5.045 Limiting sums

Limiting sums

Depending on the size and sign of the common ratio, the terms in a geometric series can converge or diverge. Here are some examples:

• $27,9,3,1,\frac{1}{3},\frac{1}{9},\ldots$27,9,3,1,13​,19​,… converge: this means the terms get smaller and smaller, approaching zero
• $5,10,20,40,80,160,\ldots$5,10,20,40,80,160,… diverge: this means the terms grow without bound, as in exponential growth
• $2,-1,\frac{1}{2},-\frac{1}{4},\frac{1}{8},\ldots$2,−1,12​,−14​,18​,… converge: although they are moving between positive and negative values, these terms are also converging to zero

The behaviour of the series as $n$n grows is determined by the common ratio, $r$r. If $|r|<1$|r|<1 the series is convergent. Each subsequent term is smaller than the previous, and this means that the value of the series only changes slightly as $n$n increases.

Investigation

Let's find the sum of the following series$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\ldots$12​+14​+18​+116​+… for $5$5 terms, $10$10 terms, and finally $50$50 terms.

Using:

$S_n=\frac{a\left(1-r^n\right)}{1-r}$Sn​=a(1−rn)1−r​ with $a=\frac{1}{2}$a=12​ and $r=\frac{1}{2}$r=12​

The sums are:

$\frac{31}{32}$3132​, $\frac{1032}{1024}$10321024​ and $\frac{1125899906842623}{1125899906842622}$11258999068426231125899906842622​respectively.

Take a look at those answers: while the numerators and denominators are getting longer, these values are getting closer and closer to $1$1. In fact, if we had tried to type that final sum into a regular scientific calculator, it would have simply rounded the answer to $1$1. It appears that as more terms are added into the sum, the value of the series is approaching $1$1. Another way of saying this is that the limit of the sum is $1$1.

This is a famous series and we can picture its converging behaviour through progressively shading areas of a square with a side length of $1$1 unit. If we shade half the square ($a=\frac{1}{2}$a=12​), then half the remaining square ($T_2=\frac{1}{4}$T2​=14​) and so forth, we get the image below. With each shading, we get closer and closer to shading the whole square, but we never quite manage to. The limit of shading the areas would be the area of the whole square or $1$1 square unit.

 

And so, mathematically we denote the limiting sum of an infinite series such as this as $S_{\infty}$S∞​. For this particular series:

$S_{\infty}$S∞​	$=$=	$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....$12​+14​+18​+116​+....
	$=$=	$1$1

Let's return to the formula for the sum of a geometric series:

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

For $|r|>1$|r|>1, we know the term $r^n$rn will grow exponentially as $n$n gets larger. This gives us ever increasing sums with no limits. But for $|r|<1$|r|<1, the term $r^n$rn will get smaller and smaller, approaching zero as $n$n becomes larger. $r^n$rn will never reach zero, but it will get so close to it just like our sum will never reach the limit we are trying to calculate. In limit notation, this can be written as:

$\lim_{n\to\infty}r^n=0$limn→∞rn=0

So to find a formula for the limiting sum, we can use this limit and let $r^n$rn equal zero, giving us:

$\frac{a\left(1-0\right)}{1-r}=\frac{a}{1-r}$a(1−0)1−r​=a1−r​

This is a very simple formula to use. We might be specifically asked to calculate a limiting sum, but in some cases, the instruction to find the sum of a series that ends with "$...$..." and no clear value of $n$n specified suggest that it is the right time to use this formula.

Worked examples

Example 1

Evaluate $100+20+4+\frac{4}{5}+\ldots$100+20+4+45​+…

Think: Here we have an infinite geometric series with $a=100$a=100 and $r=\frac{1}{5}$r=15​.

Do: Hence,

$S_{\infty}$S∞​	$=$=	$\frac{100}{1-\frac{1}{5}}$1001−15​​
	$=$=	$\frac{100}{\frac{4}{5}}$10045​​
	$=$=	$125$125

 

Converting recurring decimals to fractions

Here is a proof for the fact that $0.\overline{9}=1$0.9=1.

Let $X$X	$=$=	$0.999999....$0.999999....
Then, $10X$10X	$=$=	$9.999999....$9.999999....
Hence, $10X-X$10X−X	$=$=	$9.9999999....-0.9999999$9.9999999....−0.9999999
$9X$9X	$=$=	$9$9
$\therefore X$∴X	$=$=	$1$1

There is another method we can use to convert any recurring decimal to a fraction. It utilises the fact that every recurring decimal can be expressed as a geometric series with a limiting sum.

$0.\overline{9}$0.9 is equivalent to the infinite series $\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{9}{10000}+\ldots$910​+9100​+91000​+910000​+… which has a starting term $a=\frac{9}{10}$a=910​ and common ratio $r=\frac{1}{10}$r=110​. Substituting these values into $S_{\infty}=$S∞​= $\frac{a}{1-r}$a1−r​ we get:

$S_{\infty}$S∞​	$=$=	$\frac{\frac{9}{10}}{1-\frac{1}{10}}$910​1−110​​
$S_{\infty}$S∞​	$=$=	$\frac{\frac{9}{10}}{\frac{9}{10}}$910​910​​
$S_{\infty}$S∞​	$=$=	$1$1

 

Let's look at another example, with a more complicated repeating decimal.

 

Example 2

(b) $0.45454545....$0.45454545....

Think: Take the repeating sections of the decimal and create fractions with the same numerator.

$$ is equivalent to $\frac{45}{100}+\frac{45}{10000}+\frac{45}{1000000}+...$45100​+4510000​+451000000​+...

This is an infinite geometric series with $a=\frac{45}{100}$a=45100​ and $r=\frac{1}{100}$r=1100​.

Do: Hence,

$S_{\infty}$S∞​	$=$=	$\frac{\frac{45}{100}}{1-\frac{1}{100}}$45100​1−1100​​
$S_{\infty}$S∞​	$=$=	$\frac{\frac{45}{100}}{\frac{99}{100}}$45100​99100​​
$S_{\infty}$S∞​	$=$=	$\frac{45}{99}$4599​
$S_{\infty}$S∞​	$=$=	$\frac{5}{11}$511​

Practice questions

Question 1

Question 2