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Thread: rearrangment of series

  1. #1
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    rearrangment of series

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