Vtyjkyuk

Let $l^infty$ be the space of all bounded, complex sequences and consider the metric $$d((a_n)_n=0^infty,(b_n)_n=0^infty)=sum_n=0^infty 2^-nlvert a_n-b_nrvert$$
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 $k=0,1,2ldots$ 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 $mathbbC$ and hence converges to $c_ninmathbbC$, 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 $i<j$,
beginalign
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
endalign
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 $l^infty$.