3
$\begingroup$

I've been working on this for a bit and haven't got far :/

So the full question is:

Suppose $(a_n)_{n=1}^{\infty}$ is a bounded sequence of real numbers. Prove that

$$\liminf (a_n) \leq \lim\sup (a_n)$$

(Hint: apply the result of the previous question to a difference of two sequences.)

So the previous question was basically: $(a_n)_{n=1}^{\infty}$ is a convergent sequence and $a_n \in [0, \infty) \forall n$. Prove the limit lies in $[0, \infty)$.

How I solved it was fixed $\epsilon > 0$ then $\exists$ $N \in \mathbb{N}$ $\forall$ $n\geq N$ I let the limit $=a$.

$\therefore$ $|a_n - a| < \epsilon$ Since $a_n$ took finitely many values $\implies$ $M_1 = \max\{|a_1|,...,|a_{N-1}|\}$, let $M_2 = |a| + \epsilon$ Since it converges it's bounded by some M $\implies$ $(a_n)^{\infty}_{n=1}$ within $[0, M]$ Since $[0, M]$ is closed, it contained its limit points $\implies$ $\lim_{n\to\infty} a_n \in [0,M] \subset [0, \infty)$.

So far I've said;

$(a_n)_{n=1}^{\infty} \in [\mathbb{R}]$

$M_1 = \inf\{a_n\}, M_2 = \sup\{a_n\}$ (From previous question)

Therefore, $\sup\{a_n\} = |a_n| + \epsilon$

$\inf{a_n} = \max\{a_1, a_2, ..., a_{N-1}\}$

$|a_n|<M_1 \implies |a_n| < \inf\{a_n\}$

$|a_N| < |a| + \epsilon \implies |a_N|< \inf\{a_n\}$

That's all I have not sure if it actually means anything but yeah as I said any help would GREATLY be appreciated. Thanks ;)

$\endgroup$
2
  • 5
    $\begingroup$ Did you choose your tags by rolling some dice? $\endgroup$
    – Asaf Karagila
    Mar 27, 2017 at 10:57
  • 3
    $\begingroup$ Please use \liminf and \limsup instead of $\lim_{n\to\infty} inf$ and $\lim_{n\to\infty} sup$, which are absurd. $\endgroup$
    – Did
    Mar 27, 2017 at 11:00

1 Answer 1

3
$\begingroup$

If you know that $x_n\leq y_n$ for all $n$ implies that $\lim_{n\to \infty }x_n\leq \lim_{n\to \infty }y_n$ :

Let $x_n=\inf_{k\geq n}x_k$ and $y_n=\sup_{k\geq n}x_k$. In particular, $x_n\leq y_n$ for all $n$, and thus $$\lim_{n\to \infty }x_n\leq \lim_{n\to \infty }y_n.$$ The claim follow.

If you don't know that $x_n\leq y_n$ for all $n$ implies that $\lim_{n\to \infty }x_n\leq \lim_{n\to \infty }y_n$, prove it !

Hint : It's enough to prove that $z_n\geq 0$ for all $n$ implies that $\lim_{n\to \infty }z_n\geq 0$ (why ?)

$\endgroup$
2
  • $\begingroup$ Hi what is your $x_n$ and $y_n$? $\endgroup$
    – Sarah V.P
    Mar 24, 2021 at 14:03
  • $\begingroup$ @Negrawh well it is defined in the answer. But the bold statement hold for every sequences. $\endgroup$
    – Surb
    Mar 24, 2021 at 19:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.