^{n+1}=2(n+1)-1=2n+1\)
   So \(a_n=a_{n+1}\).
   
   If \(n\) is odd, then \(a_n=2n+(-1)^n=2n-1\) and \(a_{n+1}=2(n+1)+(-1)^{n+1}=2(n+1)+1=2n+3\)
   So \(a_n\lt a_{n+1}\).

5. \(a_n=3n+(-1)^n\)

   Strictly increasing

   We write out the first \(6\) terms of the sequence \(a_n=3n+(-1)^n\): \[\begin{aligned} a_1&=3+(-1)^1=2 \\ a_2&=6+(-1)^2=7 \\ a_3&=9+(-1)^3=8 \\ a_4&=12+(-1)^4=13 \\ a_5&=15+(-1)^5=14 \\ a_6&=18+(-1)^6=19 \end{aligned}\] So we suspect the sequence is strictly increasing.
   
   To prove this rigorously, we separately check \(a_n\lt a_{n+1}\) when \(n\) is even or odd.
   
   If \(n\) is even, then \(a_n=3n+(-1)^n=3n+1\) and \(a_{n+1}=3(n+1)+(-1)^{n+1}=3(n+1)-1=3n+2\).
   So \(a_n\lt a_{n+1}\).
   
   If \(n\) is odd, then \(a_n=3n+(-1)^n=3n-1\) and \(a_{n+1}=3(n+1)+(-1)^{n+1}=3(n+1)+1=3n+4\). So \(a_n\lt a_{n+1}\).

If the terms, \(a_n\), are a continuous function of \(n\), then you can differentiate the terms and determine if the terms are increasing or decreasing.