Limiting Cases Math Example 1

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

Example 1

easy
The formula for the sum of a geometric series S=a(1โˆ’rn)1โˆ’rS = \frac{a(1-r^n)}{1-r} (rโ‰ 1r\ne 1). Find the limiting case as nโ†’โˆžn \to \infty when โˆฃrโˆฃ<1|r| < 1.

Solution

  1. 1
    As nโ†’โˆžn \to \infty with โˆฃrโˆฃ<1|r| < 1: rnโ†’0r^n \to 0.
  2. 2
    Therefore S=a(1โˆ’rn)1โˆ’rโ†’a(1โˆ’0)1โˆ’r=a1โˆ’rS = \frac{a(1-r^n)}{1-r} \to \frac{a(1-0)}{1-r} = \frac{a}{1-r}.
  3. 3
    This is the sum of an infinite geometric series with โˆฃrโˆฃ<1|r|<1.

Answer

Sโˆž=a1โˆ’r(โˆฃrโˆฃ<1)S_{\infty} = \frac{a}{1-r} \quad (|r|<1)
Taking a limiting case (nโ†’โˆžn \to \infty) of a finite formula often yields a simpler infinite-series formula. The limit rnโ†’0r^n \to 0 for โˆฃrโˆฃ<1|r|<1 is the key step.

About Limiting Cases

Extreme values of a parameter (approaching zero, infinity, or a critical threshold) used to check formulas and reveal simplified behavior.

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