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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.

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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.

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