IGCSE Mathematics - Additional (0606) - 2021 Edition 9.06 Infinite geometric series
Lesson

Recall that the sum to $n$n terms of a GP is given by:

$S_n=\frac{t_1\left(1-r^n\right)}{1-r}$Sn=t1(1rn)1r Image -Nicolas Reusens/Barcroft from

http://www.telegraph.co.uk

Suppose a rare species of frog can jump up to $2$2 metres in one bound. One such frog with an interest in mathematics sits $4$4 metres from a wall. The curious amphibian decides to jump $2$2 metres toward it in a single bound. After it completes the feat, it jumps again, but this time only a distance of $1$1 metre. Again the frog jumps, but only $\frac{1}{2}$12 a metre, then jumps again, and again, each time halving the distance it jumps. Will the frog get to the wall?

The total distance travelled toward the wall after the frog jumps $n$n times is given by $S_n=\frac{2\left(1-\left(\frac{1}{2}\right)^n\right)}{1-\frac{1}{2}}$Sn=2(1(12)n)112 .

So after the tenth jump, the total distance is given by $S_{10}=4\left(1-\left(\frac{1}{2}\right)^{10}\right)=4\left(1-\frac{1}{1024}\right)$S10=4(1(12)10)=4(111024)

If we look carefully at the last expression, we realise that, as the frog continues jumping toward the wall, the quantity inside the square brackets approaches the value of $1$1 but will always be less than $1$1. This is the key observation that needs to be made. Therefore the entire sum must remain less than $4$4, but the frog can get as close to $4$4 as it likes simply by continuing to jump according to the geometric pattern described.

Whenever we have a geometric progression with its common ratio within the interval $-11<r<1 then, no matter how many terms we add together, the sum will never exceed some number$L$L called the limiting sum. We sometimes refer to it as the infinite sum of the geometrical progression. Since for any GP,$S_n=\frac{t_1\left(1-r^n\right)}{1-r}$Sn=t1(1rn)1r , if the common ratio is within the interval$-11<r<1 then, as more terms are added, the quantity $\left(1-r^n\right)$(1rn) will become closer and closer to $1$1. This means that the sum will get closer and closer to:

$S_{\infty}=\frac{t_1}{1-r}$S=t11r

Checking our frogs progress, $S_{\infty}=\frac{2}{1-\left(\frac{1}{2}\right)}=4$S=21(12)=4 is the limiting sum.

#### Practice questions

##### QUESTION 1

Consider the infinite geometric sequence: $2$2, $\frac{1}{2}$12, $\frac{1}{8}$18, $\frac{1}{32}$132, $\ldots$

1. Determine the common ratio, $r$r, between consecutive terms.

2. Find the limiting sum of the geometric series.

##### QUESTION 2

Consider the infinite geometric sequence: $125$125, $25$25, $5$5, $1$1, $\ldots$

1. Determine the common ratio, $r$r, between consecutive terms.

2. Find the limiting sum of the geometric series.

##### QUESTION 3

Consider the infinite geometric sequence: $16$16, $-8$8, $4$4, $-2$2, $\ldots$

1. Determine the common ratio between consecutive terms.

2. Find the limiting sum of the geometric series.

### Outcomes

#### 0606C12.5

Use the condition for the convergence of a geometric progression, and the formula for the sum to infinity of a convergent geometric progression.