2
$\begingroup$

Let $(s_n)$ and $(t_n)$ be sequences defined on $\mathbb{R}$.

Prove that lim sup $s_n$ + lim sup $t_n \geq$ lim sup $(s_n+t_n)$

Proof (can someone please verify it?): Set $\alpha = $ lim sup $s_n$ and $\beta = $ lim sup $t_n$

Let $\epsilon > 0$. Now, $\exists N_1$ such that $\forall n > N_1$, $s_n < \alpha + \frac{\epsilon}{2}$. Also, $\exists N_2$ such that $\forall n > N_1$, $t_n < \beta + \frac{\epsilon}{2}$

Set $N = $ max$\{N_1, N_2\}$. Then, $\forall n > N$, $s_n+t_n < \alpha + \beta + \epsilon$. So,

sup $\{s_n+t_n|n>N\} \leq \alpha + \beta + \epsilon$.

Since we can find such an $N$ for each $\epsilon > 0$, we conclude that lim sup $(s_n+t_n) \leq \alpha + \beta$

$\endgroup$
  • 2
    $\begingroup$ Yes, it's all fine $\endgroup$ – Hagen von Eitzen May 17 '14 at 4:38
  • $\begingroup$ You made a small typo on $(t_n)$, but yes that should be fine. $\endgroup$ – IAmNoOne May 17 '14 at 5:11
0
$\begingroup$

Yes, your proof is fine, nice work.

Here is the one typo I think Nameless is referring to:

Also, $\exists N_2$ such that $\boldsymbol{\forall n > N_1}$, $t_n < \beta + \frac{\epsilon}{2}$

This should read $\forall n > N_2$.

$\endgroup$

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.