Multiplicative inverses of formal series with non-negative coefficients What are the formal series $f$ with non-negative integer coefficients and constant term equal to $1$ whose multiplicative inverse $1/f$ has all coefficients, apart from a finite subset, all non-positive?
In fact, assume the series converges in some disk if you want...
 A: For example, $f(x) = (1 - a x)/(1 - b x)$ with $b > a > 0$ has all coefficients nonnegative while $1/f(x)$ has all coefficients nonpositive after the constant term.
A: Here is a non-rigorous attempt at an answer, but perhaps someone can make this into a formal argument.  I will assume your series is centered at $0$.  I will also denote your function by $g$ for reasons you will see as you read on.
Suppose $g(x) = \sum_{n=0}^{\infty} a_n x^n$, where $a_n \geq0$ for every $n$.
We look for a series for $h(x) = \frac{1}{g(x)} = \sum_{n=0}^{\infty} b_n$.
By taking formal derivatives $b_n = D^n(h(0))/n!$ (as in the Taylor expansion).
And then do a Google search... Behold the Faá di Bruno's Formula for derivatives of composite functions:
http://mathworld.wolfram.com/FaadiBrunosFormula.html
Let $f(x) = 1/x$ so that $h(x) = f(g(x))$.  Comparing with the terms in this formula, we see that the signs of the $b_n$ must alternate (or be zero).  This is because the derivatives of $1/x$ alternate in sign and the derivatives of $g(x)$ are non-negative by your hypothesis.
Now, I don't trust any of this, however, since I've completely neglected matters of convergence; but when dealing with formal power series, maybe this point is moot?  
A: Assume the powerseries of f(x) as $f(x)=1+a x+bx^2+cx^3+dx^4...$ where we deal with the explication of the problem just up to the coefficient $d$. Then I have a description of the $p$'th power of $f$ with $p$ as a variable, which may be best displayed with the help of a table.    
Remark: I adopted a binomial notation, where in my original derivation is a pochammer-symbol. 
So for $p*(p-1)*(p-2)/3!$ in the original I write $\binom p3$ for shortness here. Thus for negative $p$ the binomial-expressions must respectively be interpreted!

Here is the beginning of the table for $f(x)^p$ with positive $p$ :
$$ \small \begin{array} {rllll} 
f(x)^p=x^0* & \binom{p}{0}*(1)\\
+x^1*[& \binom{p}{1}*(1a)  ] \\
+x^2*[& \binom{p}{1}*(1b)&+ \binom{p}{2}* (1a^2) ] \\
+x^3*[& \binom{p}{1}*(1c)&+ \binom{p}{2}* (2ab)&+ \binom{p}{3}* (1a^3)] \\
+x^4*[& \binom{p}{1}*(1d)&+ \binom{p}{2}* (2ac+1b^2)&+ \binom{p}{3}* (3a^2b)&+ \binom{p}{4}* &(a^4)]  \\
+\ldots&\ldots \\
\end{array}
$$

If we insert $p=1$ for $f(x)^1$ we get the obvious correct result
$ \qquad \begin{array} {rllll} 
f(x)^1=x^0* & 1*(1)\\
+x^1*[& 1*(1a)  ] \\
+x^2*[& 1*(1b)  ] \\
+x^3*[& 1*(1c)  ] \\
+x^4*[& 1*(1d)  ]  \\
+\ldots&\ldots \\
\end{array}
$

If the above table is indeed a complete description for all $p$, then using $p=-1$ for the multiplicative inverse $\frac{1}{f(x)}$ , this would give:
$ \qquad \begin{array} {rllll} 
f(x)^{-1}=x^0* & 1*(1)\\
+x^1*[& -1*(1a)  ] \\
+x^2*[& -1*(1b)&+1*(1a^2) ] \\
+x^3*[& -1*(1c)&+1*(2ab)&-1*(1a^3)] \\
+x^4*[& -1*(1d)&+1*(2ac+1b^2)&-1*(3a^2b)&+ 1*(a^4)]  \\
+\ldots&\ldots \\
\end{array}
$

Then to have nonnegative coefficients in $f(x)$ as well as in $f(x)^{-1}$ means to have
     1)      $a=0$ because of  $x^1$
     2)      then $b=0$ because of $x^2$ where also $a^2=0$
     3) and so on.    

So if the above description (which was derived from positive powers $f(x)^p$ only) is sufficient also for the negative powers, then this should be acceptable as a proof, that only the constant function $f(x)=1$ has nonnegative coefficients in the formal powerseries of $f(x)$ and $f(x)^{-1}$ simultanously.

[Update]: I apply the correction according to Robert Israel, and get the following result.     

Because in &f(x)& all coefficients are nonnegative, I can, without loss of generality rewrite
$\qquad f(x)=1+ax+b^2x^2+c^3x^3+d^4x^4+...$     
Then the table for $f(x)^{-1}$ looks like     
$ \qquad \begin{array} {rllll} 
f(x)^{-1}=x^0* & 1*(1)\\
+x^1*[& -1*(1a)  ] \\
+x^2*[& -1*(1b^2)&+1*(1a^2) ] \\
+x^3*[& -1*(1c^3)&+1*(2ab^2)&-1*(1a^3)] \\
+x^4*[& -1*(1d^4)&+1*(2ac^3+1b^4)&-1*(3a^2b^2)&+ 1*(a^4)]  \\
+\ldots&\ldots \\
\end{array}
$
and from
1) $x^2$ follows that the minimal possible b equals a, so $b\ge a$
2) $x^3$ follows that the minimal possible c equals a, so $c\ge b\ge a$
3) and so on       
So we have $ 0 < a \le b \le c \le d \le \ldots $    
If all coefficients after a take their minimum, then $0<a=b=c=d=e= \ldots $ then the coefficients at all $x^k, k>1$ simplify to binomial coefficients whose sum is zero at the same power $a^k$, so the complete coefficient is zero as well. Thus  $f(x)^{-1} = 1-ax $ , then $f(x)={1 \over 1-ax}$ and we have nonnegative coefficients in $f(x)$ and nonpositive in $f(x)^{-1}$ after the constant.
If we take $b>a$ at $x^2$ then at $x^3$ follows, that $c^3=b^3$ suffices to give a negative value at $x^3$ in $f(x)^{-1}$ .    
Analoguos reasoning inherits to all following coefficients $0<a<b=c\le d \le e \le \ldots$ leading to $f(x)$ has positive coefficients only and $f(x)^{-1}$ has negative coefficients after the constant, but I didn't look yet at the coefficients $d,e, \ldots$ to arrive at Robert Israel's solution.
