Let $l^{\infty}$ be the space of all bounded, complex sequences and consider the metric $$d((a_n)_{n=0}^{\infty},(b_n)_{n=0}^{\infty})=\sum_{n=0}^{\infty} 2^{-n}\lvert a_n-b_n\rvert$$ on it. I'm trying to figure out whether this $d$ makes $l^{\infty}$ complete or not.
Let $(x_k)$ be a Cauchy sequence in $l^{\infty}$ where $k=0,1,2\ldots$ and let us denote the $n$th term of $x_k$ by $a^{(k)}_n$. Then it is easy to see that for each fixed $n$, the sequence $$a^{(1)}_n,a^{(2)}_n,a^{(3)}_n,\ldots$$ is a Cauchy sequence in $\mathbb{C}$ and hence converges to $c_n\in\mathbb{C}$, say.
We must now show that $c=(c_n)$ is a bounded sequence and that $(x_k)$ converges to $c$ in $d$-metric, but I'm not sure how to this. Is there a universal bound, $M$, bounding $a^{(k)}_n$ for all $n$ and all $k$? If that is the case, then I think I can finish the proof, but what if there is no such a bound? Please help me figure this out.
The space $l^\infty$ is not complete with that metric.
Let $a_n^{(k)}=\min(n,k)$, so $$x_1 = (1,1,1,1,\ldots)$$ $$x_2 = (1,2,2,2,\ldots)$$ $$x_3 = (1,2,3,3,\ldots)$$ The sequence $(x_k)$ is Cauchy because for $i<j$, \begin{align} d(x_i,x_j) &=\sum_{k=1}^\infty 2^{-k}|\min(i,k)-\min(j,k)| \\ &=\sum_{k=i}^j 2^{-k}(k-i) + \sum_{k=j+1}^\infty 2^{-k}(j-i) \\ &=2^{1-i}-2^{1-j}\\ &<2^{1-i} \end{align} But if $x$ is in $l^\infty$ with bound $M$, then $d(x,x_n)>2^{-n}$ whenever $n>M$. So $x$ is not a limit for the $x_n$, the $(x_n)$ have no limit in $l^\infty$, and $l^\infty$ is not complete with this metric.
The obvious (pointwise) limit is $(1,2,3,4,\ldots)$, but that is not in $l^\infty$.