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Math Help - prove that thel imsup of a sum is less than the sum of the limsup

  1. #1
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    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?
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  2. #2
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    Re: prove that thel imsup of a sum is less than the sum of the limsup

    Quote Originally Posted by wopashui View Post
    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.
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    Re: prove that thel imsup of a sum is less than the sum of the limsup

    Quote Originally Posted by Plato View Post
    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
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    Re: prove that thel imsup of a sum is less than the sum of the limsup

    Quote Originally Posted by wopashui View Post
    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?
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  5. #5
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    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.
    Last edited by konna; October 1st 2011 at 10:46 PM.
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  6. #6
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    Re: prove that thel imsup of a sum is less than the sum of the limsup

    Quote Originally Posted by wopashui View Post
    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).
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