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.


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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