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Math Help - rearrangment of series

  1. #1
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    rearrangment of series

    Let  \sum_{n=0}^{\infty} a_{n} be a convergent series of non-negative real numbers, and let  f: \bold{N} \to \bold{N} be a bijection. Prove that  \sum_{m=0}^{\infty} a_{f(m)} is also convergent, and has the same sum:  \sum_{n=0}^{\infty} a_{n} = \sum_{m=0}^{\infty} a_{f(m)} .

    Is this the general idea of the proof: Let the partial sums be  S_{N} := \sum_{n=0}^{N} a_{n} and   T_{M} := \sum_{m=0}^{M} a_{f(m)} . Then write  L := \sup(S_{N})_{n=0}^{\infty} and  L' = \sup(T_{M})_{m=0}^{\infty} . Show that  L = L' or in other words, show that the sums are equal? And to show equality, we show the following two inequalities to hold simultaneously:   L \leq L' and  L' \leq L  ?
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  2. #2
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    Quote Originally Posted by particlejohn View Post
    Let  \sum_{n=0}^{\infty} a_{n} be a convergent series of non-negative real numbers, and let  f: \bold{N} \to \bold{N} be a bijection. Prove that  \sum_{m=0}^{\infty} a_{f(m)} is also convergent, and has the same sum:  \sum_{n=0}^{\infty} a_{n} = \sum_{m=0}^{\infty} a_{f(m)} .
    ?
    Let S_1 = \Sigma_{n=0}^{\infty}a_n and b_n = a_{f(n)}.
    The partial sums of \Sigma_{n=0}^{\infty} b_n are all \leq S_1. Since each b_n\geq 0 the partial sums form a non-decreasing sequence which is bounded by S_1. Thus, \Sigma_{n=0}^{\infty}b_n exists, call it S_2, and furthermore S_2\leq S_1.
    By repeating the argument again we see that S_1\leq S_2.
    Thus, S_1=S_2.
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  3. #3
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    Or let  Z := \{m \in \bold{N}: m \leq M \} . So can we use the fact that  f is a bijection between  Z and  f(Z) ? Then  T_{N} = \sum_{m=0}^{M} a_{f(m)} = \sum_{n \in f(Z)} a_{n} . So  ((f(m))_{0}^{M} is finite and bounded. But  T_{M} \leq S_{N} hence  T_{M} \leq L \leq L' . And use  f^{-1} for opposite direction?
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