Infinite Geometric Series Math Example 3

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

Example 3

medium
Find the sum of the infinite series 12+4+43+49+⋯12 + 4 + \frac{4}{3} + \frac{4}{9} + \cdots

Solution

  1. 1
    Identify aa and rr: first term a=12a = 12. Common ratio r=412=13r = \frac{4}{12} = \frac{1}{3}. Check: 4×13=434 \times \frac{1}{3} = \frac{4}{3} ✓.
  2. 2
    Verify convergence: ∣r∣=13<1|r| = \frac{1}{3} < 1, so the series converges.
  3. 3
    Apply the sum formula: S=a1−r=121−13=1223=12×32=18S = \frac{a}{1-r} = \frac{12}{1 - \frac{1}{3}} = \frac{12}{\frac{2}{3}} = 12 \times \frac{3}{2} = 18.

Answer

S=18S = 18
An infinite geometric series with ∣r∣<1|r| < 1 converges to a1−r\frac{a}{1-r}. Each successive term contributes less and less, so the partial sums approach a finite limit. If ∣r∣≥1|r| \geq 1, the series diverges.

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 1 easy

Find the sum of the infinite geometric series [formula].

Example 2 medium

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

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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