# rearrangment of series

• Aug 3rd 2008, 05:21 PM
particlejohn
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 $L' \leq L$?
• Aug 3rd 2008, 05:44 PM
ThePerfectHacker
Quote:

Originally Posted by particlejohn
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, $S_1=S_2$.
• Aug 3rd 2008, 05:56 PM
particlejohn
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?