# Formal power series: $F_i(x)$ converges if and only if $\lim_{i\to\infty}deg(F_{i+1}(x)- F_i(x))=\infty.$

The following notations and definitions are taken from Richard Stanley's book Enumerative Combinatorics Volume $1,$ second edition.

Recall that a formal power series $F(x)$ is of the form $$\sum_{n\geq 0} a_n x^n$$ where $x$ cannot take numerical value.

If $F_1(x), F_2(x),...$ is a sequence of formal power series, and if $F(x) = \sum_{n\geq 0}a_n x^n$ then we say that $F_i(x)$ converges to $F(x)$ as $i\to\infty$ provided for every $n\geq 0,$ there is a number $\delta(n)$ such that the coefficient of $x^n$ in $F_i(x)$ is $a^n$ whenever $I\geq \delta(n).$

The degree of a nonzero formal power series $F(x) = \sum_{n\geq 0}a_nx^n$ is the least integer $n$ such that $a_n\neq 0.$ It is denoted by $deg(F(x)).$

Statement: $F_i(x)$ converges if and only if $$\lim_{i\to\infty}deg(F_{i+1}(x)- F_i(x))=\infty.$$

If all $F_i(x) = F(x),$ wouldn't the statement false? Because $F(x), F(x),...$ clearly converges to $F(x)$ but $$\lim_{i\to\infty}deg(F(x)-F(x)) = 0 \neq \infty.$$ If we assume that all $F_i(x)$ are distinct from $F(x),$ then I do not know how to prove the statement.

• What is the degree of the zero power series by definition? – dan_fulea Jun 21 '18 at 22:52
• Undefined? So my example is false? – Idonknow Jun 21 '18 at 22:55
• The above definition should be correspondingly be adapted. There is no least integer, just set it to $\deg 0 =\infty$. Then there is not counterexample using the constant sequence $F,F,F,\dots$ . – dan_fulea Jun 21 '18 at 22:57

Suppose that $\text{deg}(F_{i+1}-F_{i})\to \infty$. So, given $m$, there exists $N$ such that $i\ge N$ implies that $\text{deg}(F_{i+1}-F_{i})>m$. So $[x^m]F_{i}=[x^m]F_{i+1}$, so the coefficent of $x^m$ stabilizes as required.
Now suppose that $F_{i}\to F$. Let $m$ be given. Since $F_{i}\to F$, there exists an $M$ such that $i\geq M$ implies that $[x^{k}]F_{i}=[x^{k}]F_{i+1}$ for $k=0, 1,\dotsc, m$. Then $\deg (F_{i+1}-F_{i})>m$ as desired.
• I assume that the $m$ in second paragraph refers to natural number, right? – Idonknow Jun 22 '18 at 1:57