Geometric Series

# Geometric Series

Definition: A Geometric Series is a series in the form $\sum_{n=1}^{\infty} ar^{n-1} = a + ar + ar^2 + ar^3 + ...$ whose $n^{\mathrm{th}}$ term can be obtained by the formula $a_n = ar^{n-1}$. The value $r$ is called the Common Ratio of the geometric series as $\frac{a_{n+1}}{a_{n}} = \frac{ar^n}{ar^{n-1}} = r$ for any $n \in \mathbb{R}$. The $n^{\mathrm{th}}$ partial sum of a geometric series is $s_n = a + ar + ar^2 + ... + ar^{n-1} = \frac{a(1 - r^n)}{1 - r}$. |

One such example of a geometric series $\sum_{n=1}^{\infty} 2\left (\frac{1}{3} \right )^{n-1} = 2 + \frac{2}{3} + \frac{2}{9} + \frac{2}{27} + ...$.

Before we begin examining geometric series, let's first confirm that the $n^{\mathrm{th}}$ partial sum of a geometric series can be given by the formula $s_n = \frac{a(1 - r^n)}{(1 - r)}$ which we asserted in the definition above.

Lemma 1: The $n^{\mathrm{th}}$ partial sum of a geometric series with a common ratio $r$ is given by $s_n = \frac{a(1 - r^n)}{(1 - r)}$. |

We will show two proofs of lemma 1. The first proof is a simple direct proof, while the second proof uses the principle of mathematical induction.

**Proof 1 of Lemma 1:**Let $s_n = a + ar + ar^2 + ... + ar^{n-1}$ be the $n^{\mathrm{th}}$ partial sum of the geometric series $\sum_{n=1}^{\infty} ar^{n-1}$. Now multiply this equation by the common ratio to get $rs_n = ar + ar^2 + ... + ar^{n-1} + ar^n$. Subtracting these two equations we get that:

\begin{align} \quad s_n - rs_n = (a + ar + ar^2 + ... + ar^{n-1}) - (ar + ar^2 + ... + ar^{n-1} + ar^n) \\ \quad (1-r)s_n = a - ar^n \\ \quad s_n = \frac{a(1 - r^n)}{1 - r} \blacksquare \end{align}

**Proof 2 of Lemma 1:**We note that the $n^{\mathrm{th}}$ partial sum of a geometric series can be obtained also by the formula $s_n = a + ar + ar^2 + ... + ar^{n-1}$. Therefore we want to show that $a + ar + ar^2 + ... + ar^{n-1} = \frac{a(1 - r^n)}{(1 - r)}$. We will first factor out an $a$ so that it becomes sufficient that we need to only prove that $1 + r + r^2 + ... + r^{n-1} = \frac{(1 - r^n)}{(1 - r)}$. We will show this by induction.

- Let $P(n)$ be the statement that $1 + r + r^2 + ... + r^{n-1} = \frac{1 - r^n}{1 - r}$. First consider the base step when $n = 1$. Then $1 = \frac{1 - r}{1 - r} = 1$ and so $P(1)$ is true.

- Now suppose that for some $k \in \mathbb{N}$ that the statement $P(k) : 1 + r + r^2 + ... + r^{k-1} = \frac{1 + r^k}{1 - r}$ is true. We want to show that the truth of $P(k)$ implies the truth of the statement $P(k +1) = 1 + r + r^2 + ... + r^{k-1} + r^k = \frac{1 + r^{k+1}}{1 - r}$.

\begin{align} \underbrace{1 + r + r^2 + ... + r^{k-1}} + r^k \\ \overset{IH} = \frac{1 - r^k}{1 - r} + r^k \\ = \frac{1 - r^k + (1 - r)r^k}{1 - r} \\ = \frac{1 - r^k + r^k - r^{k+1}}{1 - r} \\ = \frac{1 - r^{k+1}}{1 - r} \end{align}

- Therefore $P(k+1)$ is true. So for $n \in \mathbb{N}$, $P(n)$ is true.

- Now if we multiply all terms by $a$ we get that $a + ar + ar^2 + ... + ar^{n-1} = \frac{a(1 - r^n)}{1 - r}$, and our proof is complete $\blacksquare$

## Convergence and Divergence of Geometric Series

We will now look at some very important properties of geometric series regarding whether they converge or diverge which will allow us to compute the sums of geometric series.

Lemma 2: The geometric series $\sum_{n=1}^{\infty} ar^{n-1}$ converges to $0$ if $a = 0$. |

**Proof:**Suppose that $a = 0$ and let $r$ be any common ratio. Then the sequence $\{ ar^{n-1} \} = \{ 0, 0, 0, ... \}$, and it is obvious that $\sum_{n=1}^{\infty} ar^{n-1}$ converges to $0$ since this is simply the zero sequence. $\blacksquare$

Lemma 3: The geometric series $\sum_{n=1}^{\infty} ar^{n-1}$ converges to $\frac{a}{1 - r}$ if $\mid r \mid < 1$. |

**Proof:**Suppose that $\mid r \mid ≤ 1$. Recall that a series is said to be convergent if the sequence of partial sums $\{ s_n \}$ converges. We note that the sequence of partial sums is $\left \{ \frac{a(1 - r^n)}{1 - r} \right \}$ by lemma 1 and we know that this sequence converges since $\mid r \mid ≤ 1$ and as $n \to \infty$, the numerator of the general term for the $n^{\mathrm{th}}$ partial sum $a(1 - r^n) \to a$ and the denominator stays fixed at $1 - r$. We now must evaluate $\lim_{n \to \infty} \frac{1 - r^n}{1 - r} = L$.

\begin{align} \quad \lim_{n \to \infty} \frac{1 - r^n}{1 - r} = \lim_{n \to \infty} \frac{a}{1 - r} - \lim_{n \to \infty} \frac{ar^n}{1 - r} = \frac{a}{1 - r} - \lim_{n \to \infty} \frac{ar^n}{1 - r} \end{align}

- Now since $\mid r \mid < 1$, we note that as $n \to \infty$, $ar^n \to 0$ and so $\lim_{n \to \infty} \frac{ar^n}{1 - r} = 0$, therefore:

\begin{align} \lim_{n \to \infty} \frac{a(1 - r^n)}{1 - r} = \frac{a}{1 - r} \quad \blacksquare \end{align}

Lemma 4: The geometric series $\sum_{n=1}^{\infty} ar^{n-1}$ diverges to positive infinity if $r ≥ 1$ and $a > 0$. |

**Proof:**Suppose that $r ≥ 1$ and $a > 0$. Therefore $\frac{1 - r^n}{1 - r} > 0$, so every term in the sequence of partial sums $\{ s_n \}$ will be positive. Now since $r ≥ 1$, then as $n \to \infty$, $\frac{a(1 - r^n)}{1 - r} \to \infty$. Therefore $\lim_{n \to \infty} \frac{a(1 - r^n)}{1 - r} = \infty$. $\blacksquare$

Lemma 5: The geometric series $\sum_{n=1}^{\infty} ar^{n-1}$ diverges to negative infinity if $r ≥ 1$ and $a < 0$. |

**Proof:**Suppose that $r ≥ 1$ and $a < 0$. Once again $\frac{1 - r^n}{1 - r} > 0$, so every term in the sequence of partial sums $\{ s_n \}$ will be negative. Now since $r ≥ 1$ then as $n \to \infty$, $\frac{a(1 - r^n)}{1 - r} \to -\infty$. Therefore $\lim_{n \to \infty} \frac{a(1 - r^n)}{1 - r} = -\infty$. $\blacksquare$

Lemma 6: The geometric series $\sum_{n=1}^{\infty} ar^{n-1}$ diverges if $r ≤ -1$ and $a \neq 0$. |

**Proof:**Suppose that $r ≤ -1$ and $a ≠ 0$. We note that $\frac{1 - r^n}{1 - r}$ will have terms alternating between being positive and negative since the denominator is always positive and the numerator is positive (if $n$ is even) or negative (if $n$ is odd). Furthermore we note that $\mid 1 - r^n \mid$ is increasing as $n \to \infty$, so $\lim_{n \to \infty} s_n$ does not exist. $\blacksquare$