Question

Find the supremum and infimum of the set {(-1)^n(1+1/n):n is natural

Answer 1

Answer on Question #44204 – Math - Real Analysis

Find the supremum and infimum of the set $\left\{(-1)^{n}(1+1/n)\right\}$: $n$ is natural

Solution

Under even $n$ the terms of this sequence become $\left(1 + \frac{1}{n}\right), n=2, 4, 6, \ldots$

Under odd $n$ the terms of this sequence become $-\left(1 + \frac{1}{n}\right), n=1, 3, 5, \ldots$

We note that every element of the set is less than 2 since

$$(-1)^n \left(1 + \frac{1}{n}\right) \leq \left(1 + \frac{1}{n}\right) < 2.$$

We claim that the supremum (the least upper bound) is 2.

Assume that 2 is not the least upper bound. Then there is an $\varepsilon > 0$ such that $2 - \varepsilon$ is also an upper bound. However, we claim that there is a natural number $n$ such that

$$2 - \varepsilon < 1 + \frac{1}{n},$$

This inequality is equivalent with the following sequence of inequalities

$$1 - \varepsilon < \frac{1}{n}, \quad \frac{1}{n} > 1 - \varepsilon.$$

If $\varepsilon < 1$ then $0 < n < \frac{1}{1 - \varepsilon}$.

If $\varepsilon \geq 1$ then $n > 0$.

There is a natural number $n$ such that

$$2 - \varepsilon < 1 + \frac{1}{n},$$

So, $2 - \varepsilon$ is not an upper bound for the set. This verifies our answer.

We note that every element of the set is greater than or equal to -2:

$$(-1)^n \left(1 + \frac{1}{n}\right) \geq -\left(1 + \frac{1}{n}\right) \geq -2$$

If a set has a minimum, then the minimum will also be an infimum. Infimum of the set is -2.

Answer: 2 and -2.

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