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

Question

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 1

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 $|q|<1$.

Answer 2

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.

Answer 3

Let $$ S=1+q+q^2+..\inf $$ $$ S=1+q(1+q+q^2+..\inf) $$ $$ S=1+qS $$ $$ S=\frac{1}{1-q} $$