1. ## 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 $\displaystyle L' \leq L$?

2. Originally Posted by particlejohn
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, $\displaystyle S_1=S_2$.

3. 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?