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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

1 Answers

One Answer

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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