# Why are these sufficient conditions for $\liminf$?

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

If

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

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

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

then

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

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 $$\lim$$.

• $\lim a_n = A$ implies $\liminf a_n = A$, thus the third condition is all you need. – AlohaSine Sep 27 '19 at 3:12
• @MathematicsStudent1122 That is a typo, it should be $c_k$ – DMH16 Sep 27 '19 at 3:13
• What does $a_{n_k}$ mean? – AlohaSine Sep 27 '19 at 3:15
• @MathematicsStudent1122 it is a subsequence of $a_n$ – 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 $$D=A$$.
• Perfect, thank you. The same conditions would be sufficient for $\limsup$ correct? Except, with the inequality signs reversed. – DMH16 Sep 27 '19 at 3:27