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

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}