Show that $\ell^1$ is non complete under $\ell^2$ norm

Question

I'm trying to show that the metric space $ \ell^1$ is not complete under $\left(\sum_{k=1}^\infty |x_k - y_k|^2\right)^{1/2}.$

So I'm trying find a non convergent cauchy sequence in $\ell^1$ w.r.t $\left(\sum_{k=1}^\infty |x_k - y_k|^2\right)^{1/2}.$

An example was given here Non-convergent Cauchy sequence in $\ell^1$ with respect to the $\ell^2$ norm , but I dont seem to understand why it is not convergent in the $\ell ^1 space $.

My understanding is the sequence goes something like $(1 , \frac{1}{2} , \frac{1}{3}, ...... \frac{1}{k} , 0 , 0 , 0 , 0 , . . . )$

The $d_{2}(x^n,x^m)$ ---> 0 for some large m,n where m > n so the sequence is obviously Cauchy.

However the sequence seems to converge to $0$ $\in \ell^1$. So I dont understand why this sequence is non convergent in $\ell^1$

Is this sequence correct? If not what is another example of a non convergent cauchy sequence in $\ell ^1$

If someone could clarify that would be appreciated

Answer 1

The example works. Indeed the limit is the harmonic sequence which does famously not belong to $\ell^1$.

For the sake of conventions, $\ell^p$ is used for sequences and $L^p$ for functions.

Answer 2

$(1,\frac 1 2 ,\frac 1 3,...,\frac 1 n,0,0...) \to (1,\frac 1 2 ,\frac 1 3,...)$ in $l^{2}$ so this sequence is Cauchy in $l^{1}$ with the $l^{2}$ norm. To show that it does not converge in $l^{1}$ use proof by contradiction: suppose it converges in $l^{2}$ norm to some point $y$ in $l^{1}$. Then, for each $k$, the k-th coordinate of $x^{(n)}$ converges to the k-th coordinate of $y$. This implies that $y_k=\lim_{n \to \infty} \frac 1 k =\frac 1 k$. But then $\sum |y_k|=\sum \frac 1 k =\infty$ which is a contradiction.