Let $(v_n)_{n\in\mathbb N}\subseteq\mathbb R$ be a sequence with $1/2\leq v_n\leq 2$ for all $n$. We assume that $l^1_v=\{(x_n):x_n\in\mathbb R, \sum^{\infty}_{n=1}v_n \vert x_n\vert<\infty \}$ is a vector space. We have defined also the norm $$\vert\vert x \vert\vert = \sum^{\infty}_{n=1}v_n \vert x_n\vert $$ on $l^1_v$. Now let $(x^{(k)})\subseteq l^1_v$ be a Cauchy sequence (under the norm above) with $x^{(k)}=(x^{(k)}_1,x^{(k)}_2,\dots ,x^{(k)}_j,\cdots)$. How can we show that for every $j\in\mathbb N$, $(x^{(k)}_j)$ converges in $\mathbb R$ as $k\rightarrow \infty$ ?
If $(x^{(k)}_j)$ converges than there exists a $x_j\in \mathbb R$ such that $\vert\vert (x^{(k)}_j)-x_j\vert\vert \rightarrow 0$ as $k\rightarrow \infty$. But I really don't know how to use the hint that the sequence above is a Cauchy sequence...any hint?
I assume in your course you have already proved that $\ell^1(\mathbb{R})$ is complete with respect to the norm $\|\cdot\|_1$ defined by $$\|x\|_1 := \sum\limits_{n=1}^\infty |x_n|$$ for $x = (x_n)_{n\in\mathbb{N}}\in\ell^1(\mathbb{R})$. The norm $\|\cdot\|_v$ you are given is equivalent to $\|\cdot\|_1$, since $\frac{1}{2}\leq v_n \leq 2$ for all $n\in\mathbb{N}$, meaning that for any $x\in\ell^1(\mathbb{R})$ one has $$\frac{1}{2}\|x\|_1 = \frac{1}{2}\sum\limits_{n=1}^\infty |x_n|\leq\underbrace{\sum\limits_{n=1}^\infty v_n|x_n|}_{=\|x\|_v } \leq 2\sum\limits_{n=1}^\infty |x_n| = 2\|x\|_1.$$
So $\ell_v^1 = \ell^1$ and any Cauchy sequence with respect to $\|\cdot\|_v$ is also one with respect to $\|\cdot\|_1$, meaning it also has to converge.