Geometric series - Knowlpedia

For complex numbers $a,r$ , the geometric series with initial term $a$ and ratio $r$ is the series

\sum_{n=0}^{\infty} ar^n.

Its partial sum through index $N$ is

\sum_{n=0}^{N} ar^n= \begin{cases} a(1-r^{N+1})/(1-r),&r\ne1,\\ a(N+1),&r=1. \end{cases}

For $r\ne1$ , subtracting $r$ times the finite sum from the sum proves this identity by cancellation.

Convergence and remainder

For $0<|r|<1$ , the nonnegative decreasing sequence $|r|^n$ has a limit $L$ by the monotone sequence convergence theorem. Passing to the limit in $|r|^{n+1}=|r|\,|r|^n$ gives $L=|r|L$ , hence $L=0$ . The case $r=0$ is immediate. Thus if $|r|<1$ , then $r^{N+1}\to0$ , and the series converges to $a/(1-r)$ . Its tail satisfies

\sum_{n=N+1}^{\infty}ar^n=\frac{ar^{N+1}}{1-r},\qquad \left|\sum_{n=N+1}^{\infty}ar^n\right| \le\frac{|a|\,|r|^{N+1}}{1-|r|}.

If $a\ne0$ and $|r|\ge1$ , the terms do not tend to zero, so the series diverges. When $a=0$ , every term is zero for every $r$ , taking the zeroth power as $1$ .

Example

The series $1+\tfrac12+\tfrac14+\cdots$ has sum $2$ . More generally, if $|u_n|\le Cr^n$ for $C\ge0$ and $0\le r<1$ , the comparison test gives $\sum_n|u_n|\le C/(1-r)$ .