${\displaystyle a_{n}=a\,r^{n-1}}$

${\displaystyle a_{n}=r\,a_{n-1}}$

${\displaystyle {\frac {a_{4}}{a_{3}}},{\frac {a_{3}}{a_{2}}}}$

${\displaystyle 20,10,5,2.5,1.25,...}$

${\displaystyle {\frac {a_{4}}{a_{3}}}={\frac {2.5}{5}}=0.5}$

${\displaystyle {\frac {a_{3}}{a_{2}}}={\frac {5}{10}}=0.5}$

${\displaystyle \sum _{k=0}^{n}ar^{k}=ar^{0}+ar^{1}+ar^{2}+ar^{3}+\cdots +ar^{n}.\,}$

${\displaystyle {\begin{aligned}(1-r)\sum _{k=0}^{n}ar^{k}&=(1-r)(ar^{0}+ar^{1}+ar^{2}+ar^{3}+\cdots +ar^{n})\\&=ar^{0}+ar^{1}+ar^{2}+ar^{3}+\cdots +ar^{n}\\&{\color {White}{}=ar^{0}}-ar^{1}-ar^{2}-ar^{3}-\cdots -ar^{n}-ar^{n+1}\\&=a-ar^{n+1}\end{aligned}}}$

${\displaystyle \sum _{k=0}^{n}ar^{k}={\frac {a(1-r^{n+1})}{1-r}}.}$

${\displaystyle \sum _{k=m}^{n}ar^{k}={\frac {a(r^{m}-r^{n+1})}{1-r}}.}$

${\displaystyle \sum _{k=0}^{n}k^{s}r^{k}.}$

${\displaystyle {\frac {d}{dr}}\sum _{k=0}^{n}r^{k}=\sum _{k=1}^{n}kr^{k-1}={\frac {1-r^{n+1}}{(1-r)^{2}}}-{\frac {(n+1)r^{n}}{1-r}}.}$

${\displaystyle (1-r^{2})\sum _{k=0}^{n}ar^{2k}=a-ar^{2n+2}.}$

${\displaystyle \sum _{k=0}^{n}ar^{2k}={\frac {a(1-r^{2n+2})}{1-r^{2}}}.}$

${\displaystyle (1-r^{2})\sum _{k=0}^{n}ar^{2k+1}=ar-ar^{2n+3}}$

${\displaystyle \sum _{k=0}^{n}ar^{2k+1}={\frac {ar(1-r^{2n+2})}{1-r^{2}}}.}$

${\displaystyle \sum _{k=0}^{\infty }ar^{k}=\lim _{n\to \infty }{\sum _{k=0}^{n}ar^{k}}=\lim _{n\to \infty }{\frac {a(1-r^{n+1})}{1-r}}=\lim _{n\to \infty }{\frac {a}{1-r}}-\lim _{n\to \infty }{\frac {ar^{n+1}}{1-r}}}$

${\displaystyle r^{n+1}\to 0{\mbox{ as }}n\to \infty {\mbox{ when }}|r|<1.}$

${\displaystyle \sum _{k=0}^{\infty }ar^{k}={\frac {a}{1-r}}-0={\frac {a}{1-r}}}$

${\displaystyle \sum _{k=0}^{\infty }ar^{2k}={\frac {a}{1-r^{2}}}}$

${\displaystyle \sum _{k=0}^{\infty }ar^{2k+1}={\frac {ar}{1-r^{2}}}}$

${\displaystyle \sum _{k=m}^{\infty }ar^{k}={\frac {ar^{m}}{1-r}}}$

${\displaystyle {\frac {d}{dr}}\sum _{k=0}^{\infty }r^{k}=\sum _{k=0}^{\infty }kr^{k-1}={\frac {1}{(1-r)^{2}}}}$

${\displaystyle \sum _{k=0}^{\infty }kr^{k}={\frac {r}{\left(1-r\right)^{2}}}\,;\,\sum _{k=0}^{\infty }k^{2}r^{k}={\frac {r\left(1+r\right)}{\left(1-r\right)^{3}}}\,;\,\sum _{k=0}^{\infty }k^{3}r^{k}={\frac {r\left(1+4r+r^{2}\right)}{\left(1-r\right)^{4}}}}$

${\displaystyle {\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{16}}+\cdots ={\frac {1/2}{1-(+1/2)}}=1.}$

${\displaystyle {\frac {1}{2}}-{\frac {1}{4}}+{\frac {1}{8}}-{\frac {1}{16}}+\cdots ={\frac {1/2}{1-(-1/2)}}={\frac {1}{3}}.}$

${\displaystyle \sum _{k=0}^{\infty }{\frac {\sin(kx)}{r^{k}}}={\frac {r\sin(x)}{1+r^{2}-2r\cos(x)}}}$

${\displaystyle \sin(kx)={\frac {e^{ikx}-e^{-ikx}}{2i}},}$

${\displaystyle \sum _{k=0}^{\infty }{\frac {\sin(kx)}{r^{k}}}={\frac {1}{2i}}\left[\sum _{k=0}^{\infty }\left({\frac {e^{ix}}{r}}\right)^{k}-\sum _{k=0}^{\infty }\left({\frac {e^{-ix}}{r}}\right)^{k}\right]}$.

${\displaystyle \prod _{i=0}^{n}ar^{i}=\left({\sqrt {a_{1}\cdot a_{n+1}}}\right)^{n+1}}$ (if ${\displaystyle a,r>0}$).

${\displaystyle P=a\cdot ar\cdot ar^{2}\cdots ar^{n-1}\cdot ar^{n}}$.

${\displaystyle P=a^{n+1}r^{1+2+3+\cdots +(n-1)+(n)}}$.

${\displaystyle P=a^{n+1}r^{\frac {n(n+1)}{2}}}$.

${\displaystyle P=(ar^{\frac {n}{2}})^{n+1}}$.

${\displaystyle P^{2}=(a^{2}r^{n})^{n+1}=(a\cdot ar^{n})^{n+1}}$.

${\displaystyle 4,x,16}$

${\displaystyle x={\sqrt {4\times 16}}=8}$

• Hall & Knight, Higher Algebra, p.  ۳۹، ISBN 81-8116-000-2
• Weisstein, Eric W. "Geometric Series". MathWorld.