Supremum and infimum of a certain sequence

Question

What is the supremum and infimum of $$\{(-1)^n-\dfrac1n\}$$ And $n$ belongs to natural numbers .

Answer 1

The members of the sequence can be seen as follows: $(-1)^n - \dfrac{1}{n}= \Bigg\{\begin{matrix}\big(1-\dfrac{1}{n}\big),& \text{when n is odd}\\ \big(-1-\dfrac{1}{n}\big),&\text{when n is even} \end{matrix}$

When $n$ is odd, the value of $\big(1-\dfrac{1}{n}\big)$ increases as $n$ increases and thus the supremum (for odd $n$) is obtained as $n \to \infty$, and the limit is $lim_{n\to \infty}\big(1-\dfrac{1}{n}\big) = 1$. The infimum (for odd $n$) is obtained for $n = 1$, that is, $1 - 1 = 0$.

Similarly, when $n$ is even, the value of $\big(-1-\dfrac{1}{n}\big)$ increases as $n$ increases and supremum (for even $n$) is obtained as $n\to \infty$, with the limit being $lim_{n\to \infty}\big(-1-\dfrac{1}{n}\big) = -1$ and infimum (for even $n$) being $-2$ obtained for $n = 1$.

So, for the given sequence $\big\{(-1)^n - \dfrac{1}{n}\big\}$, the infimum is $-2$ and the supremum is $1$.

Answer 2

Hints. Look at the two subsequences that take every other number.

$$\{(-1)^{2n}-\frac{1}{2n}\} = \{1-\frac{1}{2n}\} $$ This subsequence has first element $1/2$ and is increasing towards a number (which?).

The other subsequence $$\{(-1)^{2n-1}-\frac{1}{2n-1}\} = \{-1-\frac{1}{2n-1}\}$$

This subsequence has first element $-2$ and is increasing towards a number (which?).