This question already has an answer here:
By definition you have
$$
\sum_{k=0}^{+\infty}q^k=\lim_{n\to+\infty}\underbrace{\sum_{k=0}^{n}q^k}_{=:S_n}
$$
Notice now that $(1-q)S_n=(1-q)(1+q+q^2+\dots+q^n)=1-q^{n+1}$; so dividing both sides by $1-q$ (in order to do this, you must be careful only to have $1-q\neq0$, i.e. $q\neq1$) we immediately get
$$
S_n=\frac{1-q^{n+1}}{1-q}.
$$
If you now pass to the limit in the above expression, when $|q|<1$, it’s clear that
$$
S_n\stackrel{n\to+\infty}{\longrightarrow}\frac1{1-q}\;\;,
$$
as requested. To get this last result, you should be confident with limits, and know that $\lim_{n\to+\infty}q^n=0$ when $|q|<1$.
Let
$$
S=1+q+q^2+..\inf
$$
$$
S=1+q(1+q+q^2+..\inf)
$$
$$
S=1+qS
$$
$$
S=\frac{1}{1-q}
$$
Without worrying about convergence:
Let $s = 1 + q + q^2 + q^3 + \dotsb = 1 + q(1 + q + q^2 + \dotsb) = 1 + qs$.
So we’ve got $s = 1 + qs$. Rearranging that gives $s = \frac 1 {1 – q}$.
A word of caution: If $|q| \geq 1$, then the series doesn’t converge to any value and the algebra above is nonsense. If $|q| < 1$ then the series converges and the algebra is valid.