How can I fix my proof that $\ell^1$ is complete? In an exercise, my teacher asked us to prove that $\ell^1$ is a Banach space. I was able to do so, but there are two steps in my proof that I'm not quite so sure they are correct. This is what I came up with:

let $x^n=(x^n_1, x^n_2, ...)$ be a cauchy sequence of points in $\ell ^1$. Then, for all $\varepsilon>0$ exists a natural number $N$ such that:
$$n,m\geq N \implies\sum_{i=1}^\infty|x^n_i-x^m_i| \leq \varepsilon$$
$$\implies |x^n_i-x^m_i| \leq \varepsilon$$
So, for every $i$ the sequence $(x^n_i)_{n\in \mathbb N}$ is a Cauchy sequence on $\mathbb C$ and therefore it converges, let's call it's limit $y_i \in \mathbb C$.
We want to prove that $x^n \to y = (y_1,y_2,...)$

This is the first step of the proof that I think isn't correct. I tried but failed to prove that $y$ is indeed a point in $\ell^1$, this is:
$$\sum_{i=1}^\infty |y_i|\leq\infty$$

To show that $\lim x^n = y$, let $\varepsilon>0$. Because $\lim_nx^n_i=y_i$, for all $i$ we have that there is a natural number $N_i$ such that: $$n\geq N_i \implies |x_i^n - y_i|\leq \varepsilon\frac{1}{2^i}$$
so $$\sum_{i=1}^k|x_i^n - y_i|\leq\varepsilon\sum_{i=1}^k \frac{1}{2^i}\leq \varepsilon$$
for $n \geq \max\{N_1,...,N_k\}.$
Because this is a bounded increasing sum, we have that it converges and $$||x^n - y|| = \lim_{k\to\infty}\sum_{i=1}^k|x_i^n - y_i|\leq \varepsilon$$
So $\lim x^n = y$

This last part is the one I'm most unsure of. While I had a finite sum, the condition $$\sum_{i=1}^k|x_i^n - y_i|\leq \varepsilon$$
Was true for $n \geq \max\{N_1,...,N_k\}$ but I don't know if this is the case when $k\to\infty$.
I thought about changing the definition of $n$ to $n \geq \sup\{N_1,...,N_k\}$ to accommodate the case where $k\to\infty$, but this solves nothing because either the set is unbounded, or it eventually stays the same at some point for large values of $k$.
How can I fix these two issues with my proof?

The last part is wrong essentially because of the reason that you stated. One way to proceed would be: Let $\epsilon>0$ be given. You know there exists $N$ so that for any $n\geq m\geq N$ such that $||x^n-x^m||<\epsilon/2$. Now if you consider for some $k\geq N$:
\begin{align*}
0\leq \sum^k_{i=1} |x^n_i-y_i|, k \geq N
\end{align*}
Now for each of these components till the $k^{th}$ one, there are $N_i's>N$ so that $|x_i^{N_i} - y_i| < \epsilon/2k$. Now by triangle inequality, we have:
\begin{align*}
 \sum^k_{i=1} |x^n_i-y_i|\leq \sum^k_{i=1} |x^n_i - x^{N_i}_i| + |x_i^{N_i} - y_i| <\epsilon
\end{align*}
Hence we have:
\begin{align*}
\sum^k_{i=1} |y_i| \leq ||x^N|| + \epsilon, \forall k \geq N \implies y \in l_1
\end{align*}
and the previous part means $||x^n-y|| \rightarrow 0$ as $n\rightarrow \infty$
