Are there any non-obvious colimits of finite abelian groups? Does the forgetful functor $U : \mathsf{FinAb} \to \mathsf{Ab}$ from finite abelian groups to abelian groups  preserve colimits?
Morally this should be true, but it is not so easy (for me) to come up with a nice proof. If $(A_i)$ is a diagram of finite abelian groups, assuming that their colimit $\mathrm{colim}_i A_i$ in $\mathsf{FinAb}$ exists, we have to show that the canonical homomorphism
$$\mathrm{colim}_i U(A_i) \to U(\mathrm{colim}_i A_i)$$
is an isomorphism. Ignoring forgetful functors (which is a source of many confusions, by the way), this may be seen as the statement that the colimit in $\mathsf{FinAb}$ "is" the usual one of $\mathsf{Ab}$. And this is obviously equivalent to the statement that $\mathrm{colim}_i U(A_i)$ "is" a finite abelian group. So my question roughly says: Are there any non-obvious colimits in $\mathsf{FinAb}$? Probably not.
For example: Let $I$ be an infinite set and $(A_i)$ a family of finite abelian groups, each being non-trivial without loss of generality. Assume that $\bigoplus_{i \in I} A_i$ exists in $\mathsf{FinAb}$. Using the universal property, we may construct a split epimorphism from $\bigoplus_{i \in I} A_i$  to every coproduct $\bigoplus_{i \in F} A_i$, where $F \subseteq I$ is finite, which surely exist and are preserved by $U$. But then $\bigoplus_{i \in I} A_i$ has at least as many elements as every $\bigoplus_{i \in F} A_i$, which has at least $2^{\# F}$ elements, a contradiction. So $I$ must be finite, and clearly finite coproducts are preserved by $U$.
Coequalizers exist in $\mathsf{FinAb}$ and are preserved by $U$. Now one might use the following (which I believed to be some) "well-known Theorem":

Let $F : \mathcal{C} \to \mathcal{D}$ be a functor which preserves coequalizers and coproducts. Then $F$ preserves colimits.

But is this really true? In order to decompose a colimit  into a coequalizer of a suitable pair of maps between coproducts, we have to assume that these coproducts exist, which is not automatically the case. That theorem is certainly true when $\mathcal{C}$ has coproducts, but otherwise it is unclear to me. (Does someone have a counterexample?)
So for $U : \mathsf{FinAb} \to \mathsf{Ab}$ we have to find some different argument. I am also curious what happens for the forgetful functor $\mathsf{FinGrp} \to \mathsf{Grp}$.
 A: I claim that the colimit of the sequence of inclusions $\frac{1}{1!}\mathbb{Z}\,/\,\mathbb{Z} \hookrightarrow \frac{1}{2!}\mathbb{Z}\,/\,\mathbb{Z} \hookrightarrow \frac{1}{3!}\mathbb{Z}\,/\,\mathbb{Z} \hookrightarrow \dotsc$ exists in $\mathsf{FinAb}$. Namely, it is zero, in contrast to the colimit in $\mathsf{Ab}$, which is $\mathbb{Q}/\mathbb{Z}$. This is equivalent to the following statement:

Let $A$ be a finite abelian group containing a sequence of elements $x_1,x_2,\dotsc$ such that $n! x_n = 0$ and $(n+1) x_{n+1} = x_n$ for all $n \geq 1$. Then $x_n=0$ for all $n \geq 1$.

Proof. Because of $x_{n+1}=0 \Rightarrow x_n=0$, it suffices to prove $x_n=0$ for infinitely many $n$. So we may assume that $\mathrm{ord}(A)$ divides $n+1$. But then $x_n=(n+1) x_{n+1} = 0$ by Lagrange's Theorem. $\square$
It follows that the forgetful functor $U : \mathsf{FinAb} \to \mathsf{Ab}$ preserves coproducts and coequalizers, but it does not preserve colimits of sequences, and hence it is not cocontinuous.
(The statement that a functor, which preserves coproducts and coequalizers, is cocontinuous, can be found in many books and online references. These should be corrected.)
PS: Sorry for answering my own question.
