My professor today stated that to show that $\liminf a_n=A$ it is sufficient to show


$1.\,a_n\geq b_n,\,\,\forall n\in\mathbb{N}$

$2.\,a_{n_k}\leq c_k,\,\,\forall k\in\mathbb{N}$

$3.\lim c_k=\lim b_n=A,\,\,n,k\to\infty$


$\liminf a_n=A,\,\,n\to\infty$

What is the idea or intuition behind this? I somewhat see a squeeze theorem and maybe(?) the use of Cauchy sequences, but it is not clear to me how we can go from needing to evaluate $\liminf$ to simply evaluating $\lim$.

  • $\begingroup$ $\lim a_n = A$ implies $\liminf a_n = A$, thus the third condition is all you need. $\endgroup$ – AlohaSine Sep 27 '19 at 3:12
  • $\begingroup$ @MathematicsStudent1122 That is a typo, it should be $c_k$ $\endgroup$ – DMH16 Sep 27 '19 at 3:13
  • $\begingroup$ What does $a_{n_k}$ mean? $\endgroup$ – AlohaSine Sep 27 '19 at 3:15
  • $\begingroup$ @MathematicsStudent1122 it is a subsequence of $a_n$ $\endgroup$ – DMH16 Sep 27 '19 at 3:16

Let $D = \liminf a_n$. We have that $D \leq \lim \inf a_{n_k}$, since the $\lim \inf$ of the full sequence is at most the $\liminf$ of any subsequence. Since $\lim \inf a_{n_k} \leq \lim \inf c_k = \lim c_k = A$, we get that $D \leq A$.

On the other hand, since $a_n \geq b_n$ for each $n$, $D \geq \lim \inf b_n = \lim b_n = A$.

Thus $D=A$.

  • $\begingroup$ Perfect, thank you. The same conditions would be sufficient for $\limsup$ correct? Except, with the inequality signs reversed. $\endgroup$ – DMH16 Sep 27 '19 at 3:27

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