Stop the presses: In fact this is trivial. It holds for any norm on the space of all complex sequences. If you can't find an example that's just because you can't "find" an example of a norm on that space to begin with - it seems clear the existence of a norm must depend on the axiom of choice.
Sheer Triviality. Let $S$ be the space of all complex sequences. If $||\cdot||$ is a norm on $S$ there exists a sequence of vectors in $S$ which converges to $0$ in norm but not pointwise.
Proof. As below, define $$\Lambda_j x=x_j\quad(x=(x_1,x_2,\dots))$$and note that what we need to prove is that there exists $j$ such that $\Lambda_j$ is not bounded. Suppose each $\Lambda_j$ is bounded, and define $x=(x_1,\dots)$ by $$x_j=j||\Lambda_j||.$$Then $$\Lambda_jx=j||\Lambda_j||,$$hence $$||x||\ge j$$for every $j$, contradiction.
(Of course this doesn't quite answer the question unless we know that there is a norm on $S$. In case it's not clear, if $V$ is any real or complex vector space there exists a norm on $V$. For example, say $B$ is a basis, and define $||\sum_{j=1}^n\alpha_jb_j||=\sum|\alpha_j|$ for any $b_1,\dots b_n\in B$ and scalars $\alpha_1,\dots,\alpha_n$.)
I'm leaving the original version here, because to my way of thinking a Banach-space norm on $\ell_\infty$ such that convergence in norm does not imply pointwise convergence is more interesting/surprising than the above:
Unimportant comment on jargon: I may well be mistaken, but if I'm not then the phrase "sequence space" is often defined in a way that explicitly rules out this behavior.
(Of course that comment applies in a context where "a sequence space" is a particular sort of Banach space)
Anyway, one can "construct" such a thing using AC - it's an interesting question whether AC is needed.
Let $X=\ell_\infty$. Say $T:X\to X$ is linear and bijective, and $T^{-1}$ is not bounded (with respect to the usual norm). (You can easily construct such $T$ if you have a Hamel basis for $\ell_\infty$.)
Define $$||x||_X=||Tx||_\infty.$$
Define $$\Lambda_jx=x_j\quad(x=(x_1,x_2,\dots)).$$
We're done if there exists $j$ such that $\Lambda_j$ is not bounded (with respect to $||\cdot||_X$). Suppose to the contrary that every $\Lambda_j$ is bounded. Now, for every $x\in X$ we have $$\sup_j|\Lambda _j x|=||x||_\infty<\infty.$$So Banach-Steinhaus (uniform boundedness) shows that $||\Lambda_j||$ is bounded; say $||\Lambda_j||\le C$ for every $j$. This says precisely that $$||x||_\infty\le C||x||_X=C||Tx||_\infty,$$so $T^{-1}$ is bounded (with respect to the sup norm), contradiction.
By Cauchy-Schwarz, we have that $|\langle g,f\rangle_H|\leq\|g\|_H\|f\|_H\leq \|g\|_H$ if $\|f\|_H\leq 1$. On the other hand, every Hilbert spaces are reflexive. By Kakutani's Theorem, the closed unit ball is weakly compact, hence ($\star$) says that the function $f\mapsto\langle g,f\rangle_H$ attains its maximum over all $f$ with norm $\leq 1$. There is no square as in the question.
Best Answer
Actually, the assertion “sequence is convergent in a normed space if there exists an element in space whose difference with sequence has norm equal to zero” makes no sense. What is the difference between an element of a normed space and a sequence of elements of that space?
But I agree that, if you want to prove that $\sum_{n=1}^\infty\frac{(-t)^n}n=-\log(1+t)$ in $\bigl(C[0,1],\|\cdot\|_\infty\bigr)$, then what you have to prove is that$$\lim_{N\to\infty}\left\|-\log(1+t)-\sum_{n=1}^N\frac{(-t)^n}n\right\|_\infty=0.\tag1$$But you did not prove it. This follows from Abel's theorem: since the series $\sum_{n=1}^\infty\frac{(-1)^n}n$ converges, then the series $\sum_{n=1}^\infty\frac{(-1)^n}nt^n$ converges uniformly on $[0,1]$. Now, use the fact that$$(\forall t\in[0,1)):\sum_{n=1}^\infty\frac{(-1)^n}nt^n=-\log(1+t).$$Since the convergence is uniform, we have that the equality $\sum_{n=1}^\infty\frac{(-1)^n}nt^n=-\log(1+t)$ also holds when $t=1$, and so $(1)$ is indeed true.