# Show that $sup_{k geq 1 }inf_{n geq k} a_n = inf_{k geq 1} sup_{n geq k} a_n$ for an alternating sequence

The question I’m having trouble with is the following:

Let $$a_n$$ be defined for all $$n in mathbb{N}$$ by $$a_n = frac{1}{n^3}$$ if $$n$$ is even, and $$a_n = frac{-1}{n^2}$$ if $$n$$ is odd. Define $$A = {a_n | n in mathbb{N} }$$ and $$sup_{n geq k} a_n = sup {a_n | n geq k }$$.

I need to prove that
$$sup_{k geq 1 }inf_{n geq k} a_n = inf_{k geq 1} sup_{n geq k} a_n.$$

I’m incredibly confused as to how I’m supposed to take the supremum of an infimum or vice versa. I’d like to say that $$inf_{n geq k} a_n = a_k$$ if $$k$$ is odd, or $$a_{k+1}$$ if $$k$$ is even, with the opposite true for the supremum. However, I am unsure about how to proceed with finding $$sup_{k geq 1 }inf_{n geq k} a_n$$, much less proving the inequality.

Any help would be appreciated.

Mathematics Asked by FratSourced on November 12, 2021

It is well known limit inferior and limit superior for sequences $$limsuplimits_{n to infty}a_n = inflimits_{n} suplimits_{k geqslant n}a_k$$ $$liminflimits_{n to infty}a_n = suplimits_{n} inflimits_{k geqslant n}a_k$$ In another words they are $$sup$$ and $$inf$$ of limit points for given sequences. As your sequence is converged and $$limlimits_{n to infty}a_n=0$$, then it have only one limit point, which equals as to limit superior, so limit inferior.

Answered by zkutch on November 12, 2021

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