TITLE: Equivalence of metrics in the space of absolutely summable sequences.
QUESTION [5 upvotes]: Let $S=\{\{a_n\}_{n\in\mathbb{N}}: a_n\in\mathbb{R}, \sum|a_n|<\infty\}$. Determine if the metrics $d(\{a_n\},\{b_n\})=\sum|a_n-b_n|$ and $\rho(\{a_n\},\{b_n\})=\sup|a_n-b_n|$ are equivalent metrics (equivalent in sense of convergence of sequences).
I proved that both are metrics and are well defined. Also I proved that if $\{\{x_n\}_k\}_{k\in\mathbb{N}}$ is a sequence in $S$ such that converges in $d$ then converges in $\rho$ (this part is trivial because $\rho\leq d$). I need to determine if converse is true or false.

REPLY [5 votes]: It is not true,
Define $a_n$ a sequence of sequences as follow
$$a_1 = (1,0,0, \cdots,0, \cdots $$
$$a_2 = (\frac{1}{2},\frac{1}{2},0,0,0 \cdots,0, \cdots $$
$$a_3 = (\frac{1}{3},\frac{1}{3},\frac{1}{3},0,0 \cdots,0, \cdots $$
$$\cdots$$
$$a_n =(\frac{1}{n},\frac{1}{n},\cdots, \frac{1}{n},\frac{1}{n},0,0 \cdots,0, \cdots$$
where in the previous $\frac{1}{n}$ are $n$ times.
So $ a_{n}$ is a sequence where the first $n$ ordinates are $ \frac{1}{n}$and the other $0$.
Then take $a_{n}$ and $b=0$ where $b=(0,0, \cdots ,0, \cdots)$.
So $$ \rho( a_{n} ,  b  )= \frac{1}{n} \rightarrow 0$$
that means $a_n$ converges to $b$ in respect to $ \rho$
$$d( a_n , b )= \sum _{i=1}^{n} \frac{1}{n} =1$$
That means that $a_n$ (a sequence of sequences) does not converge to $b$ (the zero sequence) in respect to $d$