Making an infinite generating function a finite one If we have some generating function $G(x)$ that generates terms indefinitely, is there a way to translate it to be a finite generating function?
For example if I only want to generate the first $k$ terms of a sequence, can I do $G(x) - x^kG(x)$ or something similar? This isn't the right answer but it's where my thought process is. Trying to find some way to "start" the recurrence at a later point so that when I subtract one infinite generating function from the other, all the terms past $k$ drop out.
 A: The answer from @MarkoRiedel is nice and inspiring. It presents a useful technique to select the first $n$ terms of a series. Here I'd like to provide a small supplement and take a somewhat closer look at his addendum.

Situation: Given a series $G(x)=\sum_{k=0}^\infty g_kx^k$. Find a formula to extract the first $n$ terms of $G(x)$:   \begin{align*}
\sum_{k=0}^ng_kx^k\tag{1}
\end{align*}

In the following we use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series. We obtain a formula in some steps. We start with
Summing up elements
Multiplication of a series $G(x)$ with $\frac{1}{1-x}$ results in summing up the coefficients of $G(x)$.
\begin{align*}
\frac{1}{1-x}G(x)&=\left(\sum_{k=0}^\infty g_kx^k\right)\left(\sum_{l=0}^\infty x^l\right)\\
&=\sum_{n=0}^\infty\left(\sum_{{k+l=n}\atop{k,l\geq 0}}g_k\right)x^n\\
&=\sum_{n=0}^\infty\left(\sum_{k=0}^n g_k\right) x^n\tag{2}
\end{align*}
Shift of variable
The coefficient in (2) is a first step, but it is separated from $x$. Since we want to obtain the  expression stated in (1), we use a trick. We shift the meaning of the variable $x$ and make it part of the coefficient. In order to do so we introduce a new variable $t$ and consider
\begin{align*}
G(tx)=\sum_{k=0}^\infty g_k x^k t^k
\end{align*}
The shift is not fully done, but it is a step in a useful direction. Now we sum up the coefficients, but  by multiplying with $
\frac{1}{1-t}$ instead of multiplying it with $\frac{1}{1-x}$. We obtain
\begin{align*}
\frac{1}{1-t}G(xt)=\sum_{n=0}^\infty\left(\sum_{k=0}^n g_k x^k\right)t^n\tag{3}
\end{align*}
Coefficient extraction
Now the downgrade of $x$ in (3) is complete. We can extract the coefficient of $t^n$ and obtain finally
\begin{align*}
[t^n]\frac{1}{1-t}G(xt)&=[t^n]\sum_{n=0}^\infty\left(\sum_{k=0}^n g_k x^k\right)t^n\\
&=\sum_{k=0}^n g_k x^k
\end{align*}
A: Edited Jan 27 2018. Answer by M.Scheuer is sufficient.
