More Series

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More problems about series. Mostly geometric.

A series is the $\answer [format=string]{sum}$ of consecutive terms from a(n) $\answer [format=string]{sequence}$ .

An arithmetic series is the $\answer [format=string]{sum}$ of consecutive terms from a(n) $\answer [format=string]{arithmetic sequence}$ (two words).

A geometric series is the $\answer [format=string]{sum}$ of consecutive terms from a(n) $\answer [format=string]{geometric sequence}$ (two words).

Consider the geometric series: $\frac {1}{2}+\frac {1}{4}+\frac {1}{8}+\dots +\frac {1}{2048}$ First compute the sequence of partial sums: $\begin {array}{c|c} \hline \text {No. of terms} & \text {Value} \\ \hline 1 & \answer {\frac {1}{2}} \\ 2 & \answer {\frac {3}{4}} \\ 3 & \answer {\frac {7}{8}}\\ 4 & \answer {\frac {15}{16}}\\ 5 & \answer {\frac {31}{32}}\\ 6 & \answer {\frac {63}{64}}\\ \end {array}$

Make a conjecture: The sum of the first $n$ terms of this series is $\answer {1-\frac {1}{2^n}}$ .

In the series $\frac {1}{2}+\frac {1}{4}+\frac {1}{8}+\dots +\frac {1}{2048}$ there are $\answer {11}$ terms, so the sum is $\answer {1-\frac {1}{2^{11}}}$ .

Here is picture that can help explain why this particular series works out so nicely:

Begin with a unit square, and then shade an area equal to each term. At each step in the process, the area remaining unshaded is exactly equal to the area of the region last shaded. For example, after shading $\frac {1}{2}$ and then $\frac {1}{4}$ , $\answer {\frac {1}{4}}$ remains unshaded.

Furthermore, it appears that as more and more terms are added, the sum should get closer and closer to $\answer {1}$ .

Here is GeoGebra sketch that allows you to explore similar series:

The following problems demonstrate a general technique for geometric series.

First, let $S=\frac {1}{2}+\frac {1}{4}+\frac {1}{8}+\dots +\frac {1}{2048}$ Multiply this equation by the common ratio $\answer {\frac {1}{2}}$ and place the two equations above/below one another so that identical terms align.

$\begin {array}{ccccccccccccc} S &=& \frac {1}{2} &+& \frac {1}{4} &+& \frac {1}{8} &+& \dots &+& \frac {1}{2048} \\ \\ \answer {\frac {S}{2}} &=& & & \frac {1}{4} &+& \frac {1}{8} &+& \dots &+& \frac {1}{2048} &+& \answer {\frac {1}{4096}} \end {array}$

Then we $\answer [format=string]{subtract}$ the two equations so that most of the terms on the right vanish. The result is a simpler equation: $S - \answer {\frac {S}{2}} = \frac {1}{2} - \answer {\frac {1}{4096}},$ which we can solve for $S$ . $S = \frac {\frac {1}{2} - \answer {\frac {1}{4096}}}{1-\answer {\frac {1}{2}}},$ which simplifies to $S=\answer {1-\frac {1}{2^{11}}}$ as before.

Remember to factor out the $S$ on the left before dividing.

In this problem, we use geometric series to convert a repeating decimal to a fraction in order to show it is a $\answer [format=string]{rational}$ number.

First express $0.\overline {23}=0.232323...$ as geometric series. $0.\overline {23}=\frac {23}{100}+\frac {23}{100^2}+\frac {23}{100^3}+\dots$ The series is geometric because the $\answer [format=string]{ratio}$ between consecutive terms is constant. It equals $\answer {1/100}$ .

The series is infinite because to be equal to $0.\overline {23}$ , we need to sum an infinite number of terms. Any finite number of terms will be only a(n) $\answer [format=string]{approximation}$ .

To demonstrate the technique rigorously, we begin with a finite geometric series and find a closed expression (without decimals) for the exact sum.

First, let $S=\frac {23}{100}+\frac {23}{100^2}+\frac {23}{100^3}+\dots +\frac {23}{100^n}$ Note that this is the finite decimal $0.2323 \dots 23$ , with exactly $\answer {n}$ copies of the two repeating digits.

We multiply this equation by the common ratio $\answer {1/100}$ and place the two equations above/below one another so that identical terms align.

$\begin {array}{ccccccccccccc} S &=& \frac {23}{100} &+& \frac {23}{100^2} &+& \frac {23}{100^3} &+& \dots &+& \frac {23}{100^n} \\ \\ \answer {\frac {S}{100}} &=& & & \frac {23}{100^2} &+& \frac {23}{100^3} &+& \dots &+& \frac {23}{100^n} &+& \answer {\frac {23}{100^{n+1}}} \end {array}$

Then we $\answer [format=string]{subtract}$ the two equations so that most of the terms on the right vanish. The result is a simpler equation: $S - \answer {\frac {S}{100}} = \frac {23}{100} - \answer {\frac {23}{100^{n+1}}},$ which we can solve for $S$ . $S = \frac {\frac {23}{100} - \answer {\frac {23}{100^{n+1}}}}{1-\answer {\frac {1}{100}}}$

Remember to factor out the $S$ on the left before dividing.

As $n\rightarrow \infty$ , $\frac {23}{100^{n+1}} \rightarrow \answer {0}$ , so $S \rightarrow \frac {\answer {\frac {23}{100}}}{1-\frac {1}{100}}\text {, which simplifies to }\frac {\answer {23}}{\answer {99}}.$ From the original definition of $S$ above, when $n\rightarrow \infty$ , we also have $S\rightarrow 0.\overline {23}$ . Therefore, $0.\overline {23}=\frac {\answer {23}}{\answer {99}},$ demonstrating that $0.\overline {23}$ is indeed a $\answer [format=string]{rational}$ number.