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# 1.6 Infinite series

## 1.6 Infinite series (EMCF3)

So far we have been working only with finite sums, meaning that whenever we determined the sum of a series, we only considered the sum of the first $$n$$ terms. We now consider what happens when we add an infinite number of terms together. Surely if we sum infinitely many numbers, no matter how small they are, the answer goes to infinity? In some cases the answer does indeed go to infinity (like when we sum all the positive integers), but surprisingly there are some cases where the answer is a finite real number.

## Sum of an infinite series

1. Cut a piece of string $$\text{1}$$ $$\text{m}$$ in length.
2. Now cut the piece of string in half and place one half on the desk.
3. Cut the other half in half again and put one of the pieces on the desk.
4. Repeat this process until the piece of string is too short to cut easily.
5. Draw a diagram to illustrate the sequence of lengths of the pieces of string.
6. Can this sequence be expressed mathematically? Hint: express the shorter lengths of string as a fraction of the original length of string.
7. What is the sum of the lengths of all the pieces of string?
8. Predict what would happen if these steps could be repeated infinitely many times.
9. Will the sum of the lengths of string ever be greater than $$\text{1}$$?
10. What can you conclude?

## Worked example 14: Sum to infinity

Complete the table below for the geometric series $$T_{n} = \left( \frac{1}{2} \right)^{n}$$ and answer the questions that follow:

 Terms $$S_{n}$$ $$1 - S_{n}$$ $$T_{1}$$ $$\frac{1}{2}$$ $$\frac{1}{2}$$ $$\frac{1}{2}$$ $$T_{1} + T_{2}$$ $$T_{1} + T_{2} + T_{3}$$ $$T_{1} + T_{2} + T_{3} + T_{4}$$
1. As more and more terms are added, what happens to the value of $$S_{n}$$?
2. As more more and more terms are added, what happens to the value of $$1 - S_{n}$$?
3. Predict the maximum value of $$S_{n}$$ for the sum of infinitely many terms in the series.

### Complete the table

 Terms $$S_{n}$$ $$1 - S_{n}$$ $$T_{1}$$ $$\frac{1}{2}$$ $$\frac{1}{2}$$ $$\frac{1}{2}$$ $$T_{1} + T_{2}$$ $$\frac{1}{2} + \frac{1}{4}$$ $$\frac{3}{4}$$ $$\frac{1}{4}$$ $$T_{1} + T_{2} + T_{3}$$ $$\frac{1}{2} + \frac{1}{4} + \frac{1}{8}$$ $$\frac{7}{8}$$ $$\frac{1}{8}$$ $$T_{1} + T_{2} + T_{3} + T_{4}$$ $$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16}$$ $$\frac{15}{16}$$ $$\frac{1}{16}$$

### Consider the value of $$S_{n}$$ and $$1 - S_{n}$$

As more terms in the series are added together, the value of $$S_{n}$$ increases:

$\frac{1}{2} \quad < \quad \frac{3}{4} \quad < \quad \frac{7}{8} \quad < \quad \cdots$

However, by considering $$1 - S_{n}$$, we notice that the amount by which $$S_{n}$$ increases gets smaller and smaller as more terms are added:

$\frac{1}{2} \quad > \quad \frac{1}{4} \quad > \quad \frac{1}{8} \quad > \quad \cdots$

We can therefore conclude that the value of $$S_n$$ is approaching a maximum value of $$\text{1}$$; it is converging to $$\text{1}$$.

### Write conclusion mathematically

We can conclude that the sum of the series

$\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots$

gets closer to 1 ($$S_{n} \rightarrow 1$$) as the number of terms approaches infinity ($$n \rightarrow \infty$$), therefore the series converges.

$\sum _{i = 1}^{\infty}{ \left( \frac{1}{2} \right)^{i}} = 1$
temp text

We express the sum of an infinite number of terms of a series as

${S}_{\infty }=\sum _{i=1}^{\infty }{T}_{i}$

Convergence and divergence

If the sum of a series gets closer and closer to a certain value as we increase the number of terms in the sum, we say that the series converges. In other words, there is a limit to the sum of a converging series. If a series does not converge, we say that it diverges. The sum of an infinite series usually tends to infinity, but there are some special cases where it does not.

## Convergent and divergent series

Textbook Exercise 1.10

For each of the general terms below:

• Determine if it forms an arithmetic or geometric series.
• Calculate $$S_{1}, S_{2}, S_{10} \text{ and } S_{100}$$.
• Determine if the series is convergent or divergent.

$$T_{n} = 2n$$

$$2+4+6+8+ \ldots$$

\begin{align*} a &= 2 \\ d &= 2 \\ \therefore & \text{ this is an arithmetic series } \\ S_{1} &= 2 \\ S_{2} &= 2 + 4 = 6 \\ S_{n} & = \frac{n}{2}(2a + [n-1]d) \\ S_{10} &= 5(2(2) + 9(2)) = 110 \\ S_{100} &= 50(2(2) + 99(2)) = 10100 \end{align*} \begin{align*} a &= 2 \\ S_{n} &= \frac{n}{2}(2(2) + [n - 1]2) \\ &= \frac{n}{2}(2 + 2n) \\ &= n(1 + n) \\ &= n + n^{2} \\ \therefore S_{n} \to \infty & \text{ as } n \to \infty \end{align*}

Therefore, this is a divergent series.

$$T_{n} = (-n)$$

$$(-1) + (-2) + (-3) + (-4)+\ldots$$

\begin{align*} a &= -1 \\ d &= -1 \\ \therefore & \text{ this is an arithmetic series } \\ S_{1} &= -1 \\ S_{2} &= -1 -2 = -3 \\ S_{n} & = \frac{n}{2}(2a + [n-1]d) \\ S_{10} &= 5(2(-1) + 9(-1)) = -55 \\ \\ S_{100} &= 50(2(-1) + 99(-1)) = -5050 \\ & \\ S_{n} & = \frac{n}{2}(2(-1) + [n-1](-1)) \\ & = \frac{n}{2}(-2 - n + 1) \\ & = - \frac{n}{2}(n + 1) \\ \therefore S_{n} \to - \infty & \text{ as } n \to \infty \end{align*}

This is a divergent series.

$$T_{n} = (\frac{2}{3})^{n}$$

$$\frac{2}{3}+\frac{4}{9}+\frac{8}{27}+\frac{16}{81}+\ldots$$

\begin{align*} a &= \frac{2}{3} \\ r &= \frac{2}{3} \\ \therefore & \text{ this is a geometric series (with } r < 1) \\ S_{1} &= \frac{2}{3} = \text{0,666}\ldots \\ S_{2} &= \frac{2}{3} +\frac{4}{9} = \frac{10}{9} = \text{1,111}\ldots \\ S_{n} & = \frac{a(1-r^n)}{1-r} \\ S_{10} &= \frac{\frac{2}{3}\left( 1 - \left( \frac{2}{3} \right)^{10} \right)}{1 - \frac{2}{3}} \\ \\ &= 2\left( 1 - \left( \frac{2}{3} \right)^{10} \right) \\ &= \text{1,965}\ldots \\ S_{100} &= \frac{\frac{2}{3}\left( 1 - \left( \frac{2}{3} \right)^{100} \right)}{1 - \frac{2}{3}} \\ \\ &= 2\left( 1 - \left( \frac{2}{3} \right)^{100} \right) \\ &= \text{2,00} \ldots \\ & \\ S_{n} & = \frac{\frac{2}{3} (1- \left( \frac{2}{3} \right)^n)}{1 - \frac{2}{3}} \\ & = \frac{\frac{2}{3} \left(1- \left( \frac{2}{3} \right)^n \right)}{ \frac{1}{3} } \\ & = 2 \left(1 - \left( \frac{2}{3} \right)^{n} \right) \\ & = 2 - 2 \left( \frac{2}{3} \right)^{n} \\ \text{ Therefore, as } n \to \infty & \enspace S_{n} \to 2 \end{align*}

This series is convergent (since the $$r < 1$$) and converges to $$\text{2}$$.

$$T_{n} = 2^{n}$$

$$2 + 4 + 8 + 16 +\ldots$$

\begin{align*} a &= 2 \\ r &= 2 \\ \therefore & \text{ this is a geometric series (with r } > 1) \\ S_{1} &= 2\\ S_{2} &= 2 + 4 = 6 \\ S_{n} & = \frac{a(r^n - 1)}{r - 1} \\ S_{10} &= \frac{2 \left( \left( 2 \right)^{10} - 1 \right)}{2 - 1} \\ \\ &= 2 \left( \left( 2 \right)^{10} - 1 \right) \\ &= \text{2 046} \\ S_{100} &= \frac{2\left( \left( 2 \right)^{100} - 1 \right)}{2 - 1} \\ \\ &= 2\left( \left( 2 \right)^{100} - 1 \right) \\ &= \text{2,5} \times \text{10}^{\text{30}} \\ & \\ S_{n} & = \frac{2((2)^{n} - 1)}{2 - 1} \\ & = (2)^{n + 1} - 2 \\ \therefore S_{n} \to \infty & \text{ as } n \to \infty \end{align*}

This is a divergent series.

Note the following:

• An arithmetic series never converges: as $$n$$ tends to infinity, the series will always tend to positive or negative infinity.
• Some geometric series converge (have a limit) and some diverge (as $$n$$ tends to infinity, the series does not tend to any limit or it tends to infinity).

### Infinite geometric series (EMCF4)

There is a simple test for determining whether a geometric series converges or diverges; if $$-1 < r < 1$$, then the infinite series will converge. If $$r$$ lies outside this interval, then the infinite series will diverge.

Test for convergence:

• If $$-1 < r < 1$$, then the infinite geometric series converges.
• If $$r < -1 \text{ or } r > 1$$, then the infinite geometric series diverges.

We derive the formula for calculating the value to which a geometric series converges as follows:

${S}_{n}=\sum _{i=1}^{n}{a}{r}^{i-1}=\frac{{a} \left(1 - {r}^{n}\right)}{1 - r}$

Now consider the behaviour of $${r}^{n}$$ for $$-1<r<1$$ as $$n$$ becomes larger.

Let $$r=\frac{1}{2}$$:

\begin{align*} n=1& : {r}^{n}={r}^{1}={\left(\frac{1}{2}\right)}^{1}=\frac{1}{2} \\ n=2& : {r}^{n}={r}^{2}={\left(\frac{1}{2}\right)}^{2}=\frac{1}{2} \cdot \frac{1}{2}=\frac{1}{4}<\frac{1}{2} \\ n=3& : {r}^{n}={r}^{3}={\left(\frac{1}{2}\right)}^{3}=\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}=\frac{1}{8} <\frac{1}{4} \end{align*}

Since $$r$$ is in the range $$-1<r<1$$, we see that $${r}^{n}$$ gets closer to $$\text{0}$$ as $$n$$ gets larger. Therefore $$(1 - r^{n})$$ gets closer to $$\text{1}$$.

Therefore,

\begin{align*} {S}_{n}& = \frac{{a}\left(1 - {r}^{n} \right)}{1 - r} \\ \text{If } -1 < r < 1, \quad & \text{then } r^{n} \rightarrow 0 \text{ as } n \rightarrow \infty \\ \therefore {S}_{\infty }& = \frac{{a}\left(1 - 0 \right)}{1 - r} \\ & = \frac{{a}}{1-r} \end{align*}

The sum of an infinite geometric series is given by the formula

$\therefore {S}_{\infty }=\sum _{i=1}^{\infty }{a}{r}^{i-1}=\frac{{a}}{1-r} \qquad (-1<r<1)$

where

• $${a}$$ is the first term of the series;
• $$r$$ is the constant ratio.

Alternative notation:

$\underbrace{{S}_{n}}_{n \rightarrow \infty} \rightarrow \frac{a}{1 - r} \quad \text{ if }-1<r<1$

In words: as the number of terms $$(n)$$ tends to infinity, the sum of a converging geometric series $$(S_{n})$$ tends to the value $$\frac{a}{1 - r}$$.

## Worked example 15: Sum to infinity of a geometric series

Given the series $$18 + 6 + 2 + \cdots$$. Find the sum to infinity if it exists.

### Determine the value of $$r$$

We need to know the value of $$r$$ to determine whether the series converges or diverges.

\begin{align*} \frac{T_{2}}{T_{1}} &= \frac{6}{18} \\ &= \frac{1}{3} \\ \frac{T_{3}}{T_{2}} &= \frac{2}{6} \\ &= \frac{1}{3} \\ \therefore r &= \frac{1}{3} \end{align*}

Since $$- 1 < r < 1$$, we can conclude that this is a convergent geometric series.

### Determine the sum to infinity

Write down the formula for the sum to infinity and substitute the known values:

$a = 18; \qquad r = \frac{1}{3}$

\begin{align*} S_{\infty} &= \frac{a}{1 - r} \\ &= \frac{18}{ 1 - \frac{1}{3}} \\ &= \frac{18}{\frac{2}{3}} \\ &= 18 \times \frac{3}{2} \\ &= 27 \end{align*}

As $$n$$ tends to infinity, the sum of this series tends to $$\text{27}$$; no matter how many terms are added together, the value of the sum will never be greater than $$\text{27}$$.

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## Worked example 16: Using the sum to infinity to convert recurring decimals to fractions

Use two different methods to convert the recurring decimal $$0,\dot{5}$$ to a proper fraction.

### Convert the recurring decimal to a fraction using equations

\begin{align*} \text{Let } x &= \text{0,}\dot{\text{5}}\\ \therefore x &= \text{0,555} \ldots \ldots (1) \\ 10x &= \text{5,55} \ldots \ldots (2) \\ (2) - (1): \quad 9x &= 5 \\ \therefore x &= \frac{5}{9} \end{align*}

### Convert the recurring decimal to a fraction using the sum to infinity

\begin{align*} \text{0,}\dot{\text{5}} &= \text{0,5} + \text{0,05} + \text{0,005} + \ldots \\ \text{or } \quad \text{0,}\dot{\text{5}} &= \frac{5}{10} + \frac{5}{100} + \frac{5}{1000} + \ldots \end{align*}

This is a geometric series with $$r = \text{0,1} = \frac{1}{10}$$. And since $$- 1 < r < 1$$, we can conclude that the series is convergent.

\begin{align*} S_{\infty} &= \frac{a}{1 - r} \\ &= \frac{\frac{5}{10}}{1 - \frac{1}{10}} \\ &= \dfrac{\frac{5}{10}}{\frac{9}{10}} \\ &= \frac{5}{9} \end{align*}

## Worked example 17: Sum to infinity

Determine the possible values of $$a$$ and $$r$$ if

$\sum _{n=1}^{\infty}{ar^{n-1}} = 5$

### Write down the sum to infinity formula and substitute known values

\begin{align*} S_{\infty} &= \frac{a}{1 - r} \\ \therefore 5 &= \frac{a}{1 - r} \\ a &= 5(1 - r) \\ \therefore a &= 5 - 5r \\ \text{And } 5r &= 5 - a \\ \therefore r &= \frac{5 - a}{5} \end{align*}

### Apply the condition for convergence to determine possible values of $$a$$

For a series to converge: $$- 1 < r < 1$$

\begin{align*} -1 & < r < 1 \\ -1 & < \frac{5 - a}{5} < 1 \\ -5 & < 5 - a < 5 \\ -10 & < - a < 0 \\ 0 & < a < 10 \end{align*}

For the series to converge, $$0 < a < 10$$ and $$- 1 < r < 1$$.

## Sum to infinity

Textbook Exercise 1.11

What value does $${\left(\frac{2}{5}\right)}^{n}$$ approach as $$n$$ tends towards $$\infty$$?

\begin{align*} S_{n} &= \frac{2}{5} + \frac{4}{25} + \frac{8}{125} + \ldots \\ \therefore a &= \frac{2}{5} \\ \text{And } r &= \dfrac{\frac{4}{25}}{\frac{2}{5}} \\ &= \frac{2}{5} \qquad (- 1 < r < 1) \\ \text{So then } S_{\infty} &= \frac{a}{1-r} \\ &= \frac{\frac{2}{5}}{1-\frac{2}{5}} \\ &= \frac{\frac{2}{5}}{\frac{3}{5}} \\ &= \frac{2}{3} \end{align*}

Find the sum to infinity of the geometric series $$3+1+\frac{1}{3}+\frac{1}{9}+ \cdots$$

\begin{align*} a &= 3 \\ r &= \frac{1}{3} \\ S_{\infty} &= \frac{a}{1 -r} \\ &= \frac{3}{1 - \frac{1}{3}} \\ &= \frac{3}{\frac{2}{3}} \\ &= \frac{9}{2} \end{align*}

Determine for which values of $$x$$, the geometric series $$2+\frac{2}{3} \left(x+1\right)+\frac{2}{9} {\left(x+1\right)}^{2}+ \cdots$$ will converge.

\begin{align*} a &= 2 \\ r &= \frac{\frac{2}{3} \left(x+1\right)}{2} \\ &= \frac{1}{3} \left(x+1\right) \end{align*}

For the series to converge, $$-1 < r < 1$$, therefore:

\begin{align*} -1&< r < 1 \\ -1 &< \frac{1}{3} \left(x+1\right) < 1 \\ -3 &< \left(x+1\right) < 3 \\ -3-1 &< x < 3-1 \\ -4 &< x < 2 \end{align*}

The sum to infinity of a geometric series with positive terms is $$4\frac{1}{6}$$ and the sum of the first two terms is $$2\frac{2}{3}$$. Find $$a$$, the first term, and $$r$$, the constant ratio between consecutive terms.

\begin{align*} T_{1} + T_{2} &= \frac{8}{3} \\ \therefore a + ar &= \frac{8}{3} \\ a(1 + r) &= \frac{8}{3} \\ \therefore a &= \frac{8}{3(1 + r)} \ldots \ldots (1) \\ S_{\infty} &= 4\frac{1}{6} = \frac{25}{6} \\ \therefore \frac{a}{1 -r}&= \frac{25}{6} \\ 6a &= 25(1 -r) \ldots \ldots (2) \\ \text{Substitute eqn. } (1) \rightarrow (2): \quad 6 \left( \frac{8}{3(1 + r)} \right)&= 25(1 -r) \\ 16 &= 25(1 -r)(1 + r)\\ 16 &= 25(1 - r^{2}) \\ 16 &= 25 - 25r^{2} \\ 25r^{2} &= 25 - 16 \\ 25r^{2} &= 9 \\ r^{2} &= \frac{9}{25} \\ \therefore r &= \pm \frac{3}{5} \\ \text{But } T_{n} & > 0 \\ \therefore r &= \frac{3}{5} \\ \text{And } a &= \frac{8}{3(1 + r)} \\ &= \frac{8}{3 + 3r} \\ &= \frac{8}{3 + 3 \left(\frac{3}{5} \right)} \\ &= \frac{8}{\frac{15}{5} + \frac{9}{5}} \\ &= \frac{8}{\frac{24}{5}} \\ \therefore a &= \frac{5}{3} \end{align*}

Use the sum to infinity to show that $$\text{0,}\dot{\text{9}} = 1$$.

Rewrite the recurring decimal:

$\text{0,}\dot{\text{9}} = \frac{9}{10} + \frac{9}{100} + + \frac{9}{1000} + \ldots$

This is a geometric series with $$a = \frac{9}{10}$$ and $$r = \frac{1}{10}$$.

\begin{align*} S_{\infty} &= \frac{a}{1 - r} \\ &= \frac{\frac{9}{10}}{1 - \frac{1}{10}} \\ &= \frac{\frac{9}{10}}{\frac{9}{10}} \\ &= 1 \end{align*}

A shrub $$\text{110}$$ $$\text{cm}$$ high is planted in a garden. At the end of the first year, the shrub is $$\text{120}$$ $$\text{cm}$$ tall. Thereafter the growth of the shrub each year is half of it's growth in the previous year. Show that the height of the shrub will never exceed $$\text{130}$$ $$\text{cm}$$. Draw a graph of the relationship between time and growth.

[IEB, Nov 2003]

Write the annual growth of the shrub as a series:

$10 + 5 + \frac{5}{2} + \frac{5}{4} + \ldots$

This is a geometric series with $$a = 10$$ and $$r = \frac{1}{2}$$.

\begin{align*} S_{\infty} &= \frac{a}{1 - r} \\ &= \frac{10}{1 - \frac{1}{2}} \\ &= \frac{10}{\frac{1}{2}} \\ &= 20 \end{align*}

Therefore the growth of the shrub is limited to $$\text{20}$$ $$\text{cm}$$, and the maximum height of the shrub is therefore $$\text{110}$$ $$\text{cm}$$ +$$\text{20}$$ $$\text{cm}$$ = $$\text{130}$$ $$\text{cm}$$.

Note: we may join the points on the graph because the growth is continuous.

Find $$p$$:

$\sum_{k = 1}^{\infty}{27p^{k}} = \sum_{t=1}^{12}{(24 - 3t)}$

Write out the series on the RHS:

$\sum_{t=1}^{12}{(24 - 3t)} = 21 + 18 + 15 + \ldots$

This is an arithmetic series with $$a = 21$$ and $$d = -3$$.

\begin{align*} S_{n} &= \frac{n}{2}[2a + (n - 1)d] \\ S_{12} &= \frac{12}{2}[2(21) + (12 - 1)(3)] \\ &= 6[42 - 33] \\ &= 54 \\ \therefore \sum_{t=1}^{12}{(24 - 3t)} = 54 \end{align*}

Write out the series on the LHS:

$\sum_{k = 1}^{\infty}{27p^{k}} = 27p + 27p^{2} + 27p^{3} + \ldots$

This is a geometric series with $$a = 27p$$ and $$r = p$$ ($$-1 < p < 1$$ for the series to converge).

\begin{align*} S_{\infty} &= \frac{a}{1 - r} \\ \therefore 54 &= \frac{27p}{1 - p} \\ 27p &= 54 - 54 p \\ 81 p &= 54 \\ \therefore p &= \frac{54}{81} \\ &= \frac{2}{3} \end{align*}