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Math Help - limit superior proof

  1. #1
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    limit superior proof

    I was wondering if someone could show me how to do this proof:

    If yn converges to L and xn is a bounded sequence, show that
    limsup(xn + yn) = limsup(xn) + L.

    Thanks for any help. MK
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  2. #2
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    Quote Originally Posted by MKLyon View Post
    I was wondering if someone could show me how to do this proof:

    If yn converges to L and xn is a bounded sequence, show that
    limsup(xn + yn) = limsup(xn) + L.

    Thanks for any help. MK
    \limsup (x_n+y_n) \leq \limsup x_n + \limsup y_n = \limsup x_n + L
    We will show that,
    \limsup (x_n+y_n) \geq \limsup x_n + \limsup y_n.
    There exists subsequence x_{n_k} so that \lim x_{n_k} = \limsup (x_n+y_n) (a theorem). Then \lim y_{n_k} = L (because a subsequence of a convergent sequence stays the same). Thus, x_{n_k}+y_{n_k} is a subsequence of x_n + y_n and so \lim x_{n_k}+y_{n_k} is a subsequencial limit so \lim (x_{n_k}+y_{n_k}) \leq \limsup (x_n + y_n) (a theorem - limsup's are largest subsequential limits). Thus, \limsup x_n + L \leq \limsup (x_n+y_n).
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  3. #3
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    Could someone help with something similar?

    I have that (xn) and (yn) are bounded sequences. How would I show that liminf(xn) + limsup(yn) <= limsup(xn + yn)?

    It seems easy enough, but I can't figure it out.
    Thanks for any help.
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  4. #4
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    Quote Originally Posted by MKLyon View Post
    Could someone help with something similar?

    I have that (xn) and (yn) are bounded sequences. How would I show that liminf(xn) + limsup(yn) <= limsup(xn + yn)?

    It seems easy enough, but I can't figure it out.
    Thanks for any help.
    Here is a result that you should know.

    Lemmon: If x_n \mbox{ and }y_n are convergent sequences and x_n\leq y_n then x\leq y where x is the limit of x_n and y is the limit of y_n.

    Now we can prove that \limsup (x_n+y_n)\leq \limsup (x_n)+\limsup(y_n). The most important thing here is to understand what \limsup means. It means the limit of the superior sequence, i.e. \limsup (x_n) = \lim (\sup \{x_k| k\geq n\})*. Now \sup\{ x_k + y_k | k\geq n\}\leq \sup\{ x_k|k\geq n\} + \sup\{ x_k|k\geq n\}. Since bounded sequences always have limit superiors it means these superior sequences have limits. So by the lemmon: \lim (\sup\{ x_k+y_k|k\geq n\} ) \leq \lim (\sup\{ x_k |k\geq n\}) + \lim (\sup\{ x_k|k\geq n\} ). Thus, \limsup (x_n+y_n) \leq \limsup (x_n)+\limsup (y_n).



    *)Example. Say x_n = \frac{1}{n}+(-1)^n. Then x_1 = 1 + (-1) = 0, x_2 = \frac{1}{2} + (-1)^2 = \frac{3}{2}, x_3 = \frac{1}{3} + (-1)^3 = - \frac{2}{3}, .... So \sup\{ x_k| k\geq 1\} = \sup \{ x_1,x_2,x_3,... \} = \frac{3}{2}, \sup \{ x_k |k\geq 2\} = \sup \{ x_2,x_3,...\} = \frac{3}{2}, \sup \{x_k | k\geq 3\} = \sup \{x_3,x_4,...\} = \frac{5}{4} .... So in general the n-th term in the superior sequence is: \frac{3}{2},\frac{3}{2},\frac{5}{4},\frac{5}{4},\f  rac{7}{6},\frac{7}{6},..., thus the limit is 1. This means \limsup \left( \frac{1}{n} + (-1)^n \right) = 1.
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  5. #5
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    Thanks, but how does this relate to liminf(xn) + limsup(yn) <= limsup(xn + yn)?

    I'm not sure if you misread what I wrote or if I'm missing something, but I need the liminf(xn) not the limsup(xn).
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