# Thread: prove that thel imsup of a sum is less than the sum of the limsup

1. ## prove that thel imsup of a sum is less than the sum of the limsup

Let $(a_n)$n from 1 to infinity and $(b_n)$n from 1 to infinity be bounded sequences of real numbers. Prove that $sup_{n\inN}$ $(a_n+b_n)$ <= $sup_{n\inN}$ $a_n$ $+ sup_{n\inN}$ $b_n$.

n $\in$ N

Conclude that $limsup_{n--> infinity}$ $(a_n +b_n)$ <= $limsup_{n--> infinity}$ $a_n$ + $limsup_{n--> infinity}$ $b_n$. Formulate analogus results for inf and liminf (no need to prove those).

The example i could think of is $a_n = (-1)^n$ and $b_n = (-1)^{n+1}$. 0<2. [tex] How to prove it in general?

2. ## Re: prove that thel imsup of a sum is less than the sum of the limsup

Originally Posted by wopashui
Let $(a_n)$n from 1 to infinity and $(b_n)$n from 1 to infinity be bounded sequences of real numbers. Prove that $sup_{n\inN}$ $(a_n+b_n)$ <= $sup_{n\inN}$ $a_n$ $+ sup_{n\inN}$ $b_n$.
Let $\alpha=\sup(a_n)~\&~\beta=\sup(b_n)$.
$\left( {\forall n} \right)\left[ {a_n + b_n \leqslant \alpha + \beta } \right]$ thus $(a_n+b_n)$ is bounded.
Let $\gamma=sup(a_n+b_n)\text{ so }\gamma\le \alpha+\beta.$

3. ## Re: prove that thel imsup of a sum is less than the sum of the limsup

Originally Posted by Plato
Let $\alpha=\sup(a_n)~\&~\beta=\sup(b_n)$.
$\left( {\forall n} \right)\left[ {a_n + b_n \leqslant \alpha + \beta } \right]$ thus $(a_n+b_n)$ is bounded.
Let $\gamma=sup(a_n+b_n)\text{ so }\gamma\le \alpha+\beta.$
Let $\gamma=sup(a_n+b_n)\text{ so }\gamma\le \alpha+\beta.$ How do we know this, this is what we are trying to prove, $\gamma\le \alpha+\beta$

4. ## Re: prove that thel imsup of a sum is less than the sum of the limsup

Originally Posted by wopashui
Let $\gamma=sup(a_n+b_n)\text{ so }\gamma\le \alpha+\beta.$ How do we know this, this is what we are trying to prove, $\gamma\le \alpha+\beta$
Do you understand the original question?

5. ## Re: prove that thel imsup of a sum is less than the sum of the limsup

I think it so hard for directly prove.

My definition is $\limsup_{n \longrightarrow \infty} x_n := inf_{n}sup_{n \geq k} x_k$

Proof. Let $\varepsilon > 0$ . Assume $a:=limsup_{n \longrightarrow \infty} a_n$ , $b:=limsup_{n \longrightarrow \infty} b_n$ and $c_n=a_n+b_n.\\$Then there exists $N_1,N_2 \in \mathbb{N}$ such that $\forall n> N_1 \Rightarrow a_n \leq a+\frac{\varepsilon}{2}$ and $\forall n> N_2 \Rightarrow b_n \leq b+\frac{\varepsilon}{2}.$ Choose $N=max \{N_1,N_2\}.$ For all $n \geq N$, we have $a_n+b_n \leq a+b+\varepsilon$ and so $a_n+b_n \leq a+b$ or $c_n \leq a+b$ . Since $\{c_n\}$ is bounded, the Bolzano- Weierstrass implies that there is a convergent subsequence $\{c_{n_{k}}\}$ of $\{c_n\}$ such that $c_{n_{k}} \longrightarrow c$, with $c=limsup_{n \longrightarrow \infty} c_n$. Thus, for all $k \geq N, c_{n_{k}} \leq a+b$ , take limit as $k \longrightarrow \infty,$we get $lim_{k \longrightarrow \infty} c_{n_{k}} \leq lim_{k \longrightarrow \infty} (a+b)$ and so $limsup_{n \longrightarrow \infty} c_n \leq a+b.$ Hence $limsup_{n \longrightarrow \infty} (a_n+b_n) \leq limsup_{n \longrightarrow \infty} a_n+limsup_{n \longrightarrow \infty} b_n.$

6. ## Re: prove that thel imsup of a sum is less than the sum of the limsup

Originally Posted by wopashui
Let $\gamma=sup(a_n+b_n)\text{ so }\gamma\le \alpha+\beta.$ How do we know this, this is what we are trying to prove, $\gamma\le \alpha+\beta$
For all $n$, $a_n+b_n\leq \alpha+\beta$. So $(\alpha+\beta)$ is an upper bound of $(a_n+b_n)$. By definition, $\gamma$ is the least upper bound, i.e. $\gamma\leq M$ for all upper bounds $M$. Therefore $\gamma\leq(\alpha+\beta)$.