Prove that for all bijective functions $\pi:\mathbb{N} \rightarrow \mathbb{N} \times \mathbb{N}$, $\sum_{n=1}^{\infty} a_{\pi (n)}$ is convergent. I am tasked with this question:

Let $$\sum_{n=1}^{\infty} a_{(i, n)} = a_{(i, 1)}+a_{(i, 2)}+a_{(i, 3)}+\dots$$
Such that $\sum_{n=1}^{\infty} a_{(i, n)}$ is absolutely convergent for each $i$, and $\sum_{n=1}^{\infty} |a_{(i, n)}| = b_i.$ Furthermore, $\sum_{n=1}^{\infty} b_i$ is also convergent. Prove that for all bijective functions $\pi:\mathbb{N} \rightarrow \mathbb{N} \times \mathbb{N}$, $\sum_{n=1}^{\infty} a_{\pi (n)}$ is convergent.


My attempt:
Since $\pi$ is bijective, every $a_{(i,n)}$ is represented in $\sum a_{\pi (n)}$. Thus $\sum a_{\pi (n)} = \sum a_{(i,n)}$.
We take $\sum |a_{\pi (n)}| = \sum b_i$, which is convergent. Since every absolutely convergent series is also just convergent, it follows that $\sum a_{(i,n)} = \sum a_{\pi (n)}$ is convergent.

Is that just it? The weightage of the question is rather high, and I can't help but feel that I've made a mistake in reasoning. Anyway I can further 'flesh out' the proof in case something is missing?
 A: Let $b=\sum_{n\ge 1}b_n$; the real work is in proving that
$$\sum_{n\ge 1}|a_{\pi(n)}|=b\,.\tag{1}$$
Let $\epsilon>0$; there is an $n_\epsilon\in\Bbb Z^+$ such that $\sum_{k>n_\epsilon}b_k<\frac{\epsilon}2$. For each $i\in\{1,\ldots,n_\epsilon\}$ there is an $m_i\ge n_\epsilon$ such that $\sum_{k>m_i}|a_{(i,k)}|<\frac{\epsilon}{2^{i+1}}$; let
$$m=\max\{m_i:i=1,\ldots,n_\epsilon\}\,.$$
Finally, there is an $\ell_\epsilon\in\Bbb Z^+$ such that
$$\{1,\ldots,m\}\times\{1,\ldots,m\}\subseteq\pi[\{1,\ldots,\ell_\epsilon\}]\,.$$
Then
$$\begin{align*}
\sum_{k=1}^{\ell_\epsilon}|a_{\pi(k)}|&\ge\sum_{i=1}^{\ell_\epsilon}\sum_{n=1}^{\ell_\epsilon}|a_{(i,n)}|\\
&\ge\sum_{i=1}^{\ell_\epsilon}\left(b_i-\sum_{n>\ell_\epsilon}\left|a_{(i,n)}\right|\right)\\
&>\sum_{i=1}^{\ell_\epsilon}\left(b_i-\frac{\epsilon}{2^{i+1}}\right)\\
&\ge\sum_{i=1}^{n_\epsilon}b_i-\sum_{i\ge 1}\frac{\epsilon}{2^{i+1}}\\
&>b-\frac{\epsilon}2-\sum_{i\ge 1}\frac{\epsilon}{2^{i+1}}\\
&=b-\epsilon\,.
\end{align*}$$
Clearly $\sum_{k=1}^n|a_{\pi(k)}|<b$ for all $n\in\Bbb Z^+$, so $(1)$ holds.
