Infinite sum for GP's

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

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

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)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 infinite sum of the geometrical progression.

Since for any GP, $S_n=\frac{a\left(1-r^n\right)}{1-r}$Sn​=a(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:

$S_{\infty}=\frac{a}{1-r}$S∞​=a1−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. 

Practice questions

QUESTION 1

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QUESTION 3