If $a_n < b_n$ for all $n$, then $\lim a_n \leq \lim b_n$.
I'm not sure how to prove this. In particular, why is there "$\leq$" in the limits?
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Martin Sleziak
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user221985
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1If $a_n < b_n ; \forall ; n$, then it follows that $b_n - a_n > 0 ; \forall ; n$ – Joseph Eck Oct 03 '18 at 06:42
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3For your second question, consider $a_n = -1/n$ and $b_n = 1/n$. Term-by-term, the inequality is strict, but the sequences have the same limit. – Blue Oct 03 '18 at 06:43
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Simply because $c_n=b_n-a_n$ can approach a limit without ever being equal to it - and $c_n$ can have properties like being positive and decreasing and still tend to a limit. – Mark Bennet Oct 03 '18 at 08:00
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1You need to assume that both $a_n$ and $b_n$ are convergent. – Michael Hoppe Oct 03 '18 at 10:45
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See also Suppose that $(s_n)$ converges to $s$, $(t_n)$ converges to $t$, and $s_n \leq t_n : \forall : n$. Prove that $s \leq t$. and other questions linked there. – Martin Sleziak Jul 20 '19 at 03:25
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The statement makes no sense unless we assume $a_n$ and $b_n$ converges or diverges properly. – Derek Pan Sep 25 '24 at 18:55
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If $a_n < b_n$ then $c_n=b_n-a_n>0$ for all $n$, therefore $$\lim c_n\geq0$$ otherwise let $\lim c_n=\ell<0$, there exists $N$ such that for $n\geq N$ $$|c_n-\ell|<\ell/2$$ and $c_N<\frac32\ell<0$ contradict with $c_n>0$ for all $n$.
Nosrati
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To give a proof of this I would personally go for a contradiction:
Assume that $\lim a_n > \lim b_n$, let $\epsilon = \lim a_n - \lim b_n$ ans consider a $N$ such that $|a_N-\lim a_n|<\epsilon/3$ and $|b_N-\lim b_n|<\epsilon/3$ (it exists because of the definition of the limit). Then find a contradiction with $a_N<b_N$.
As for the equality The example given in the comment by Blue is a good example.
wece
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If $a_n < b_n$ then the largest value of $a_n$ is smaller than the smallest value of $b_n$. But since these sequences are infinite, we express this in terms of $\sup$ and $\inf$, namely:
$$\sup \{a_n \} \leq \inf \{ b_n \}$$
and hence it follows that for the limit superior and limit inferior we have
$$\limsup \{a_n\} \leq \liminf \{b_n\}$$
Since $\lim \{x_n\}$ is defined as being the $x$ such that $\liminf \{x_n\} = \limsup \{x_n\} = x$ the result follows.
If this is confusing to you, I'd recommend reading the Wikipedia page.
To answer your second question why an equality is necessary, consider what happens if $b_n$ converges to $x$ from above, while $a_n$ converges to $x$ from below.
Xiaomi
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