Code Scholars Tutoring · AP Calculus BC

Geometric series and the correct first term

Unit 10 · Topics 10.2 · Detailed notes, a worked example, and a self-check with reasoning.

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Understand the idea

A geometric series has a constant ratio between successive terms. Its finite partial sum can be written explicitly, which explains why an infinite sum exists when the common ratio has magnitude less than one. Alternating signs are allowed; what matters is the ratio's magnitude.

Identify the actual first included term and the ratio before applying the sum formula. A series that begins at $\displaystyle n=2$ may have a different first term from the same expression beginning at $\displaystyle n=0$ . The index symbol itself does not determine the first term.

For a nonzero geometric series, ratios of magnitude at least one do not produce terms tending to zero. When $\displaystyle |r|<1$ , the omitted tail after a finite sum is also geometric, so it can be evaluated exactly rather than bounded indirectly.

\sum_{n=0}^{\infty}ar^n=\frac{a}{1-r}\quad(|r|<1),\qquad S_N=\frac{a(1-r^N)}{1-r}\quad(r\ne1)

A method you can apply

Calculate the first included term $\displaystyle a$ and the constant ratio $\displaystyle r$ .
Check $\displaystyle |r|<1$ before using an infinite-sum formula.
Compute $\displaystyle a/(1-r)$ and account for any excluded initial terms.

Worked example

Find $\displaystyle \sum_{n=0}^{\infty}3(-2/5)^n$ .

The first term is $\displaystyle a=3$ and the ratio is $\displaystyle r=-2/5$ .
Since $\displaystyle |r|=2/5<1$ , the series converges.
The sum is $\displaystyle 3/[1-(-2/5)]=3/(7/5)$ .

The sum is

\displaystyle 15/7

Try it yourself

Find $\displaystyle \sum_{n=2}^{\infty}(1/3)^n$ .

Show solution and reasoning

The first included term is $\displaystyle 1/9$ and the ratio is $\displaystyle 1/3$ . Therefore the sum is $\displaystyle (1/9)/(1-1/3)=1/6$ .