Let $\ell_0(\mathbb{N})$ denote the space of compactly sequences:
$$\ell_0(\mathbb{N})=\left\{\sum_{k=1}^\infty a_k \,e_k:\#\{j:a_j\neq0<\infty\}\right\}$$
Show that the map $\lambda:\ell_0(\mathbb{N})\to\mathbb{C}$: $\sum_{j=1}^N a_j \,e_j\mapsto\sum_{j=1}^N a_j$ does not satisfy the following property:
There exists $f\in\ell_2(\mathbb{N})$ such that for all $\tau\in \ell_0(\mathbb{N})$, $\lambda(\tau)=\sum_{j=1}^\infty\tau(j)\overline{f(j)}$.
I don't quite understand the notation $\sum_{k=1}^\infty a_k \,e_k$ to represent a sequence. In particular, could you clarify the difference between $\sum_{j=1}^N a_j e_j$ and $\sum_{j=1}^N a_j$?
Also can I please get a hint on how to solve this problem.
I assume the confusion around notation is already resolved, so I'll focus on the representability of $\lambda$ as the inner product against a fixed square-summable sequence $f$.
Suppose there were such $f \in \ell^2$, then, for each basis vector/sequence $e_i$, we would have to have $\lambda(e_i) = 1 = \sum_{j=1}^\infty e_i(j)\overline{f(j)} = \overline{f(i)}$. Thus, $f = (1,1,...)$ but such sequence isn't in $\ell^2$. Therefore, no such $f$ exists.