# if $f$ has a non-zero constant term, then it has an inverse as a power series

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.

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 $$fg = 1$$.