Sum of an infinite Geometric Progression

The sum of an infinite Geometric Progression whose first term 'a' and common ratio 'r' (-1 < r < 1 i.e., |r| < 1) is

S = \(\frac{a}{1 - r}\)

Proof:

A series of the form a + ar + ar\(^{2}\) + ...... + ar\(^{n}\) + ............... ∞ is called an infinite geometric series.

Let us consider an infinite Geometric Progression with first term a and common ratio r, where -1 < r < 1 i.e., |r| < 1. Therefore, the sum of n terms of this Geometric Progression in given by

S\(_{n}\) = a(\(\frac{1 - r^{n}}{1 - r}\)) = \(\frac{a}{1 - r}\)\(\frac{ar^{n}}{1 - r}\) ........................ (i)

Since - 1< r < 1, therefore r\(^{n}\)  decreases as n increases and r^n tends to zero an n tends to infinity i.e., r\(^{n}\) → 0 as n → ∞.

Therefore,

\(\frac{ar^{n}}{1 - r}\) → 0 as n → ∞.

Hence, from (i), the sum of an infinite Geometric Progression ig given by

S = \(\lim_{x \to 0}\) S\(_{n}\)  = \(\lim_{x \to \infty} (\frac{a}{ 1 - r} - \frac{ar^{2}}{1 - r})\) = \(\frac{a}{1 - r}\) if |r| < 1


Note: (i) If an infinite series has a sum, the series is said to be convergent. On the contrary, an infinite series is said to be divergent it it has no sum. The infinite geometric series a + ar + ar\(^{2}\) + ...... + ar\(^{n}\) + ............... ∞ has a sum when -1 < r < 1; so it is convergent when -1 < r < 1. But it is divergent when r > 1 or, r < -1.

(ii) If r ≥ 1, then the sum of an infinite Geometric Progression tens to infinity.


Solved examples to find the sum to infinity of the Geometric Progression:

1. Find the sum to infinity of the Geometric Progression

-\(\frac{5}{4}\), \(\frac{5}{16}\), -\(\frac{5}{64}\), \(\frac{5}{256}\), .........

Solution:

The given Geometric Progression is -\(\frac{5}{4}\), \(\frac{5}{16}\), -\(\frac{5}{64}\), \(\frac{5}{256}\), .........

It has first term a = -\(\frac{5}{4}\) and the common ratio r = -\(\frac{1}{4}\). Also, |r| < 1.

Therefore, the sum to infinity is given by

S = \(\frac{a}{1 - r}\) = \(\frac{\frac{5}{4}}{1 - (-\frac{1}{4})}\)  = -1


2. Express the recurring decimals as rational number: \(3\dot{6}\)

Solution:

\(3\dot{6}\) = 0.3636363636............... ∞

= 0.36 + 0.0036 + 0.000036 + 0.00000036 + .................. ∞

= \(\frac{36}{10^{2}}\) + \(\frac{36}{10^{4}}\) + \(\frac{36}{10^{6}}\) + \(\frac{36}{10^{8}}\) + .................. ∞, which is an infinite geometric series whose first term = \(\frac{36}{10^{2}}\) and common ratio = \(\frac{1}{10^{2}}\) < 1.

= \(\frac{\frac{36}{10^{2}}}{1 - \frac{1}{10^{2}}}\), [Using the formula S = \(\frac{a}{1 - r}\)]

= \(\frac{\frac{36}{100}}{1 - \frac{1}{100}}\)

= \(\frac{\frac{36}{100}}{\frac{100 - 1}{100}}\)

= \(\frac{\frac{36}{100}}{\frac{99}{100}}\)

= \(\frac{36}{100}\) × \(\frac{100}{99}\)

= \(\frac{4}{11}\)

 Geometric Progression




11 and 12 Grade Math 

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