sequences and series - Why $sum_{k=0}^{infty} q^k$ sum is $frac{1}{1-q}$ when $|q| < 1$

Why is the infinite sum of $\sum_{k=0}^{\infty} q^k = \frac{1}{1-q}$ when $|q| < 1$

I don't understand how the $\frac{1}{1-q}$ got calculated. I am not a math expert so I am looking for an easy to understand explanation.

Answer

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 $|q|<1$.