Infinite geometric series

The sum of an infinite series is the limit of its partial sums $S_n$ as $n\to\infty$. For a geometric sequence, $S_n=\frac{a(1-r^n)}{1-r}$ $(r\ne1)$, and when $|r|<1$ we have $r^n\to0$, so

Find the common ratio by dividing the second term by the first. For $3+1+\frac{1}{3}+\cdots$, $r=\frac{1}{3}$ and the sum is $\frac{3}{1-\frac{1}{3}}=\frac{9}{2}$.

Examples

1. 01What is the common ratio of this series?$\displaystyle 1 + \frac{1}{2} + \frac{1}{4} + \cdots$AnswerHide

   Answer$\displaystyle \frac{1}{2}$

   Steps

   1. Look at the ratio of neighbouring terms
   2. $\displaystyle \frac{\frac{1}{2}}{1}$$\displaystyle = \frac{1}{2}$$\displaystyle \frac{\frac{1}{4}}{\frac{1}{2}}$$\displaystyle = \frac{1}{2}$
2. 02What is the sum of this infinite geometric series?$\displaystyle a=6, \quad r=\frac{1}{3}$AnswerHide

   Answer$\displaystyle 9$

   Steps

   1. $\displaystyle 1 - \frac{1}{3}$$\displaystyle = \frac{2}{3}$
   2. Divide the first term by this
   3. $\displaystyle \frac{6}{\frac{2}{3}}$$\displaystyle = 9$
3. 03The common ratio is $\frac{1}{3}$. What is the sum of this infinite geometric series?$\displaystyle 3 + 1 + \frac{1}{3} + \cdots$AnswerHide

   Answer$\displaystyle \frac{9}{2}$

   Steps

   1. $\displaystyle 1 - \frac{1}{3}$$\displaystyle = \frac{2}{3}$
   2. Divide the first term by this
   3. $\displaystyle \frac{3}{\frac{2}{3}}$$\displaystyle = \frac{9}{2}$
4. 04What is the sum of this infinite geometric series?$\displaystyle a=3, \quad r=\frac{1}{4}$AnswerHide

   Answer$\displaystyle 4$

   Steps

   1. $\displaystyle 1 - \frac{1}{4}$$\displaystyle = \frac{3}{4}$
   2. Divide the first term by this
   3. $\displaystyle \frac{3}{\frac{3}{4}}$$\displaystyle = 4$
5. 05What is the common ratio of this series?$\displaystyle 1 + \frac{1}{3} + \frac{1}{9} + \cdots$AnswerHide

   Answer$\displaystyle \frac{1}{3}$

   Steps

   1. Look at the ratio of neighbouring terms
   2. $\displaystyle \frac{\frac{1}{3}}{1}$$\displaystyle = \frac{1}{3}$$\displaystyle \frac{\frac{1}{9}}{\frac{1}{3}}$$\displaystyle = \frac{1}{3}$
6. 06What is the sum of this infinite geometric series?$\displaystyle a=2, \quad r=\frac{1}{2}$AnswerHide

   Answer$\displaystyle 4$

   Steps

   1. $\displaystyle 1 - \frac{1}{2}$$\displaystyle = \frac{1}{2}$
   2. Divide the first term by this
   3. $\displaystyle \frac{2}{\frac{1}{2}}$$\displaystyle = 4$

FAQ

Q1What happens when $|r|\ge1$?AnswerHide

A

If $a\ne0$, the series does not converge (it diverges). For example, with $r=-1$ the partial sums oscillate $a,\ 0,\ a,\ 0,\ \ldots$ and have no limit. If $a=0$, the sum is $0$ for every $r$.

Q2How do I read off the common ratio reliably?AnswerHide

A

Compute the ratio of two neighboring terms. For $1+\frac{1}{2}+\frac{1}{4}+\cdots$ it is $\frac{1}{2}\div1=\frac{1}{2}$. Look at the ratio, not the difference.