Deciding if $l^1(\mathbb{Z}_+)$ over $\mathbb{F}$ is complete with the norm $\lVert x\rVert=2\left|\sum_{n\geq 1}x_n\right|+\sum_{n\geq2}(1+1/n)|x_n|$

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Consider the space $l^1(\mathbb{Z}_+)$ over a field $\mathbb{F}$ (real or complex) and take it as granted that the function

$$\lVert x\rVert=2\left|\sum_{n\geq 1}x_n\right|+\sum_{n\geq 2}(1+1/n)|x_n|$$

defines a norm in the space. I am trying to decide whether $l^1(\mathbb{Z}_+)$ is or is not complete w.r.t. this norm. That is, have failed to find an example of a conditionally convergent series which we can approximate with a sequence of $l^1$ elements. I am conjecturing that if the space is not complete, then the counterexample has to be at least conditionally convergent, for otherwise the first summand in $\lVert\cdot\rVert$'s definition would blow up to infinity.

On the other hand, I have not made any tangible progress with a direct proof, i.e. if we take any Cauchy sequence $\left(x_n\right)_{n\in\mathbb{Z}_+}\subset l^1(\mathbb{Z}_+)$, then $\lim\limits_{n\to\infty}\lVert x - x_n\rVert = 0$ for some $x\in l^1(\mathbb{Z}_+)$.

Something to keep in mind is that if $\;\lVert\cdot\rVert_{l^1}$ is the canonical $l^1$ norm, $\lVert x\rVert_{l^1} = \sum_{n\geq 1}|x_n|$, then $\lVert x\rVert\leqslant2\lVert x\rVert_{l^1}\,$. So far I have not been able to utilize this observation.

I appreciate all tips and hints (or at this point even just a plain solution).

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We have $$\|x\|\ge {1\over 2}\left |\sum_{n=1}^\infty x_n\right |+\sum_{n=2}^\infty |x_n| \\ \geq {1\over 2}\left [|x_1|-\sum_{n=2}^\infty |x_n|\right ]+\sum_{n=2}^\infty |x_n|\\ ={1\over 2}\|x\|_{\ell^1} $$ On the other hand $$\|x\|\le 2\sum_{n=1}^\infty |x_n|+2\sum_{n=2}^\infty |x_n|\\ \le 4\|x\|_{\ell^1}$$ Thus the norms are equivalent, hence $\ell^1$ is complete with the new norm, as it is complete with the norm $\|\cdot\|_{\ell^1}.$