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

Let $\displaystyle (a_n)$n from 1 to infinity and $\displaystyle (b_n)$n from 1 to infinity be bounded sequences of real numbers. Prove that $\displaystyle sup_{n\inN}$$\displaystyle (a_n+b_n) <= \displaystyle sup_{n\inN} \displaystyle a_n \displaystyle + sup_{n\inN} \displaystyle b_n. n\displaystyle \in N Conclude that \displaystyle limsup_{n--> infinity}$$\displaystyle (a_n +b_n)$ <= $\displaystyle limsup_{n--> infinity} $$\displaystyle a_n + \displaystyle limsup_{n--> infinity} \displaystyle b_n. Formulate analogus results for inf and liminf (no need to prove those). The example i could think of is \displaystyle a_n = (-1)^n and \displaystyle b_n = (-1)^{n+1}. 0<2. [tex] How to prove it in general? 2. ## Re: prove that thel imsup of a sum is less than the sum of the limsup Originally Posted by wopashui Let \displaystyle (a_n) n from 1 to infinity and \displaystyle (b_n) n from 1 to infinity be bounded sequences of real numbers. Prove that \displaystyle sup_{n\inN}$$\displaystyle (a_n+b_n)$ <= $\displaystyle sup_{n\inN}$ $\displaystyle a_n$ $\displaystyle + sup_{n\inN}$ $\displaystyle b_n$.
Let $\displaystyle \alpha=\sup(a_n)~\&~\beta=\sup(b_n)$.
$\displaystyle \left( {\forall n} \right)\left[ {a_n + b_n \leqslant \alpha + \beta } \right]$ thus $\displaystyle (a_n+b_n)$ is bounded.
Let $\displaystyle \gamma=sup(a_n+b_n)\text{ so }\gamma\le \alpha+\beta.$

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

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

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

Originally Posted by wopashui
Let $\displaystyle \gamma=sup(a_n+b_n)\text{ so }\gamma\le \alpha+\beta.$ How do we know this, this is what we are trying to prove, $\displaystyle \gamma\le \alpha+\beta$
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 $\displaystyle \limsup_{n \longrightarrow \infty} x_n := inf_{n}sup_{n \geq k} x_k$

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

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

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