I am a bit confused with how $\mathcal{l}^1$ can be complete.
So we know that the sequence space $\mathcal{l}^1$, equipped with $||\cdot||_{\mathcal{l}^1}$ is complete. But the sequence of sequences $$\left(x^{(n)}_k\right)_{k\in\mathbb{N}} = \left(\frac{1}{k^{1+1/n}}\right)_{k\in\mathbb{N}},$$ with norm $||x^{(n)}||_{\mathcal{l}^1}=\zeta(1+1/n)<\infty$ is in $\mathcal{l}^1$ for each $n\in\mathbb{N}$, but the limit $$\left(x_k\right)_{k\in\mathbb{N}} = \left(\frac{1}{k}\right)_{k\in\mathbb{N}}$$ is not in $\mathcal{l}^1$, as $||x||_{\mathcal{l}^1}=\infty$.
Could anybody point out, where I am going wrong?
Any help is very much appreciated.
It is true indeed that, for each $n\in\Bbb N$, $\lim_{n\to\infty}\dfrac1{k^{1+1/n}}=\dfrac1k$. However, it doesn't follow from this that your sequence converges to $\left(\frac1k\right)_{k\in\Bbb N}$. That doesn't even make sense, since, as you wrote, this sequence does not belong to $\ell^1$.