# limits of sequences

I can't find any faults with the logic, but still I am not a 100% sure. Is the following correct:

Let $a_n$ and $b_l$ be two sequences such that, $\exists \; L(n)\; \forall\; n$ for which $a_n \leq b_l, \;\; \forall\; l\geq L(n)$. Thus, we have, \begin{align} a_n \leq \lim_{l\rightarrow \infty} b_l \;\;\forall n \\ \implies \lim_{n\rightarrow \infty} a_n \leq \lim_{l\rightarrow \infty} b_l \end{align}

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If the limits exist, then this is correct. – David Mitra Jan 17 '13 at 13:26
No. Counterexample: Let $a_n=1$ for all $n$ and $b_1=10$, $b_n=0$ for n>1 – Amr Jan 17 '13 at 13:26
@Amr Maybe I'm reading the question wrong, but it seems the given condition is that for every $n$ we have $a_n\le b_l$ for all sufficiently large $l$. – David Mitra Jan 17 '13 at 13:29
Your order of quantors according to natural language instead of logical structure is somewhat confusing/misleading. What you mean is $\forall n\exists L\forall l\ge L\colon a_n\le b_l$. – Hagen von Eitzen Jan 17 '13 at 14:25
@HagenvonEitzen. Yes. that's what I mean – UnadulteratedImagination Jan 17 '13 at 16:13

Let $(a_n),(b_n):\mathbb{N}\to \mathbb{R}$ so that $\forall n\in \mathbb{N}$ $$\exists L_n\in \mathbb{R}\text{ so that }\forall l\in \mathbb{N}\ l\ge L_n\implies a_n\le b_l$$ Then for fixed $n\in \mathbb{N}$ $\exists L_n\in \mathbb{R}$ so that $$\forall l\in \mathbb{N}\ l\ge L_n\implies a_n\le b_l$$ If $(b_m)$ converges to $b\in \mathbb{R}$ then indeed $$a_n\le b$$ As this holds $\forall n\in \mathbb{N}$, if $(a_m)$ converges to $a\in \mathbb{R}$ then $$a\le b$$
As a counterexample consider $a_n=(-1)^n$ and $b_n=(-1)^n+2013$ and $b'_n=\frac 1n+2013$