finite geometric series and infinite geometric series

finite geometric series and infinite geometric series

This post contains details about finite geometric series and infinite geometric series. It also illustrates with an example.

Finite Geometric Series:

A Finite Geometric Series is the geometric series Which has finite number of terms in it.
e.g. 2+4+8+16+32+…..10 terms is a finite geometric series as this series has 10 terms.

3+12+48+……….+768 is a finite geometric series as this series has finite number of terms.

Sum of finite Geometric Series is given by S = \frac{a({r}^{n}-1)}{(r-1)}

Where , a = first term of the geometric series
r = Common Ratio = \frac{nthterm}{(n-1)thterm}
n = number of terms in the geometric series

Whether the Common ratio r is greater than 1 or less than 1, it remains the same formula.

InFinite Geometric Series:

An infinite geometric series is an infinite series whose successive terms have a common ratio. Such a series converges if and only if the absolute value of the common ratio is less than one (| r | < 1). If the Common ratio is (| r | > 1) , then it does not converge and it will go towards Infinity. If the common ration(r) is greater than 1, then it will go towards positive infinity and if the common ratio is less than -1 , then it will go towards negative infinity.

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

As ,|r| is less than 1 , so

{r}^{(n+1)} -> 0 as n -> \infty

So ,

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

So, Sum of Infinite Geometric Series where Common ratio |r| < 1 = \frac { a }{ 1-r }

(Q.) Find the Sum of the Series

\frac{1}{3} + \frac{1}{9}+\frac{1}{27}+\frac{1}{81}+\frac{1}{243}+.....................

Solution : Common Ratio = \frac{SecondTerm}{FirstTerm} = \frac{\frac{1}{9}}{\frac{1}{3}} = \frac{1}{3} Which is less than 1

Hence , Sum of the given infinite geometric series = \frac{a}{1-r} = \frac{\frac{1}{3}}{1-\frac{1}{3}} = \frac{1}{2} Ans.

For More Detail about Geometric Progression , Click Here Geometric Sequence