Sum of a geometric sequence, finite and infinite

Start from the definition $S_n = u_1+u_1q+u_1q^2+\cdots+u_1q^{n-1}$. Multiply both sides by $q$: $qS_n = u_1q+u_1q^2+\cdots+u_1q^{n-1}+u_1q^n$. Every term of $qS_n$ except the last, $u_1q^n$, already appears in $S_n$ shifted by one position.

Subtract: $S_n-qS_n$ cancels every shared term $u_1q,u_1q^2,\dots,u_1q^{n-1}$, leaving only the first term of $S_n$ (namely $u_1$) minus the last term of $qS_n$ (namely $u_1q^n$). This gives exactly $(1-q)S_n = u_1 - u_1 q^n$, i.e. $(1-q)S_n=u_1-u_1q^n$.

If $q\neq 1$, divide both sides by $1-q$ (legitimate since $1-q\neq 0$) to get $S_n = u_1\cdot\dfrac{1-q^n}{1-q}\quad (q\neq 1)$. If instead $q=1$, the telescoping identity reads $0\cdot S_n=0$, which carries no information, so this case must be handled directly from the definition: every term equals $u_1$, so $S_n=nu_1$.

Finally take $|q|<1$ and let $n\to\infty$ in the formula $S_n=u_1\cdot\dfrac{1-q^n}{1-q}$: since $|q|<1$ implies $q^n\to 0$ as $n$ grows, the numerator $1-q^n\to 1$, so $S_n\to \dfrac{u_1}{1-q}$. This limit is exactly $S = \dfrac{u_1}{1-q}\quad (|q|<1)$, the sum of the infinite geometric series; if instead $|q|\geq 1$ the term $q^n$ does not shrink to $0$ (it grows or stays constant in size), so the infinite sum does not exist as a finite number in that case.