0
$\begingroup$

I am having some issues with understanding a particular step in a proof given in my book of the theorem that states: "We have that $(a_n)$ is convergent if and only if $$\liminf_{n\rightarrow\infty}a_n=\limsup_{n\rightarrow\infty}a_n$$ and in this case $\lim_{n\rightarrow\infty}a_n=\liminf_{n\rightarrow\infty}a_n=\limsup_{n\rightarrow\infty}a_n$."

The problem I have is where they prove that if $\liminf_{n\rightarrow\infty}a_n=\limsup_{n\rightarrow\infty}a_n=a$, then $(a_n)$ converges to $a$.

This is their proof: Let $\varepsilon>0$. There exists an $N\in\mathbb{N}$ such that $u_n:=\sup_{k\geqslant n}a_k\in(a-\varepsilon,a+\varepsilon)$ and $l_n:=\inf_{k\geqslant n}\in(a-\varepsilon,a+\varepsilon)$ for all $n\geqslant N$. This proves that for $n\geqslant N$ we have $a_n\leqslant\sup_{k\geqslant n}a_k<a+\varepsilon$ and $a_n\geqslant\inf_{k\geqslant n}a_k>a-\epsilon$.

This is not the end of the proof, but this is where they loose me, because I don't see why this per se proves that $\sup_{k\geqslant n}a_k\geqslant a_n$ or $\inf_{k\geqslant n}\leqslant a_n$ for $n\geqslant N$. Because it could still be possible that, for instance for a decreasing series, there are elements $a_n$ that are larger than the supremum of the tail $k\geqslant n$? And likewise for the infimum. I first thought that it might be because you know that $\sup_{k\geqslant n}a_k>a-\epsilon$ and for $n\geqslant N$ you could say that $a_n>a-\varepsilon$, but I think you're already using the assumption of convergence then and that still does not really show that $\sup_{k\geqslant n}a_k>a-\epsilon$.

Any help and insight on what I am missing here would be much appreciated!

$\endgroup$
  • $\begingroup$ $\sup_{k\geq n}a_k\geq a_n$ is obvious because by definition of supremum a supremum of a set is at least as large as all of its elements. Since $\sup_{k\geq n}a_k$ is supremum of the set $\{a_k:k\geq n\}$ which contains $a_n$, the mentioned inequality holds. $\endgroup$ – Wojowu Mar 5 '16 at 21:40
  • $\begingroup$ Ah yes of course, thank you! It escaped me that $a_n$ was was also necessarily a part of this set, but of course it is. Thank you for clarifying! $\endgroup$ – Soof_fie Mar 5 '16 at 21:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.