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Given the set $$B=\left\{\frac{1}{n}+(-1)^n, n \in \mathbb N\right\}$$ I have to find $\sup B$, $\inf B$, $\max B$, $\min B$.$$$$ For $n=even:$

$$B_{even}=\left\{\frac{1}{2k}+1, k=1,2,...\right\}$$

For $n=odd:$

$$B_{odd}=\left\{\frac{1}{2k+1}-1, k=0,1,2,...\right\}$$

So, $\max B= 1+ \frac{1}{2}=\frac{3}{2}$, $\sup B =\frac{3}{2}$, $\min B=-1$, $\nexists \inf B$. Is this correct?

AsukaMinato
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Mary Star
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1 Answers1

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$$B_{even}={\frac{1}{2k}-1, k=1,2,...} $$

$$B_{odd}=\{\frac{1}{2k+1}+1, k=0,1,2,...\}$$

So $\inf B= -1$, $\sup B = \max B = 2$, $\nexists \min B$.

Yamid Yela
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  • Oh..I accidentally wrote wrong the set..It's with a $+$ and not with a $-$. I will edit my post.Is there a way to show that $\inf B=-1$?? – Mary Star Feb 10 '14 at 21:27
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    Yes, first you can see that $\frac{1}{n}+1 > 0>-1$ for $n$ being odd and $-1 + \frac{1}{n} > -1$ for $n$ being even. Now $\frac{1}{2k}-1$ is a sequence that converges to $-1$, hence there is always an element of $B$ as close as you want to the rigth to -1. – Yamid Yela Feb 10 '14 at 21:47
  • Ok!!Thank you very much for your answer!! – Mary Star Feb 11 '14 at 01:20