What if the series starts at an index other than $n=1$?

The formula $S = a / (1 - r)$ still applies. Simply evaluate the formula for the sequence at the given starting index to find your first term, $a$.

Does the finite sum formula work if the common ratio is exactly 1?

No, substituting $r=1$ causes division by zero. If $r=1$, the series is simply $n$ copies of $a$, so the sum of $n$ terms is $n imes a$.

How do I handle a series where the terms alternate signs?

An alternating geometric series simply has a negative common ratio $r$. Keep the negative sign with $r$ when substituting into the sum formulas.

How to find the sum of a geometric series

A geometric series is the sum of a sequence where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. This method applies to both finite geometric series and infinite geometric series that converge.

The setup

Identify the first term, $a$ , and the common ratio, $r$ . Determine if the series is finite (having a specific number of terms, $n$ ) or infinite.

The steps

Extract $a$ and $r$ from the given series expression. 2. For a finite series with $n$ terms, calculate the sum using $S_n = a(1 - r^n) / (1 - r)$ . 3. For an infinite series, verify that $|r| < 1$ . If so, calculate the sum using $S = a / (1 - r)$ . If $|r| \ge 1$ , the infinite series diverges and has no finite sum.

Checking the result

Manually add the first three to four terms of the series to ensure the partial sum approaches your calculated result. Verify that the sign of your final sum makes sense given the signs of $a$ and $r$ .

Common errors

A frequent mistake is using the incorrect number of terms, $n$ , in the finite formula. For example, a sum indexed from $i=0$ to $k$ has $k+1$ terms, not $k$ . Another major error is applying the infinite sum formula $a / (1 - r)$ when $|r| \ge 1$ , which yields a mathematically invalid result.

Worked example

Find the sum of the infinite geometric series: $3 - 1 + \frac{1}{3} - \frac{1}{9} + ...$

Identify the first term: $a = 3$ . Find the common ratio by dividing the second term by the first: $r = -1 / 3$ . Check convergence: $|-1/3| < 1$ , so the series converges. Apply the infinite geometric series formula: $S = a / (1 - r)$ . Substitute the values: $S = 3 / (1 - (-1/3))$ . Simplify the denominator: $1 - (-1/3) = 1 + 1/3 = 4/3$ . Calculate the final sum: $S = 3 / (4/3) = 9/4$ .

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References: Stewart Calculus 8th Edition, Chapter 11.2 · OpenStax Calculus Volume 2, Chapter 5.2 · Khan Academy, Unit: Infinite sequences and series