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NoName3

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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?

$\displaystyle x_n = (-1)^n \left(\frac{1}{4}-\frac{2}{n} \right)$ 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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