I am trying to prove that given a sequence of finite dimensional vector spaces $V_i$ (though I think we can be more general, but whatever), I have a canonical isomorphism that gives me:
$$\left( \bigoplus _i V_i \right)^* = \prod_i V_i ^*$$
where $\bigoplus _i V_i$ is the set of sequences whose $i$-th term belongs to $V_i$ that are definitely zero, and $\prod_i V_i$ the set of all sequences whose $j$-th term belongs to $V_j$.
It is trivial with a finite number of terms, but I don't get how to extend it to infinite products, and why you need sequences that are definitely zero.
If $f \in \left ( \oplus_{i \in I} V_i \right )^*$, then for each $i$, $f \upharpoonright V_i \in V_i^*$. Then the product $(f \upharpoonright V_i)_{i \in I}$ lives in $\Pi_{i \in I} V_i^*$.
Notice there is nothing saying that all but finitely many of the $f \upharpoonright V_i$ should be $0$, so $(f \upharpoonright V_i)_{i \in I}$ does not live in $\oplus_{i \in I} V_i^*$, in general!
As for the other direction, given $(f_i)_{i \in I} \in \Pi_{i \in I}V_i^*$, define (for some $J \subseteq I$ finite) $f(\sum a_j v_j) = \sum_{j \in J} a_j f_j(v_j)$ .
Since $(f_i)_{i \in I}$ tells us what to do to each $V_i$, we know (by linearity) what we have to do to a sum of elements of the $V_i$.
I leave it to you to check that these maps are mutually inverse.
Edit:
More formally, we see $\left ( \oplus_{i \in I} V_i \right )^*\cong \Pi_{i \in I} V_i^*$ since we have maps $\varphi : \left ( \oplus_{i \in I} V_i \right )^* \to \Pi_{i \in I} V_i^*$ and $\psi : \Pi_{i \in I} V_i^* \to \left ( \oplus_{i \in I} V_i \right )^*$ given by
$\varphi(f) = (f \upharpoonright V_i)_{i \in I}$
$\psi((f_i)_{i \in I}) = F$
To define $F : \oplus_{i \in I} V_i \to K$, it suffices to define it on each summand, and then extend linearly. We put $F(v_j) = f_j(v_j)$ for $v_j \in V_j$.
Again, I leave it to you to verify these maps are mutually inverse (and indeed that they are linear maps).
But why does this only work with $\oplus V_i$?
For $f \in (\Pi_{i \in I} V_i)^*$, we can also consider $f \upharpoonright V_i$, and we can also form the product $(f \upharpoonright V_i)_{i \in I} \in \Pi_{i \in I} V_i^*$.
Unfortunately, this new operation is not injective. For simplicity, say each $V_i$ is finite dimensional, so that $|V_i| = |V_i^*|$. Then when $K$ is the underlying field of the vector space:
$|(\Pi_{i \in I} V_i)^*| = |K|^{|\Pi_{i \in I} V_i|} > |\Pi_{i \in I} V_i| = |\Pi_{i \in I} V_i^*|$
Restricting ourselves to only finite sums solves this cardinality issue, and allows the function defined above to be injective (indeed bijective).
Hope this helps! ^_^