Let $\ell^1:=\{f:\mathbb{N}\to\mathbb{C}\colon \|f\|_1:=\sum_{n=1}^\infty |f(n)|<\infty\}$ provided with $\|\cdot\|_1$. Show, that $(\ell^1, \|\cdot\|_1)$ is a banach-space.
I have already shown, that $(\ell^1, \|\cdot\|_1)$ is a normed $\mathbb{C}$-vectorspace. It is missing, that $(\ell^1, \|\cdot\|_1)$ is complete. Hence every cauchy-sequence converges.
Let $(f_k)_{k}$ be a cauchy-sequence. I have to find a limit $f\in\ell^1$. Hence $\|f_k-f\|_1\leq\varepsilon$ for every $\varepsilon>0$. But I do not really know how to construct such an $f$ using that we have a cauchy-sequence.
Can you give a hint? Thanks in advance.
Since $\{f_k\}_k$ is Cauchy in $\ell^1$, for every $n \in \mathbb N$, the sequence $\{f_k(n)\}_k$ is Cauchy in $\mathbb C$. Since $\mathbb C$ is complete, each $\{f_k(n)\}$ must converge to some value. Define $f \in \ell^1$ by $$f(n) = \lim_{k\to \infty} f_k(n).$$ Thus $f_k \to f$ pointwise; we need to show that $f \in \ell^1$ and that $f_k \to f$ in $\ell^1$. We do these both in one step. Let $\varepsilon > 0$. Since $\{f_k\}$ is Cauchy, there is $K \in \mathbb N$ such that $k, \ell \ge K$ give $$\|f_k - f_\ell\|_1 \le \varepsilon.$$ In particular, for any $N \in \mathbb N$, this shows $$\sum^N_{n=1} \lvert f_k(n) - f_\ell(n) \rvert \le \varepsilon.$$ Since there are only finitely many terms in the sum, we can take the limit as $\ell \to \infty$ to see that $$\sum^N_{n=1} \lvert f_k(n) - f(n)\rvert \le \varepsilon.$$ Since this holds for all $N \in \mathbb N$, we can take the limit as $N \to \infty$ to see $$\|f_k - f\|_1 \le \varepsilon.$$ This shows that $f_k - f \in \ell^1$. But then since $f_k \in \ell^1$ and $\ell^1$ is a vector space, we must also have $f = f_k - (f_k -f) \in \ell^1$. And since we already showed that $$\|f_k - f\|_1 \le \varepsilon$$ for all $k \ge K$, we conclude that $f_k \to f$ in $\ell^1$ and so $\ell^1$ is complete.