Evaluating whether or not a particular sequence is monotone.

Question

I am given the following sequence $$a_n=\frac{1}{n^2}+\frac{(-1)^n}{3^n}$$ and asked to decide whether or not it is monotonic or not. My strategy has been to consider $a_n-a_{n+1}$ and see if I can evaluate the expression to greater or less than zero.

I broke the problem into two cases based on whether $n$ was even or odd. The first case went smooth as follows.

Suppose $n$ is even, then $$a_n-a_{n+1}=\frac{1}{n^2}+\frac{(-1)^n}{3^n}-\frac{1}{(n+1)^2}+\frac{(-1)^{n+1}}{3^{n+1}}=\frac{1}{n^2}+\frac{1}{3^n}+\frac{1}{(n+1)^2}-\frac{1}{3^{n+1}}.$$

If I combine the $3^n$ fractions I get a $\frac{2}{3^n}$ so each term is positive and I know that $a_n-a_{n+1}>0$ whenever $n$ is even.

Now I was trying to work the some proces out for $n$ being odd. When I do I get to the expression $$\frac{1}{n^2}+\frac{1}{(n+1)^2}-\frac{2}{3^{n+1}}$$

I have been trying to compare the sizes of the positive and negative terms but I am finding it difficult to reach a conclusion.

Answer 1

I found the first several terms rounded to $$\{ .6666, .36111, .07407, 07484, .03588, .029149, .0199, .0157,.0122,.0100,...\}$$

Obviously it is not monotonic because the fourth term is larger than the the third one while others follow a decreasing pattern. It seems that after the fourth term it is decreasing. $a_n-a_{n+1}>0$ According to OP's statement for odd terms we have $$a_n-a_{n+1}= \frac{1}{n^2}+\frac{1}{(n+1)^2}-\frac{2}{3^{n+1}}$$

All we have to show is that $$\frac {1}{n^2}\ge \frac {2}{3^{n+1}}$$ or $$3^{n+1} \ge 2n^2$$ which is true for all $ n.$

Answer 2

For $n=2k+1$

$$a_n-a_{n+1}=a_{2k+1}-a_{2k+2}=\frac{1}{(2k+1)^2}-\frac{1}{3^{2k+1}}-\frac{1}{(2k+2)^2}-\frac{1}{3^{2k+2}}=\\=\frac{2k+3}{(2k+1)^2(2k+2)^2}-\frac{4}{3^{2k+2}}\stackrel{?}>0$$

which is true eventually since

$$\frac{\frac{2k+3}{(2k+1)^2(2k+2)^2}}{\frac{4}{3^{2k+2}}}\to \infty$$