0
$\begingroup$

The question: Suppose there exists $N_0$ such that $s_n \leq t_n$ for all $n > N_0$. Prove that if $\lim_{n\to\infty} s_n$ and $\lim_{n\to\infty} t_n$ exists, then $\lim_{n\to\infty} s_n \leq \lim_{n\to\infty} t_n$.

My attempt at a proof:

Let $\lim_{n\to\infty} s_n = s$ and $\lim_{n\to\infty} t_n = t$. Assume, for purposes of contradiction, that $s > t$. Since $\lim_{n\to\infty} s_n$ exists, it must be true that there exists $N_1$ for all $\epsilon > 0$ such that $n > N_1$ implies $\mid s_n - s \mid < \epsilon$. Since, by premise, $s - t > 0$, we may choose $\epsilon = s - t$ and there must exist $N_1$ satisfying $\mid s_n - s \mid < \epsilon$, i.e. $\mid s_n - s \mid < s - t$, and thus by the absolute value properties, $s - (s-t) < s_n < s + (s-t)$, which directly implies $t < s_n$ for sufficiently large $n$. Now we examine the implications on $\lim_{n\to\infty} t_n$ exists, we may select any $\epsilon > 0$ to satisfy $\mid t_n - t \mid < s_n - t$. Since $s_n - t > 0$ for sufficiently large $n$, we may set $\epsilon = s_n - t$ for some $s_n$, however this would imply some $N$ exists such that $n > N$ implies $\mid t_n - t \mid < s_n - t$, i.e. $t_n < s_n - t + t$, that is, $t_n < s_n$ for all sufficiently large $n$, which is a contradiction. Thus $\lim_{n\to\infty} s_n \leq \lim_{n\to\infty} t_n$.

I'm certain I got lost in the details somewhere. Could someone point out errors that I've made?

$\endgroup$

1 Answer 1

1
$\begingroup$

Your idea of the proof is correct, though you made it a bit too complicated with the technical part. Here is a clear way to write it. Assume $s>t$. Then there exists $\epsilon>0$ such that $s-\epsilon>t+\epsilon$. Now you know there is $N_1\in\mathbb{N}$ such that $s_n>s-\epsilon$ for all $n\geq N_1$. Also there is $N_2\in\mathbb{N}$ such that $t_n<t+\epsilon$ for all $n\geq N_2$. Now let $N=max\{N_0,N_1,N_2\}$. For all $n\geq N$ you have:

$s_n>s-\epsilon>t+\epsilon>t_n$

Which is a contradiction.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .