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# 1. Geometric Series

### Definition

A geometric series is a series of the form $1 + x + x^2 + x^3 + x^4 + \dots$ for some real number $$x.$$

### Convergence and Computation

Recall that the value of a series is defined by $\sum_{i=0}^\infty a_i = \lim_{n \rightarrow \infty} \sum_{i=0}^n a_i$ We will analyze the series by looking at the partial sum.

For a geometric series, the partial sums are $\sum_{i=0}^n x^i = 1 + x + \dots + x^n$ The partial sums can be written as fractions. To see this, multiply by $$1-x.$$ $(1 + x + \dots + x^n)(1-x) = (1 + x + \dots + x^n) - (x + x^2 + \dots + x^{n+1})$ All of the terms can be paired except the $$1$$ and $$x^{n+1}.$$ $1 + (x - x) + (x^2 - x^2) + \dots + (x^n - x^n) - x^{n+1}$ All of the terms cancel, leaving $$1 - x^{n+1}.$$ So, $(1 + x + \dots + x^n)(1-x) = 1 - x^{n+1}.$ Next, divide both sides by $$1 - x$$ to get $1 + x + \dots + x^n = \frac{1-x^{n+1}}{1-x}.$ So, \begin{align} \sum_{i=0}^\infty x^i & = \lim_{n \rightarrow \infty} \sum_{i = 0}^n x^i \\ & = \lim_{n \rightarrow \infty} \frac{1-x^{n+1}}{1-x} \end{align} There are several cases:

• $$x > 1:$$ When $$x > 1,$$ $$\lim_{n \rightarrow \infty} x^n = \infty.$$ So, $$1 + x + x^2 + \dots = \infty.$$
• $$x = 1:$$ The formula $$\frac{1 - x^{n+1}}{1-x}$$ is undefined when $$x = 1,$$ but $$1 + 1^2 + 1^3 + \dots = 1 + 1 + 1 + \dots = \infty.$$
• $$-1 < x < 1:$$ In this case, $$\lim_{n \rightarrow \infty} x^n = 0,$$ so $$\sum_{i = 0}^\infty x^i = \frac{1}{1-x}$$
• $$x = -1:$$ In this case, the sum is $$1 - 1 + 1 - 1 + \dots.$$ The sequence of partial sums is \begin{align} & 1 \\ & 1 - 1 = 0 \\ & 1 - 1 + 1 = 1 \\ & 1 - 1 + 1 - 1 = 0 \end{align} The sequence alternates between $$1$$ and $$0$$ and never converges, so the limit does not exist.
• $$x < -1:$$ The partial sums alternate signs and the absolute values of the partial sums tends to infinity. So, the series does not converge.

The most important formula is when $$-1 < x < 1.$$

Example: Evaluate $\sum_{i=1}^\infty \frac{3}{4^i}$ Solution: The formula for a geometric series cannot be applied directly, but can be applied after some simple manipulation. First, we can pull out the $$3$$ in the numerator since it is not raised to the $$i$$th power. $\sum_{i=1}^\infty \frac{3}{4^i} = 3\sum_{i=1}^\infty \frac{1}{4^i}$ Next, we can change the index set from $$\{1, 2, 3, \dots\}$$ to $$\{0, 1, 2, \dots\}$$ by replacing $$i$$ with $$i+1$$ in the formula. $3\sum_{i=1}^\infty \frac{1}{4^i} = 3\sum_{i=0}^\infty \frac{1}{4^{i+1}}$ For every $$i,$$ $$\frac{1}{4^{i+1}} = \frac{1}{4} \cdot \frac{1}{4^i},$$ so we can pull out $$\frac{1}{4}$$ from every term. $3\sum_{i=0}^\infty \frac{1}{4^{i+1}} = \frac{3}{4}\sum_{i=0}^\infty \frac{1}{4^i}$ Now we can apply the formula for a geometric series and work out the algebra. \begin{align} \frac{3}{4}\sum_{i=0}^\infty \frac{1}{4^i} & = \frac{3}{4} \cdot \frac{1}{1 - \frac{1}{4}} \\ & = \frac{3}{4} \cdot \frac{4}{3} \\ & = 1 \end{align}