0

For real sequences $\{a_n\}$ and$\{b_n\}$. Prove that $\limsup_{n\to \infty} a_n+b_n\le \limsup_{n\to \infty} a_n+\limsup_{n\to \infty} b_n$

If $\limsup a_n=a$ and $\limsup b_n=b$, where $a,b\in\Bbb R$.

By Rudin theorem 3.8 of $\limsup a_n=a$, $$\forall \varepsilon>0, \exists N\in\Bbb N\ni \forall n>N, a_n<a+\varepsilon$$

So, $\forall n>N,a_n+b_n<a+b_n+\varepsilon$ and take $\limsup$ both side we get, $$\limsup_{n\to \infty}(a_n+b_n)<a+\varepsilon+\limsup_{n\to \infty}b_n=a+b+\varepsilon$$, which is true $\forall \varepsilon>0$, let $\varepsilon\to0$, we have $$\limsup_{n\to \infty}(a_n+b_n)\le a+b$$

Is my attempt correct? I have read some answers, but all of them choose some subsequence and the symbol of subsequence is not easy to read, hence I try to use another definition to prove this statement.

Steven Lu
  • 1,037
  • I should say that by Rudin theorem 3.17: Let ${x_n}$ be a sequence in $\Bbb R$ and $E$ be the set of all subsequential limits of ${x_n}$, then $\limsup_{n\to \infty}x_n\in E$ and $\forall \varepsilon>0, \exists N\in\Bbb N\ni \forall n>N, x_n<\limsup_{n\to \infty}x_n+\varepsilon$. – Steven Lu Aug 29 '20 at 03:27
  • In general if you take the $\limsup$ of a strict inequality, it could become weak. – user580918 Aug 29 '20 at 03:29

0 Answers0