# Proof that $a_n \leq b_n$ implies that $\limsup a_n \leq \limsup b_n$.

I am trying to prove that for sequences $$(a_n)$$ and $$(b_n)$$, that if $$a_n \leq b_n$$ for all $$n \geq m$$, then $$\limsup\limits_{n \to \infty} a_n \leq \limsup\limits_{n \to \infty} b_n$$. This is part three of Lemma 6.4.13. in Tao's analysis textboo, so I am a bit limited in terms of what I am allowed to use. Here is what I have so far.

By definition, we have \begin{align*} \limsup\limits_{n \to \infty} a_n = \inf(a_N^+)_{N=m}^{\infty} \limsup\limits_{n \to \infty} b_n = \inf(b_N^+)_{N=m}^{\infty}. \end{align*} But $$\limsup\limits_{n \to \infty} a_n = \inf(a_N^+)_{N=m}^{\infty}$$ and $$\limsup\limits_{n \to \infty} a_n = \inf(a_N^+)_{N=m}^{\infty}$$ are non-increasing sequences, so: \begin{align*} a_1^+ \geq a_m^+ \; \forall m > 1 \\ b_1^+ \geq b_m^+ \; \forall m > 1 \end{align*} We therefore have: \begin{align*} a_1^+ \geq \sup(a_N^+)_{N=1}^{\infty} \\ b_1^+ \geq \sup(b_N^+)_{N=1}^{\infty} \end{align*} This is the step that I am most unsure about, though I know there must be some way to tie together the two series, from which the result should follow directly.

Any helpful comments would be greatly appreciated.

## 1 Answer

There is no loss of generality taking $$m=1$$.

The sequences $$a_N^+$$ and $$b_N^+$$ are nonincreasing hence both convergent (possibly to $$\pm \infty$$).

Fix an index $$N$$. If $$n \ge N$$ then $$a_n \le b_n \le b_N^+$$ so that $$b_N^+$$ is an upper bound of the set $$\{a_n\}_{n \ge N}$$. That is, $$a_N^+ \le b_N^+$$. Now take the limit as $$N \to \infty$$ and apply the fact that limits preserve inequality.