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?
Your intuition on the problem is correct, however, the explanation you provide is not a proof. You can prove that the series $\sum_{n=1}^{\infty} d_n$ is convergent by appealing to its partial sums which are closely tied to the partial sums of $\sum_{n=1}^{\infty} c_n$.
Fix $\epsilon > 0$. Since $\sum_{n=1}^{\infty} c_n$ is absolutely convergent, its partial sums are Cauchy. That is, there exists an $N \in \mathbb{N}$ such that for all $n \geq m \geq N$, $$|c_m| + |c_{m+1}| + \dots |c_n| < \epsilon.$$ Then, choose $N_1 > N$ so that if $n \geq N_1$ then $\phi(n) \geq N_1$. Note that since $\phi : \mathbb{N} \rightarrow \mathbb{N}$ is a bijection, it is also an injection so there are only a finite number of indices $n$ so that $\phi(n) \leq N_1$. Then if $k \geq \ell \geq N_1$, we have that $$|c_{\phi(\ell)}| + |c_{\phi(\ell+1)}| + \dots |c_{\phi(k)}| < \epsilon$$ i.e., $$|d_{\ell}| + |d_{\ell+1}| + \dots + |d_{k}| < \epsilon.$$ Thus the partial sums of $\sum_{n=1}^{\infty} d_n$ are Cauchy so the sum must converge. To prove that the sums are indeed equal, you should appeal to the surjectivity of $\phi$.