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In one paper I read that for any sequence $\lim\limits_{n\to \infty} \inf \dfrac{a_n}{n}\geqslant \inf\limits_{n} \dfrac{a_n}{n}$.

I know that $\lim\limits_{n\to \infty} \inf \dfrac{a_n}{n}:=\inf E$ where $E$ is the set of all subsequential limits. How to prove that using the definition?

Raheem Najib
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1 Answers1

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Since $$ d_n=\inf_{k\ge n} c_k $$ is an increasing sequence and $\liminf_n c_n=\lim_n d_n$ and $\inf_n c_n=d_0$ one always has $$ \inf_{n\in\Bbb N} c_n\le \liminf_{n\in\Bbb N}c_n. $$ The case $c_n=\frac{a_n}n$ is just one specialization of that.

Lutz Lehmann
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  • Why $\liminf_n c_n=\lim_n d_n$? – Raheem Najib Nov 08 '15 at 15:37
  • Because $\liminf_nc_n=\lim_n\inf_{k\ge n}c_k=\sup_n\inf_{k\ge n}c_k$. – Lutz Lehmann Nov 08 '15 at 15:39
  • I see that for the first time. Can I find it somewhere in details? When I read Rudin he defines it little bit different – Raheem Najib Nov 08 '15 at 15:40
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    It seems that way, see https://math.stackexchange.com/a/50449/115115. It is a nice exercise to show the equivalence. The $d_n$ provide a lower bound for every sub-sequence, and for every $ϵ$ there are indices $n_k$ with $c_{n_k}<d_k+ϵ$ etc. – Lutz Lehmann Nov 08 '15 at 15:57