# 9.2 | Geometric Series

A series is, informally speaking, the sum of the terms of a sequence. That is, the sequence
$$\left\lbrace a_1, a_2, a_3, \cdots \right\rbrace$$
corresponds to the series
$$a_1 + a_2 + a_3 + \cdots$$
A series is denoted more compactly as:
$$\sum_{n = 1}^{\infty} a_n \text{ or } \sum a_n$$
Given a series
$$\sum_{n=1}^{\infty} a_n = a_1 + a_2 + a_3 + \cdots,$$
let $$s_n$$ denote its $$n^{th}$$ partial sum:
$$s_n = \sum_{i=1}^{n} a_i = a_1 + a_2 + a_3 + \cdots + a_n$$
If the sequence $$\{ s_n \}$$ is convergent and $$\displaystyle \lim_{n \rightarrow \infty} s_n = s$$ exists as a real number, then the series $$\sum a_n$$ is called convergent and we write
$$\sum_{n=1}^{\infty} a_n = s$$
The number $$s$$ is called the sum of the series. If the sequence $$\{ s_n \}$$ is divergent, then the series is called divergent.

A geometric series is a series of the form
$$\sum_{n = 1}^{\infty} ar^{n-1} = a + ar + ar^2 + \cdots$$
and
$$\sum_{n =1}^{\infty} ar^{n-1} = \begin{cases} \frac{a}{1-r}, & \; |r| <1 \\ \\ \text{divergent}, & \; \text{otherwise} \end{cases}$$
If the series $$\displaystyle \sum_{n = 1}^{\infty} a_n$$ is convergent, then $$\displaystyle \lim_{n \rightarrow \infty} a_n = 0$$.
The Divergence Test: If $$\displaystyle \lim_{n \rightarrow \infty} a_n$$ does not exist or $$\displaystyle \lim_{n \rightarrow \infty} a_n \not = 0$$, then the series $$\displaystyle \sum_{n = 1}^{\infty} a_n$$ is divergent.
If $$\sum a_n$$ and $$\sum b_n$$ are convergent series, then so are the series:

• $$\sum_{n =1}^{\infty}ca_n = c\sum_{n =1}^{\infty}a_n$$
• $$\sum_{n =1}^{\infty}(a_n \pm b_n) = \sum_{n =1}^{\infty}a_n \pm \sum_{n =1}^{\infty}b_n$$
Determine whether the series is convergent or divergent. If convergent, find the sum; if divergent, enter div .
$$\sum_{n=1}^{\infty} \frac{n}{\sqrt{n^2+19}}$$

The series diverges by the divergence test:

$$\lim_{n \rightarrow \infty} \frac{n}{\sqrt{n^2+19}} = \lim_{n \rightarrow \infty} \frac{n}{\sqrt{n^2}} = \lim_{n \rightarrow \infty} \frac{n}{|n|} = \lim_{n \rightarrow \infty} \frac{n}{n} = 1 \not=0$$
\begin{aligned} \lim_{n \rightarrow \infty} \frac{n}{\sqrt{n^2+19}} & = \lim_{n \rightarrow \infty} \frac{n}{\sqrt{n^2}} = \lim_{n \rightarrow \infty} \frac{n}{|n|} \\ & = \lim_{n \rightarrow \infty} \frac{n}{n} = 1 \not=0 \end{aligned}
\begin{aligned} \lim_{n \rightarrow \infty} \frac{n}{\sqrt{n^2+19}} & = \lim_{n \rightarrow \infty} \frac{n}{\sqrt{n^2}} \\ & = \lim_{n \rightarrow \infty} \frac{n}{|n|} \\ & = \lim_{n \rightarrow \infty} \frac{n}{n} = 1 \not=0 \end{aligned}
Express 9.27272727272…. as a fraction.

• Using Geometric Series:
\begin{aligned} 9.272727272… & = 9+ .27 + .0027 + .000027 + \cdots \\ & = 9 + \frac{27}{100} + \frac{27}{(100)^2} + \frac{27}{(100)^3} + \cdots \\ & = 9 + \frac{27}{100} \left(1 + \frac{1}{100} + \frac{1}{(100)^2} + \cdots \right)\\ & = 9 + \frac{27}{100} \sum_{n = 0}^{\infty} \left(\frac{1}{100}\right)^n\\ & = 9 + \frac{27}{100} \frac{1}{1-\frac{1}{100}}\\ & = 9 + \frac{27}{100} \frac{100}{99}\\ & = 9 + \frac{27}{99} = \frac{891+27}{99} = \frac{918}{99}\\ \end{aligned}
\begin{aligned} & 9.272727272… \\ & = 9+ .27 + .0027 + .000027 + \cdots \\ & = 9 + \frac{27}{100} + \frac{27}{(100)^2} + \frac{27}{(100)^3} + \cdots \\ & = 9 + \frac{27}{100} \left(1 + \frac{1}{100} + \frac{1}{(100)^2} + \cdots \right)\\ & = 9 + \frac{27}{100} \sum_{n = 0}^{\infty} \left(\frac{1}{100}\right)^n\\ & = 9 + \frac{27}{100} \frac{1}{1-\frac{1}{100}}\\ & = 9 + \frac{27}{100} \frac{100}{99}\\ & = 9 + \frac{27}{99} = \frac{891+27}{99} = \frac{918}{99}\\ \end{aligned}
\begin{aligned} & 9.272727272… \\ & = 9+ .27 + .0027 + \cdots \\ & = 9 + \frac{27}{100} + \frac{27}{(100)^2} + \cdots \\ & = 9 + \frac{27}{100} \left(1 + \frac{1}{100} + \frac{1}{(100)^2} + \cdots \right)\\ & = 9 + \frac{27}{100} \sum_{n = 0}^{\infty} \left(\frac{1}{100}\right)^n\\ & = 9 + \frac{27}{100} \frac{1}{1-\frac{1}{100}}\\ & = 9 + \frac{27}{100} \frac{100}{99}\\ & = 9 + \frac{27}{99} = \frac{891+27}{99} = \frac{918}{99}\\ \end{aligned}
• Algebraically:
Let $$x = 9.272…$$, then $$100x = 927.272…$$ and
\begin{aligned} 100x – x & = 927.272… – 9.272… \\ 99x & = 918 \\ x & = \frac{918}{99} = \frac{102}{11} \end{aligned}

The Koch snowflake can be constructed by starting with an equilateral triangle, then recursively altering each line segment as follows:

1. divide the line segment into three segments of equal length.
2. draw an equilateral triangle that has the middle segment from step 1 as its base and points outward.
3. remove the line segment that is the base of the triangle from step 2.
4. After one iteration of this process, the resulting shape is the outline of a hexagram. The Koch snowflake is the limit approached as the above steps are followed over and over again. The Koch curve originally described by Koch is constructed with only one of the three sides of the original triangle. In other words, three Koch curves make a Koch snowflake.

Find

1. the perimeter of the Koch snowflake.
2. the area of the Koch snowflake.