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On p.40 of the fourth edition of Serge Lang´s complex analysis book, he states that if $f=\sum_{n\geq 0}a_n T^n$ is a formal power series with $\mathrm{ord}(f)=0$ (i.e. $f$ has constant term $a_0\neq 0$), then $f$ has a formal inverse, which means that there is a formal power series $g=\sum_{n\geq 0} b_n T^n$ such that $fg=1$. His proof goes as follows:

Up to multiplying by $a_0^{-1}$, we are reduced to the case when $a_0=1$. Let $\varphi=-\sum_{n\geq 1} a_n T^n$ so that $f=1-\varphi$. Then we can define the inverse of $f$ as $g=\sum_{n\geq 0} \varphi^n$ and this makes sense since we have defined sums and products of power series so a finite sum $\sum_{k\leq n} \varphi^k$ makes sense. Also $\varphi^n=(-1)^n a_1^n T^n+$ higher terms, so $\mathrm{ord}(\varphi^n)\geq n$ and therefore if $k>n$, the term $\varphi^k$ has all coefficients of order $\leq n$ equal to $0$. Thus we may define the $n$-th coefficient of $g$ to be the $n$-th coefficient of $\sum_{k\leq n} \varphi^k$ and it is then easy to verify that $fg=1+$ a power series of arbitrarily high order, and consecuently is equal to $1$.

My Question

I don't understand why the last power series has an arbitrarily high order and why this implies that the power series is zero. Since $g$ is a power series and have defined the multiplication of power series, if we let $g=\sum_{n\geq 0} b_n T^n$, by the above we have that $b_n=n$-th coefficient of $\sum_{k\leq n} \varphi^k$ and that $fg=1+\sum_{n\geq 1} \left(\sum_{k\leq n} a_k b_{n-k}\right)T^n$ so in order for this series to be zero we need that $\sum_{k\leq n} a_k b_{n-k}=0$ so $b_n=-\sum_{k=1}^n a_k b_{n-k}$, ¿why is this the case? I know we could define $g$ like this from the beginning, but I want to know why Lang says what he says.

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The statement that the $n$-th coefficient of $g$ is defined to be the $n$-th coefficient of $\sum_{k\leq n} \varphi^k$ is a way of making precise the statement that $g = \sum_{n =0}^{\infty} \varphi^n$. If you want to show that $fg = 1$ here, for any $m$ you can write $g = \sum_{n =0}^{m} \varphi^n + R_m$ where $ord(R_m) \geq m + 1$. Then $$fg = f\sum_{n =0}^{m} \varphi^n + f R_m$$ $$= (1 - \varphi)\sum_{n =0}^{m} \varphi^n + (1 - \varphi)R_m$$ $$= 1 - \varphi^{m+1} + (1 - \varphi)R_m$$ Both $\varphi^{m+1}$ and $(1 - \varphi)R_m$ have zeroes of order at least $m + 1$. Since this holds for any $m$ one has $fg = 1$.

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  • $\begingroup$ thank you!! I bet this is what Lang meant. $\endgroup$
    – Nerhú
    Commented Jan 14 at 5:36

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