One can show that all coefficients with sufficiently large index are positive. Indeed, using Maple, the pole of $f$ closest to the origin is: $a:=0.543689...>0,$ and the residue at this pole is $c:=-0.3115580216...<0$. So $$\frac{c}{z-a}=-\frac{c}{a}\sum_{n=0}^\infty \left(\frac{z}{a}\right)^n$$ has positive coefficients.

It is not difficult to estimate the integer $n_0$ such that for $n>n_0$ this part of the partial fraction decomposition will dominate the rest, and the first $n_0$ coefficients can be computed using Maple.

(The second pole closest to the origin is $a_1=0.56984...$ and the residue at it $c_1=0.3383$. The other 4 poles are two complex conjugate pairs, and their absolute values are $>1$, and the residues less than 9 by absolute value, so they have no influence. On the other hand Maple computes the first 100 or 200 coefficients in no time, and they are positive.)

One can show that all coefficients with sufficiently large index are positive. Indeed, using Maple, the pole of $f$ closest to the origin is: $a:=0.543689...>0,$ and the residue at this pole is $c:=-0.3115580216...<0$. So $$\frac{c}{z-a}=-\frac{c}{a}\sum_{n=0}^\infty \left(\frac{z}{a}\right)^n$$ has positive coefficients.

It is not difficult to estimate the integer $n_0$ such that for $n>n_0$ this part of the partial fraction decomposition will dominate the rest, and the first $n_0$ coefficients can be computed using Maple.

(The second pole closest to the origin is $a_1=0.56984...$ and the residue at it $c_1=0.3383$. So the contribution from $a$ overtakes the contribution from $a_1$ already for $n\geq 2$. The other 4 poles are two complex conjugate pairs, and their absolute values are $>1$, and the residues less than 9 by absolute value, so they have no influence, say for $n>10$. On the other hand Maple computes the first 100 or 200 coefficients in no time, and they are all positive.)

One can show that all coefficients with sufficiently large index are positive. Indeed, using Maple, the pole of $f$ closest to the origin is: $a:=0.543689...,$ and the residue at this pole is $c:=-0.3115580216$. So $$\frac{c}{z-a}=-\frac{c}{a}\sum_{n=0}^\infty \left(\frac{z}{a}\right)^n$$ has positive coefficients.

It is not difficult to estimate the integer $n_0$ such that for $n>n_0$ this part of the partial fraction decomposition will dominate the rest, and the first $n_0$ coefficients can be computed using Maple.

(The second pole closest to the origin is $a_1=0.56984...$ and the residue at it $c_1=0.3383$. The other 4 poles are two complex conjugate pairs, and their absolute values are $>1$, and the residues less than 9 by absolute value, so they have no influence. On the other hand Maple computes the first 100 or 200 coefficients in no time, and they are positive.)

One can show that all coefficients with sufficiently large index are positive. Indeed, using Maple, the pole of $f$ closest to the origin is: $a:=0.543689...>0,$ and the residue at this pole is $c:=-0.3115580216...<0$. So $$\frac{c}{z-a}=-\frac{c}{a}\sum_{n=0}^\infty \left(\frac{z}{a}\right)^n$$ has positive coefficients.

It is not difficult to estimate the integer $n_0$ such that for $n>n_0$ this part of the partial fraction decomposition will dominate the rest, and the first $n_0$ coefficients can be computed using Maple.

(The second pole closest to the origin is $a_1=0.56984...$ and the residue at it $c_1=0.3383$. The other 4 poles are two complex conjugate pairs, and their absolute values are $>1$, and the residues less than 9 by absolute value, so they have no influence. On the other hand Maple computes the first 100 or 200 coefficients in no time, and they are positive.)