Question about convergence of of sequence Define 
$$
X := \left \{ (a_n)_{n=0}^\infty :\sum_{n =0}^\infty |a_n| < \infty \right \}
$$ with metrics 
$$
d_{l^1}((a_n)_{n=0}^\infty,(b_n)_{n=0}^\infty) := \sum_{n=0}^\infty |a_n - b_n|
$$
$$
 d_\infty((a_n)_{n=0}^\infty,(b_n)_{n=0}^\infty) := \sup_{n\geq 0} |a_n-b_n|
$$
I want to show that these metrics are not equivalent. And here is my "proof" of that. I am not sure if this is correct.
Let $(x^{(n)})_{n=0}^\infty$ be a sequence in $X$ such that 
$$
 x^{(n)}_k := \begin{cases}
  \frac 1 n & \text{ if } k \leq n \\
  0 & \text{ if k > n }
\end{cases}
$$ 
Define further $\tilde{x} \in X$ with $\tilde{x}_k = 0 $ for all $k \in \mathbb N$. Then
$$
 \lim_{n \rightarrow \infty} d_\infty (x^{(n)},\tilde{x}) = \lim_{n \rightarrow \infty} |1/n| = 0 
$$ such that $(x^{(n)})_{n=0}^\infty$ converges to $\tilde{x}$ with respect to $d_\infty$.
Assume now that there exists some (another) sequence $\tilde{x} \in X$ such that $\lim_{n \rightarrow \infty} d_{l^1}(x^{(n)},\tilde{x}) = 0$. Then we have
$$
 0=\lim_{n \rightarrow \infty} \sum_{k=0}^n |1/n - \tilde{x}_k| + \sum_{k=n+1}^\infty |\tilde{x}_k|
$$
Now we can split the limit and the right term will be $0$ because $\tilde{x} \in X$ and thus by the zero-test the claim follows. But I am not sure how to show that the left term is not $0$.
 A: Let me slightly redefine your $x^{(n)}$ by letting $x^{(n)}_k=\frac1n$ if $k<n$ and $0$ otherwise. Then 
$$d_{\ell^1}\left(x^{(n)},\tilde x\right)=\sum_{k\ge 0}\left|x^{(n)}_k\right|=n\cdot\frac1n=1$$
for all $n\in\Bbb N$, so $\left\langle x^{(n)}:n\in\Bbb N\right\rangle\not\to\tilde x$ in the topology generated by $d_{\ell^1}$. Thus, $d_{\ell^1}$ and $d_\infty$ generate different topologies on $X$, since (as you showed) $\left\langle x^{(n)}:n\in\Bbb N\right\rangle\to\tilde x$ in the topology generated by $d_\infty$. It’s not necessary to show that $\left\langle x^{(n)}:n\in\Bbb N\right\rangle$ isn’t convergent at all with respect to $d_{\ell^1}$, which is what you seem to be trying to do.
In fact it’s true that $\left\langle x^{(n)}:n\in\Bbb N\right\rangle$ isn’t convergent with respect to $d_\infty$. Suppose that $\tilde x\ne\hat x\in X$, so that $\hat x_m\ne 0$ for some $m\in\Bbb N$. Then for all $n\ge m$ we have
$$d_{\ell^1}\left(x^{(n)},\hat x\right)\ge\left|\frac1n-\hat x_m\right|\;,$$
and $$\left|\frac1n-\hat x_m\right|~\underset{n\to\infty}\longrightarrow~\left|\hat x_m\right|>0\;,$$
so the sequence does not converge to any non-zero sequence, either.
A: You are on the right track until near the end.  Note that $|x^{(n)}_k-\tilde{x}_k|\leq d_{\ell^1}(x^{(n)},\tilde{x})$ for each $k$, so $x^{(n)}_k\to \tilde x_k$ for each $k$, which implies $\tilde x_k=0$ for each $k$.  So what is $d_{\ell^1}(x^{(n)},\tilde x)$?
