Infinite Geometric Series Math Example 1

Follow the full solution, then compare it with the other examples linked below.

Example 1

easy
Find the sum of the infinite geometric series ∑n=0∞(13)n\displaystyle\sum_{n=0}^{\infty} \left(\frac{1}{3}\right)^n.

Solution

  1. 1
    Identify: first term a=1a = 1 (when n=0n=0), common ratio r=13r = \frac{1}{3}.
  2. 2
    Check: ∣r∣=13<1|r| = \frac{1}{3} < 1, so the series converges.
  3. 3
    Sum formula: S=a1−r=11−13=123=32S = \frac{a}{1-r} = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2}.

Answer

32\frac{3}{2}
Always verify ∣r∣<1|r| < 1 before applying the formula. The first term aa is the value at n=0n=0, which is (13)0=1\left(\frac{1}{3}\right)^0 = 1.

About Infinite Geometric Series

The sum of all terms of a geometric sequence with common ratio ∣r∣<1|r| < 1. The infinite sum converges to a1−r\frac{a}{1-r}, where aa is the first term.

Learn more about Infinite Geometric Series →

More Infinite Geometric Series Examples

Example 2 medium

Convert the repeating decimal [formula] to a fraction using an infinite geometric series.

Example 3 medium

Find the sum of the infinite series [formula]

Example 4 easy

Find the sum: [formula]

Example 5 hard

For what values of [formula] does [formula] converge, and what is the sum?

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