Sequences and Series

Hong Kong

Stage 4 - Stage 5

Lesson

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

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

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 $S_n=\frac{2\left(1-\left(\frac{1}{2}\right)^n\right)}{1-\frac{1}{2}}$`S``n`=2(1−(12)`n`)1−12 .

So after the tenth jump, the total distance is given by $S_{10}=4\left(1-\left(\frac{1}{2}\right)^{10}\right)=4\left(1-\frac{1}{1024}\right)$`S`10=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`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, $S_n=\frac{a\left(1-r^n\right)}{1-r}$`S``n`=`a`(1−`r``n`)1−`r` , if the common ratio is within the interval $-1`r`<1 then, as more terms are added, the quantity $\left(1-r^n\right)$(1−`r``n`) will become closer and closer to $1$1. This means that the sum will get closer and closer to:

$S_{\infty}=\frac{a}{1-r}$`S`∞=`a`1−`r`

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

As another example, the limiting sum of the geometric series $108+36+12\dots$108+36+12… is simply $S_{\infty}=\frac{108}{1-\left(\frac{1}{3}\right)}=162$`S`∞=1081−(13)=162.

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

Determine the common ratio, $r$

`r`, between consecutive terms.Find the limiting sum of the geometric series.

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

Determine the common ratio, $r$

`r`, between consecutive terms.Find the limiting sum of the geometric series.

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

Determine the common ratio between consecutive terms.

Find the limiting sum of the geometric series.