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Math Help - sequence, convergence

  1. #1
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    sequence, convergence

    Suppose that (x_n) is a sequence of real numbers. Define a sequence (y_n) by y_n=\frac{x_n+x_{n+1}}{2} \forall n \in \mathbb{N}
    (a) Prove that (y_n) converges to a real number x if (x_n) converges to x.
    (b) If (y_n) converges, does (x_n) also converge?
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    Quote Originally Posted by poincare4223 View Post
    Suppose that (x_n) is a sequence of real numbers. Define a sequence (y_n) by y_n=\frac{x_n+x_{n+1}}{2} \forall n \in \mathbb{N}
    (a) Prove that (y_n) converges to a real number x if (x_n) converges to x.
    (b) If (y_n) converges, does (x_n) also converge?
    I think that (a) can be proven fairly easy within the definition of convergent sequences.

    (b) on the other hand I think is false. A possible counter example is

    x_n = \frac{1}{n} + (-1)^n,\;\;\;y_n = \frac{1}{2} \left( \frac{1}{n} + \frac{1}{n+1} \right)
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  3. #3
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    Another counter example for (b) is x_n = (-1)^n.
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  4. #4
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    Oh Ya - much easier.
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    I figured it out :-)
    Last edited by poincare4223; January 14th 2009 at 06:46 PM.
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    Quote Originally Posted by poincare4223 View Post
    Suppose that (x_n) is a sequence of real numbers. Define a sequence (y_n) by y_n=\frac{x_n+x_{n+1}}{2} \forall n \in \mathbb{N}
    (a) Prove that (y_n) converges to a real number x if (x_n) converges to x.
    (b) If (y_n) converges, does (x_n) also converge?
    For the first one we can easily prove that if x_n\to x then x_{n+1}\to x

    Proof: We know that since x_n\to x we may choose N such that n\leqslant N then d(x_n,x)<\varepsilon and since N\leqslant n\implies N\leqslant n+1 we must have that d(x_{n+1},x)<\varepsilon

    This implies that \lim x_n,\lim x_{n+1} and consequently \lim y_n exist. So because of this existence \lim y_n=\frac{1}{2}\left[\lim x_n+\lim x_{n+1}\right]=\frac{1}{2}[x+x]=x
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