Science:Infinite Series Module/Units/Unit 1/1.3 Infinite Series/1.3.03 The Geometric Series

There are certain forms of infinite series that are frequently encountered in mathematics. The following example

$\sum _{k=1}^{\infty }ar^{k-1}=a+ar+ar^{2}+ar^{3}+\ldots ,\ a\neq 0$

for constants $a$ and $r$ is known as the geometric series. The convergence of this series is determined by the constant $r\in \mathbb {R}$, which is the common ratio.

Theorem: Convergence of the Geometric Series
Let $r$ and $a$ be real numbers. Then the geometric series  $\sum _{k=1}^{\infty }ar^{k-1}={\frac {a}{1-r}}$ converges if $|r|<1$, and otherwise diverges.

Proof

To prove the above theorem and hence develop an understanding the convergence of this infinite series, we will find an expression for the partial sum, $s_{n}$, and determine if the limit as $n$ tends to infinity exists. We will further break down our analysis into two cases.

Case 1: $r=1$

If $r=1$, then the partial sum $s_{n}$ becomes

$s_{n}=\sum _{k=1}^{n}ar^{k-1}=a+a+a+a+\ldots =na.$

So as

$n\rightarrow \infty$

we have that

$s_{n}\rightarrow \pm \infty$.

Hence, the geometric series diverges if r = 1.

Case 2: $r\neq 1$

A short derivation for a compact expression for $s_{n}$ will be useful. First note that

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} s_n &= a + ar+ar^2+\ldots+ar^{n-1} \\  rs_n &= ar + ar^2+ar^3+\ldots + ar^{n-1} + ar^n. \end{align}}

The second equation is the first equation multiplied by $r$. Subtracting these two equations yields

Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} s_n - rs_n &= a - ar^{n} \\ s_n &= \frac{a(1 - r^{n})}{1-r}, \ r\ne 1. \end{align} }

Using this result, we see that:

• if $|r|<1$, then as n → ∞, s_n → a/(1-r)
• if $|r|>1$, then as n → ∞, s_n → ∞
• if $|r|=-1$, then the series is divergent by the Divergence Test (which we cover in a lesson in Unit 2)

Summary

The above results can be summarized in the following figure.

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