For $x > 1$, the geometric series $\sum\limits_{i = 0}^n{x}^i$ is equal to $\frac{x^{n+1}-1}{x-1}$.
By getting the limit,
$$ \lim\limits_{n \rightarrow \infty} \sum\limits_{i = 0}^n{x}^i = \lim\limits_{n \rightarrow \infty} \frac{x^{n+1}-1}{x-1} $$
$$ = \infty $$
However, the generating function $\sum\limits_{n \geq 0}x^n$ is equal to $\frac{1}{1-x}$ since
$$ \sum\limits_{n \geq 0}{x}^n - x\sum\limits_{n \geq 0}{x}^n= (1 + x + x^2 + x^3 + \dots) - (x + x^2 + x^3 + \dots) = 1 $$
$$ \Rightarrow (1-x)\sum\limits_{n \geq 0}{x}^n= 1 $$
$$ \Rightarrow \sum\limits_{n \geq 0}{x}^n= \frac{1}{1-x} $$
Obviously, $\frac{1}{1-x}$ is a finite number.
What is the underlying concept that makes them yield different results?
Although this is not the intended interpretation, one could also make sense of the identity $$ \sum_{n \geq 0} x^n = (1-x)^{-1} \tag{$\dagger$} $$ formally; i.e., in the ring of formal power series $\mathbf C[[x]]$. The user Bill Dubuque explains the idea in several posts in this site; see, e.g., his answer to the post Product of two power series.
The advantage in interpreting $(\dagger)$ formally is that since $x$ is an indeterminate rather than a number, there are no longer any convergence issues. In this sense, the series $\sum_{n \geq 0} x^n$ is the same as the finite thing $(1-x)^{-1}$.
The flip side of this is that one can no longer plug in values into the series: it does not make sense to talk about the sum of the series at $x = \frac{1}{2}$. Thus the statement "the series converges for $|x| < 1$ and diverges otherwise" is meaningless under this interpretation.