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 $(s_n+t_n) \leq \alpha + \beta$

  • 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

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 $\forall n > N_2$.


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