Assume that $\sum_{n=1}^{\infty} c_n$ is absolutely convergent. Let $\phi$ : $\mathbb{N}$ → $\mathbb{N}$ be a bijection. Set $d_n = c_{\phi(n)}$ for $n\in \mathbb{N}.$ Show that $\sum_{n=1}^{\infty} d_n$ is convergent, and
$\sum_{n=1}^{\infty} d_n$ = $\sum_{n=1}^{\infty} c_n.$
Basically, this problem means that, for an absolutely convergent series, it does not matter what order of summation we are taking as we always get the same result.
I completely understand the intuition here, but it seems that the result is trivial...if $\phi$ has the same cardinality as $\mathbb{N}$, isn't it really just equivalent to $\mathbb{N}$ in a different order? Thus, instead of summing $c_1, c_2, c_3, ...$, we're summing, for example, $c_5, c_{11}, c_{81}, ...$, so that $c_5 + c_{11} + c_{81} + ...$ = $d_1 + d_2 + d_3 + ...$? Is this a sufficient proof? Or, am I making some crucial mistake?
