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)1−12 .

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(1−11024)

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 $-1−1<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 sum to infinity of the geometrical series.

Since for any GP, $S_n=\frac{t_1\left(1-r^n\right)}{1-r}$Sn=t1(1−rn)1−r , if the common ratio is within the interval $-1−1<r<1 then, as more terms are added, the quantity $\left(1-r^n\right)$(1−rn) 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∞=t11−r

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

Geometric series sum to infinity

For any geometric sequence with starting value $t_1$t1 and common ratio within the interval $-1−1<r<1 we can find the limiting sum $S_{\infty}$S∞ using:

$S_{\infty}=\frac{t_1}{1-r}$S∞=t11−r

Practice questions

QUESTION 1

Consider the infinite geometric sequence: $1$1, $\frac{1}{4}$14, $\frac{1}{16}$116, $\frac{1}{64}$164, $\ldots$…

Determine the common ratio, $r$r, between consecutive terms.
Find the limiting sum of the geometric series.

QUESTION 2

The recurring decimal $0.2222\dots$0.2222… can be expressed as a fraction when viewed as an infinite geometric series.

Express the first decimal place, $0.2$0.2 as an unsimplified fraction.
Express the second decimal place, $0.02$0.02 as an unsimplified fraction.
Hence write, using fractions, the first five terms of the geometric sequence representing $0.2222\dots$0.2222…
State the values of $t_1$t1, the first term, and $r$r, the common ratio, of this sequence.

$t_1$t1$=$=$\editable{}$

$r$r$=$=$\editable{}$
If we add up infinitely many terms of this sequence, we will have the fraction equivalent of our recurring decimal. Calculate the infinite sum of the sequence as a fraction.

QUESTION 3

The limiting sum of the infinite sequence $1$1, $5x$5x, $25x^2$25x2, $\ldots$… is $\frac{9}{4}$94. Solve for the value of $x$x.

QUESTION 4

Find $\sum_{k=1}^{\infty}4^{-k}$∞∑k=14−k.

Outcomes

2.2.7

understand the limiting behaviour as n→∞ of the terms t_n in a geometric sequence and its dependence on the value of the common ratio r

8.06 Infinite geometric series

Sum to infinity