Why Definition of decreasing Sequence not Applicable here?

Question

I have read the definition of Decreasing Sequence $\left\{a_n\right\}$ as:

$$a_{n+1}\le a_n$$ $\forall$ $n \in \mathbb{N}$

Now consider the sequence:

$$a_n=\frac{(-1)^n}{n}$$

We have:

$$\frac{a_{n+1}}{a_n}=\frac{-n}{n+1} \lt 1$$

hence

$$a_{n+1}\lt a_n$$

But Actually its not a Monotone Sequence.

What is the correction in the definition?

Answer 1

Your claim

$$\frac{a_{n+1}}{a_n}=\frac{-n}{n+1} \lt 1$$

is false since

$$-\frac{n}{n+1}>1 ~~\text{if $n\in(-1,-\frac 12)$}$$

Furthermore, the definition of "monotonic decreasing" applied to a sequence $a_n$ means specifically that $a_{n+1} < a_n$ for every $n.$

This is not the same as $\frac{a_{n+1}}{a_n} < 1.$ In particular, if $a_n < 0$ and $a_{n+1} < a_n,$ then $\frac{a_{n+1}}{a_n} > 1.$ That is, when you divide by a negative number, you reverse the sign of the inequality.

Answer 2

Just because the inequality $\frac{a_{n+1}}{a_n} < 1$ is satisfied does not in and of itself imply that the inequality $a_{n+1} < a_n$. It does if the $a_n$'s; $n \in \mathbb{Z}^+$ are nonnegative.

For example, also consider the sequence $\{a_n\}; n \in \mathbb{Z}^+$; $a_n= (-2)^n$.