Let
$$
F=\sum\_{i\ge0}\frac1{(T+2)^i}\left(\frac T{T+1}\right)^{3^i}\in\mathbb F\_3\left(\!\!\left(\frac1T\right)\!\!\right).$$
Does $F$ belong to $\mathbb F\_3(T)$?
Here, truncations of the series do not give a good approximation of $F$.



Is this function rational?
It is not rational.


First of all we denote $1/T=x$, then $$F=\sum\_i \left(\frac{x}{1+2x}\right)^i(1+x)^{-3^i}\in \mathbb{F}\_3((x)).$$
Now denote $x/(1+2x)=y$. Note that rationality in $x$ is equivalent to rationality in $y$ (and also that $\mathbb{F}\_3((x))=\mathbb{F}\_3((y))$). We have $x=y/(1-2y)$; $1+x=(1-y)/(1-2y)$; $$(1+x)^{-1}=\frac{1-2y}{1-y}=\frac{1+y}{1-y}=1-(y+y^2+y^3+\ldots).$$
Thus $$F=\sum\_i y^i(1-y^{3^i}-y^{2\cdot 3^i}-\ldots)=\frac1{1-y}-\sum\_{i\geqslant 0, k\geqslant 1}y^{i+k\cdot 3^i}.$$We should prove that the sequence of coefficients $a\_n$ of the power series $$\sum\_{i\geqslant 0, k\geqslant 1}y^{i+k\cdot 3^i}=\sum a\_ny^n$$
is not eventually periodic modulo 3 (since rationality of a power series with coefficients in a finite field is equivalent to eventual periodicity of its coefficients).


We have $a\_n=|A(n)|$, where $A(n)$ is the set of all non-negative integers $i<n$, for which $3^i$ divides $n-i$. It is straightforward that $A(k)=A(k+3^{k-1})$ for all $k>1$ and $A(k+3^{n-1})=A(k)\sqcup \{k\}$ for all $n>k>1$. This yields $a\_{k+3^{k-1}}=a\_k$ and $a\_{k+3^{n-1}}=a\_k+1$ for all $n>k>1$.


Now assume that, on the contrary, $a\_{n+T}\equiv a\_{n} \pmod 3$ for certain integer $T>0$ and all $n\geqslant n\_0$. Consider the powers of 3 of the form $3^{m-1}$, $m>n\_0$. By pigeonhole principle, there exist two of them which are congruent modulo $T$, say, $T$ divides $3^{n-1}-3^{k-1}$ for certain $n>k>n\_0$. Then $T$ divides $(k+3^{n-1})-(k+3^{k-1})$ but from above we have $a\_{k+3^{n-1}}=a\_k+1=a\_{k+3^{k-1}} +1$. This contradicts to our periodicity assumptions.


