Geometric Series

Much like for arithmetic series, the word ‘series‘ indicates that we should be adding a set of given numbers together. However, the word ‘geometric‘ indicates that the numbers in the series are increasing (or decreasing) by the same factor. We call this the common ratio and is usually denoted $r$ . For example, 1+2+4+8+16+32 is a geometric series with common ratio $r=2$ . We refer to the numbers in the series as ‘terms‘. So, this geometric series has 6 terms and the first term is 1. We usually denote the number of terms $n$ and the first term $a$ . It follows that the general geometric series is as follows:

$a+ar+ar^2+ar^3+…+ar^{n-1}$

Note that the final or nth term in this series is the first term multiplied by the common ratio just $n-1$ times. If there are more terms, we may refer to the nth term in any geometric series as $u_n=ar^{n-1}$ . You should memorise this. It is also possible to add all of the terms (or just the first n terms) of any geometric series using the formula

$S_n=\frac{a\left(1-r^n\right)}{1-r}$

The ‘ $S$ ’ stands for ‘sum’. This formula is given in the Edexcel formula booklet. See below for proof of the formula and see Examples 1 and 2 for some geometric series and their summations.

Note that we can also use sigma notation to write a geometric series in shorthand. $a+ar+ar^2+ar^3+…+ar^{n-1}$ can be written as $\sum_{k=1}^nar^{k-1}$ . It follows that $\sum_{k=1}^nar^{k-1}=\frac{a\left(1-r^n\right)}{1-r}$ and we can use the summation formula to find the sum of any geometric series given in sigma notation. See Example 4 or see more on how to use sigma notation.

Proof of the summation formula for geometric series

The proof of the formula is started off by writing out $S_n$ so the terms are visible. The … indicates that there are some terms in between that follow the pattern as expected. We then multiply both sides by the common ratio $r$ :

$S_n=a+ar+ar^2+ar^3+…+ar^{n-1}$

$rS_n=ar+ar^2+ar^3+ar^4+…+ar^{n}$

noting that both lines have most terms in common except there is an $a$ in the top and an $ar^n$ in the bottom. We proceed by subtracting the bottom from the top giving:

$S_n-rS_n=a-ar^{n}$

We then factorise to get:

$S_n(1-r)=a(1-r^n)$ .

Hence, obtaining $S_n=\frac{a\left(1-r^n\right)}{1-r}$ as required by dividing both sides by $(1-r)$ .

Sum to infinity for Geometric Series

Unlike with arithmetic series, it is possible to take the sum to infinity with a geometric series. This means that we may allow the terms to continue to be added forever. This is only possible, however, if the terms in the series are decreasing in size. It follows that it is possible to take the sum to infinity when the common ratio is between -1 and 1 but (not inclusive). We write $-1<r<1$ or $\vert r\vert <1$ (see more on modding). The common ratio cannot be 1 because the terms will remain the same (or alternate in sign when $r=-1$ ) and adding on the same number forever will result in an infinite answer. We say that it is possible to sum to infinity when the series converges and this happens when $\vert r\vert <1$ . The result when summing to infinity is given by

$S_\infty=\frac{a}{1-r}, \hspace{30pt} \vert r\vert <1$

Note that this is the same as the formula for $S_n$ if we let $n\rightarrow\infty$ . This is because $r^n\rightarrow 0$ as $n\rightarrow\infty$ if $\vert r\vert<1$ . See Example 4 to see how to use the sum to infinity.

Examples

Find the 12th term and the sum of the first 12 terms for the following geometric series:

$3+6+12+24+48+…$

Solution:

The first term of the geometric series is $a=3$ and the common ratio is $r=2$ . Hence the 12th term of the sequence is $u_{12}=3\times 2^{12-1}=6144$ and the sum of the first 12 terms is $S_{12}=\frac{3\left(1-2^{12}\right)}{1-2}=12285$ .

The 3rd term of a geometric sequence is $3$ and the 10th term is $-6561$ . Find the first term $a$ and the common ratio $r$ .

Solution:

We write the 3rd term as $u_3=ar^2=3$ and the 10th term as $u_{10}=ar^9=-6561$ . Hence, we have simultaneous equations. It follows that:

$\frac{u_{10}}{u_3}=\frac{ar^9}{ar^2}=\frac{-6561}{3}=-2187$

That is $r^7=-2187$ and so the common ratio is $r=\sqrt[7]{-2187}=-3$ . It follows from $u_3$ that $9a=3$ and so the first term is $a=\frac{1}{3}$ .

The sum of the first $k$ terms of a geometric series whose nth term is defined by $u_n=4\times (0.75)^{n-1}$ is 15.97146 to 5 decimal places. Find $k$ .

Solution:

We use the summation formula with first term $a=4$ and common ratio $r=0.75$ :

$S_k=\frac{4\left(1-0.75^k\right)}{1-0.75}=15.97146$

which we solve for $k$ . The expression simplifies to $16\left(1-0.75^k\right)=15.97146$ . Dividing both sides by 16 gives $1-0.75^k=0.99822$ or $0.75^k=1.78375\times 10^{-3}$ . We can solve for $k$ using logs: $\ln{\left(0.75^k\right)}=\ln{\left(1.78375\times 10^{-3}\right)}$ – this works for logs to any base. Using the log laws we can write $k\ln{0.75}=\ln{\left(1.78375\times 10^{-3}\right)}$ and so $k=\frac{\ln{\left(1.78375\times 10^{-3}\right)}}{\ln{0.75}}=22.00011$ to 5 decimal places. This suggests that $k=22$ which we check very simply by putting back into the formula for $S_{22}$ .

Evaluate the following:

$\sum_{i=1}^{15}7\left(\frac{2}{3}\right)^i$
$\sum_{i=1}^{\infty}7\left(\frac{2}{3}\right)^i$

Solution:

We use the $S_n$ formula to find the sum of the first 15 terms of the geometric series with first term $a=\frac{14}{3}$ and $r=\frac{2}{3}$ : $S_{15}=\frac{\frac{14}{3}\left(1-\left(\frac{2}{3}\right)^{15}\right)}{1-\frac{2}{3}}=13.968$ to 3 decimal places.
We can take the sum to infinity since $r=\frac{2}{3}$ which satisfies $\vert r\vert<1$ . So, we use the $S_\infty$ formula to find the sum to infinity with the same $a$ and $r$ : $S_\infty=\frac{\frac{14}{3}}{1-\frac{2}{3}}=14$ .

Geometric Series

Proof of the summation formula for geometric series

Sum to infinity for Geometric Series

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