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To prove that functions $f_1(x), \dots, f_n(x)$ with $x \in \mathbb R$ are linearly independent, we only need to show that the Wronskian of these functions is non-zero at a certain value of $x$. Now suppose that $f_1(x), \dots, f_n(x)$ are formal power series. Is there a systematic way to show that these functions are linearly independent over the field of rational functions, i.e. there doesn't exist polynomials $p_1(x), \dots, p_n(x)$ such that $$ p_1(x)f_1(x) + \dots + p_n(x)f_n(x) = 0? $$

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    $\begingroup$ In the smallest case $n=2$ this is equivalent to $f_1/f_2$ to be rational, and for a power series there is a rationality criterion in terms of Hankel determinants (all sufficiently large ones vanish). Is this a satisfactory answer to your question for this special case? $\endgroup$
    – Alexandre Eremenko
    Commented May 7 at 13:54

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There is such a criterion, in principle, of the same type as mentioned in my comment. Let us write $$f_k=\sum_{j=0}^\infty a_{k,j}z^j,\quad p_k=\sum_{j=0}^\infty c_{k,j}z^j.$$ Then your relation defines a homogeneous system of linear equations with respect to $c_{k,j}$ with coefficients depending on $a_{k,j}$. Each equation of this system contains only finitely many of the unknowns $c_{k,j}$. So you can write a sequence of determinants of truncated systems. For linear dependence of your functions, it is necessary and sufficient that all sufficiently large determinants of this sequence vanish.

I don't think you can do substantially better even in the case $n=2$.

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