We have seen for a finite number of terms the sum of a geometric series is given by:
S_n=\frac{t_1\left(1-r^n\right)}{1-r}
We have also seen terms in a geometric series can converge, diverge or oscillate. For example the terms of:
- 27, 9, 3, 1, \frac{1}{3}, \frac{1}{9},... converge - getting smaller and smaller approaching zero
- 5, 10, 20, 40, 80, 160,.... diverge - growing without bound
- 2, -2, 2, -2, 2, -2,.... oscillate - changes between the value of 2 and -2 rather than settling to either value
What determined the behaviour in each case? Since t_n=t_1r^n, the size of the terms as n gets larger is determined by the common ratio r.
Complete the following sentences with the appropriate word (diverge, converge, oscillate)
- If |r|<1, then the terms t_n ________ as n grows larger.
- If |r|>1, then the terms t_n ________ as n grows larger.
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If r=-1, then the terms t_n ________ as n grows larger.
What happens as we sum more and more terms in geometric series? Do we see similar behaviour? If the series converges, can we calculate the limiting value?
Find the sum of the following series for the increasing number of terms indicated and see if you notice any trends.
A. \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+.... for 5 terms, 10 terms, 50 terms.
B. 3+6+12+24+48+.... for 5 terms, 10 terms, 20 terms.
C. \frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+.... for 5 terms, 10 terms, 50 terms.
D. 5-10+20-40+....... for 5 terms, 10 terms, 15 terms, 20 terms.
E. 2-2+2-2+....... for 5 terms, 10 terms, 15 terms, 20 terms.
F. \frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{9}{10000}+.... for 5 terms, 10 terms, 50 terms.
What did you find? Just as with terms in a sequences, as the number of terms grows a series can be convergent (A, C, and F), divergent (B, D and E) and a series can alternate and be convergent (C) or divergent(D and E).
The behaviour of the series as n grows is determined by the common ratio, r:
- If |r|<1 the series is convergent
- If |r|\ge1 the series is divergent
When a series in convergent, what is the limiting value?
For the series A above \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...., what did you find the series was converging to?
This is a famous series and we can picture its converging behaviour as progressively shading areas of a square with side length 1 unit. If we shade half the square, then half the remaining square and so forth we get a picture like the one illustrated below. The limit of shading the areas would be the area of the whole square or 1 square unit.
Hence, if we denote the limiting value of an infinite series as S_\infinity, then for the series:
| S_\infinity | = | \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+.... |
| = | 1 |
We can also see this from the formula for the partial sum of a geometric series, S_n=\frac{t_1\left(1-r^n\right)}{1-r}. For |r|>1, the term r^n will grow without bound as n gets larger and hence give a divergent series. For |r|<1, the term r^n will get smaller and smaller approaching zero as n becomes larger. Hence, the sum will approach the limiting value of \frac{t_1\left(1-0\right)}{1-r}=\frac{t_1}{1-r}.
For any geometric series with first term t_1 and common ratio r, if - 1 < r < 1 the series will converge and the sum to infinity is given by:
S_\infinity=\frac{t_1}{1-r}
Let's see how this formula works in a few examples.
Example 1
Find the sum of the following infinite series.
a) \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+....
Think: We have already seen that this series sums to 1 but let's see how the formula works in this case. We have an infinite geometric series with t_1=\frac{1}{2} and r=\frac{1}{2}.
Do: Hence,
| S_\infinity | = | \frac{\frac{1}{2}}{1-\frac{1}{2}} |
| = | \frac{\frac{1}{2}}{\frac{1}{2}} | |
| = | 1 |
b) \frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+....
Think: Here we have an infinite geometric series with t_1=\frac{1}{2} and r=-\frac{1}{2}.
Do: Hence,
| S_\infinity | = | \frac{\frac{1}{2}}{1+\frac{1}{2}} |
| = | \frac{\frac{1}{2}}{\frac{3}{2}} | |
| = | \frac{1}{3} |
c) 100+20+4+\frac{4}{5}+....
Think: Here we have an infinite geometric series with t_1=100 and r=\frac{1}{5}.
Do: Hence,
| S_\infinity | = | \frac{100}{1-\frac{1}{5}} |
| = | \frac{100}{\frac{4}{5}} | |
| = | 125 |
Applications
That we can add an infinite number of terms and get a finite sum is pretty cool. Let's look at how we can apply this to recurring decimals and some interesting geometric shapes.
Recurring decimals
Have you seen a proof for the fact that 0.\bar{9}=1?
| Let X | = | 0.999999.... |
| Then, 10X | = | 9.999999.... |
| Hence, 10X-X | = | 9.9999999....-0.9999999 |
| 9X | = | 9 |
| \therefore X | = | 1 |
This uses a similar trick that we used when finding the formula for a geometric series, which is not surprising since 0.\bar{9} is equivalent to the infinite series \frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+\frac{9}{10000}+...., which has a starting term t_1=\frac{9}{10} and common ratio r=\frac{1}{10}. So we should be able to use our formula for an infinite series to write any recurring decimal as a fraction.
Lets's look at two examples to see how this works.
Example 2
Use the formula for an infinite series to write the following recurring decimals as fractions.
a) 0.66666...
Think: 0.\bar{6} is equivalent to the infinite geometric series with t_1=\frac{6}{10} and r=\frac{1}{10}.
Do: Hence,
| S_\infinity | = | \frac{\frac{6}{10}}{1-\frac{1}{10}} |
| = | \frac{\frac{6}{10}}{\frac{9}{10}} | |
| = | \frac{2}{3} |
b) 0.45454545....
Think: 0.\bar{45} is equivalent to the infinite geometric series with t_1=\frac{45}{100} and r=\frac{1}{100}
Do: Hence,
| S_\infinity | = | \frac{\frac{45}{100}}{1-\frac{1}{100}} |
| = | \frac{\frac{45}{100}}{\frac{99}{100}} | |
| = | \frac{45}{99} | |
| = | \frac{5}{11} |
Try for yourself to find fraction equivalents for the following:
a) 0.25252525...
b) 0.58333333...
c) 2.727272...
Create a number with a recurring decimal, swap with a friend and rewrite as a fraction. Look for patterns and questions to ponder such as:
- Have you noticed all the decimal expansions of ninths have just a single repeated number? \frac{1}{9}=0.\bar{1},\frac{2}{9}=0.\bar{2},\frac{3}{9}=0.\bar{3}, ...
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Have you noticed all the decimal expansions of sevenths are permutations of the same repeated numbers? \frac{1}{7}=0.\bar{142857},\frac{2}{7}=0.\bar{2857142},\frac{3}{7}=0.\bar{428571}, ...
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The decimal expansion of \frac{1}{81}=0.012345679012345679..., I feel like something is missing.
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The decimal expansion of \frac{1}{23} has 22 digits in a repeated pattern. For a given number n, what is the largest number of digits that can be in a repeated pattern of the decimal expansion of \frac{1}{n}?
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How do we know the decimal expansion of \pi does not ever fall into a repeated pattern?
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\pi can be written as the infinite series \pi=4\times\left(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-...\right), are there other series that result in \pi?
Geometry
Can you imagine a shape with an infinite perimeter but finite area? Let's look at creating such a shape known as a Koch Snowflake.
The snowflake is an example of a fractal curve built from an iterative process. Here is the snowflake after just 4 iterations.
The recipe for making a Koch snowflake is as follows;
- Draw an equilateral triangle with area A as shown in figure (a) below.
- To each of the three sides of figure (a) remove the inner third and build an equilateral triangle with sides of length \frac{1}{3} that of the original triangle, as shown in figure (b).
- To each of the 12 sides of figure (b), remove the inner third and build an equilateral triangle with sides of length \frac{1}{3} that of the triangles formed in step 2.
- To each of the 48 sides of figure (c) repeat the process of removing the inner third and building a new equilateral triangle with side lengths \frac{1}{3} that of the previous step.
- Continue in this manner indefinitely to form what becomes the Koch snowflake.
Let's now show that our emerging shape has an infinite perimeter yet a finite area.
If the length of one side of the original triangle is l, then then length of a single side after n iterations is l_n=\left(\frac{1}{3}\right)^{n}. How many sides are there at each iteration? Create a table and see if you can see a pattern:
| Iteration | Sides |
|---|---|
| 0 | 3 |
| 1 | 12 |
| 2 | |
| 3 | |
| 4 | |
| n |
Can you come up with a rule for how many sides there are after n iterations?
Hence, the perimeter after n iterations is P_n=\text{Number of sides}\times l_n. Show this is a geometric sequence with r>1. Thus we have an infinite perimeter.
The area can be found using a geometric series. If the area of the first triangle is A, what is the area of the 3 smaller triangles at stage (b)? Since the side lengths were scaled down by \frac{1}{3} the area will be scaled down by \frac{1}{9}. Write out the first few terms of the series and try to find the pattern. Other than the first term you should find a geometric series with |r|<1 and hence, the series will converge to a finite total area.
To check your answer and find more information about the Koch Snowflake visit this site.
Further things to ponder include:
- Are there other shapes that have an infinite perimeter but finite area?
- What are some well known fractals? Try researching Mandelbrot set, Sierpinski gasket, Apollonian gasket.
- Can you form a curve similar to the Koch Snowflake by starting with a different shape or adding a different shape to the sides? What happens if you use a non-equilateral triangle? Try here for some inspiration.
- Can you draw an infinite number of elephants across your page? Check out this video for some fun sketches involving series.