# Greatest lower bound

• MHB
NoName3
Find $\text{glb}(A)$ if $A = \left\{(-1)^n \left(\frac{1}{4}-\frac{2}{n} \right): n \in \mathbb{N}\right\}$.

$\displaystyle x_n = (-1)^n \left(\frac{1}{4}-\frac{2}{n} \right)$ then $\displaystyle x_{2k} = \frac{1}{4}-\dfrac{1}{k} = \frac{k-4}{4k}$ and $\displaystyle x_{2k+1} = -\left(\frac{1}{4}-\frac{2}{2k+1}\right) = \frac{7-2k}{4(2k+1)}$.

Now we analyse $x_{2k}$ Let $j = 2k$ then $x_{j} = \dfrac{j-8}{4j}$ and $x_{j+1} = \dfrac{j-7}{4(j+1)}$. Which is bigger: $x_j$ or $x_{j+1}$?

That's where I'm stuck - I can't analyse $x_{2k}, x_{2k+1}$. Is there a trick to it or we have to analyse cases?

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Gold Member
MHB
$\displaystyle x_{2k} = \frac{1}{4}-\dfrac{1}{k} = \frac{k-4}{4k}$.

we analyse $x_{2k}$ Let $j = 2k$ then $x_{j} = \dfrac{j-8}{8j}$

Your formula for $x_j$ is not correct.

We have the following:
$\displaystyle x_{2k} = \frac{k-4}{4k}=\frac{\frac{2k}{2}-4}{2(2k)} \ \overset{j=2k}{=} \ \frac{\frac{j}{2}-4}{2j}=\frac{j-8}{4j}$

Gold Member
MHB
Which is bigger: $x_j$ or $x_{j+1}$?

Suppose that $\frac{j-8}{4j}\leq \frac{j-7}{4(j+1)}$, then $$4(j+1)(j-8)\leq 4j(j-7) \\ \Rightarrow 4(j^2-7j-8)\leq 4j^2-28j \\ \Rightarrow 4j^2-28j-32\leq 4j^2-28j \\ \Rightarrow -32\leq 0$$ which is true.

Therefore, $x_j\leq x_{j+1}$.

NoName3
Thanks, mathmari. Since the equality is never attained, we have $x_j < x_{j+1}$ thus $x_j$ is strictly increasing from $-3/4$ to $1/4$. Letting $i= 2k+1$ we have $x_i = \dfrac{8-i}{4i}$ and $x_{i+1} = \dfrac{7-i}{4(i+1)}$. Therefore $x_{i+1} > x_i$. Hence $x_i$ is decreasing from $7/4$ to $-1/4$. That's the same thing as saying it's increasing from $-1/4$ to $7/4$. Therefore $\min(-3/4, -1/4) < x_n < \max(1/4, 7/4) \iff -3/4 < x_n < 7/4$ Therefore we have $\text{lub}(A) = -3/4.$ Is this correct analysis? I've seen a solution that considers the point $x_2 = 5/12$, which makes me doubt this.

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NoName3
It appears that I was having an off-day here. The $\epsilon-\delta$ proof is far simpler than that.

Let $x_n = \dfrac{(-1)^n(n-8)}{4n}.$ A lower bound for $x_n$ is $-\dfrac{3}{4}$. We want to show that $-\dfrac{3}{4}+\epsilon$
is not a lower bound for any $\epsilon > 0$. If $\epsilon > 0$, then $\exists ~ (2n+1)$ s.t. $\dfrac{2(2n+1)+8}{4(2n+1)}< \epsilon$.

Then $x_{2n+1} = \dfrac{7n-2n}{4(2n+1)} = \dfrac{-3(2n+1)+4n+10}{4(2n+1)} = -\dfrac{3}{4}+\dfrac{2(2n+1)+8}{4(2k+1)} < -\dfrac{3}{4}+\epsilon.$

Hence $-\dfrac{3}{4}+\epsilon$ is not a lower bound for $x_n$ for any $\epsilon > 0$, and therefore $\text{glb}(A) = -\dfrac{3}{4}.$

NoName3
Actually my proof above is not valid for $\epsilon \le \frac{1}{2}$.

Let $x_n = \dfrac{(-1)^n(n-8)}{4n}.$ Clearly a lower bound for $x_n$ is $-\dfrac{3}{4}$.
We need to show that $-\dfrac{3}{4}+\epsilon$ is not a lower bound for $\epsilon > 0$.
If $\epsilon > 0$, there exists an even integer $2n$ s.t. $\dfrac{2n-2}{2n}< \epsilon$, then:
$$x_{2n} = \dfrac{2n-8}{4(2n)} = \dfrac{8(n-1)-6n}{4(2n)} = -\dfrac{3}{4}+\dfrac{2n-2}{2n}< -\dfrac{3}{4}+\epsilon.$$ Thus $-\dfrac{3}{4}$ is the greatest lower bound of $A$. I believe it works now. (Happy)