Why are sequentially small modules finitely presented? I am trying to verify the claim in Remark 7.15 of Dwyer-Spalinski that the sequentially small objects in the category of $R$-modules are the finitely presented modules; the part I'm stuck on is showing that a sequentially small $R$-module is finitely presented. "Sequentially small" means something similar to "compact," except that instead of filtered colimits we only use sequential colimits:

An object $A$ of $\mathbf{C}$ is said to be sequentially small if for every functor $B : \mathbf{Z}^+ \to \mathbf{C}$ the canonical map $\mathrm{colim}_n \hom_{\mathbf{C}}(A, B(n)) \to \hom_{\mathbf{C}}(A, \mathrm{colim}_n B(n))$ is a bijection. (Dwyer and Spalinski, Definition 7.14.)

Here, $\mathbf{Z}^+$ is the category $1 \to 2 \to 3 \to \ldots$. This answer comes very close to answering my question, by making use of the fact that any module is a filtered colimit of finitely presented modules, but they're assuming the module is compact rather than sequentially small. To make a similar argument work here, I would need to know that any module is a sequential colimit of finitely presented modules, but I haven't been able to find a proof of this.
Is it in fact true that any module is a sequential colimit of finitely presented modules, or should I be taking a different approach to showing that a sequentially small module is finitely presented?

EDIT: It appears my question was based on a misreading of Dwyer and Spalinski; the original text is:

A set is sequentially small if and only if it is finite. An $R$-module is sequentially small if it has a finite presentation, i.e., it is isomorphic to the cokernel of a map between two finitely generated free $R$-modules.

I must have been primed by the "if and only if" in the first sentence (about sets) and read a non-existent "and only if" into the second sentence (about $R$-modules). As this answer indicates, the converse does not hold.
 A: It is not true that sequentially small modules are always finitely presented.  For instance, let $A$ be an ordered abelian group with the property that every countable set of positive elements has a positive strict lower bound.  Let $R$ be a valuation ring with value group $A$ and let $k$ be the residue field of $R$.  Then $k$ is not finitely presented as an $R$-module, since the maximal ideal of $R$ is not finitely generated (indeed, it is not even countably generated, by our assumption on $A$).  However, I claim $k$ is sequentially small.
Indeed, suppose $$M_0\to M_1\to M_2\to\dots$$ is a sequential diagram of $R$-modules with colimit $M$.  Since $k$ is finitely generated, if a homomorphism $f:k\to M_n$ becomes $0$ after mapping into $B$, then it becomes $0$ after mapping into $M_m$ for some $m>n$, since we just need the images of the finitely many generators of $k$ to become $0$.  So, the canonical map $\operatorname{colim}\operatorname{Hom}(k,M_n)\to\operatorname{Hom}(k,M)$ is injective.
To show it is surjective, let $f:k\to M$ be a homomorphism; we wish to show $f$ can be lifted to $M_n$ for some $n$.  First, $f(1)$ can be lifted to to an element $x\in M_i$ for some $i$.  Let $x_n$ be the image of $x$ in $M_n$ for each $n\geq i$, and let $I_n\subseteq R$ be the annihilator of $x_n$.  If $I_n$ contains the maximal ideal for some $n$, then we can lift $f$ to a homomorphism $k\to M_n$ by mapping $1$ to $x_n$ and we're done, so we may assume $I_n$ does not contain the maximal ideal for each $n$.  Then for each $n$, we can pick some $a_n>0$ in $A$ such that every element of $I_n$ has valuation at least $a_n$.  By our choice of $A$, there exists $a>0$ such that $a<a_n$ for all $n$.  If $r\in R$ has valuation $a$, then, the image of $ax$ in $M_n$ is nonzero for all $n$, and thus the image of $ax$ in $M$ is nonzero.  But the image of $ax$ in $M$ is $af(1)=0$ since $a$ annihilates $k$.  This is a contradiction.
