Sum of infinite terms of a geometric progression - SS2 Mathematics Lesson Note

In some instances of a geometric progression, as the number of term \(n\) increases the series or sum of terms tends to a definite value. The larger \(n\) gets the more the series \(S_{n}\) tends to a definite value. This is written mathematically as, \(\lim_{n \rightarrow \infty}S_{n} = m.\) This is only true if and only if \(r < 1\).

\(\lim_{n \rightarrow \infty}S_{n} = m\) \(becomes\) \(S_{\infty} = \frac{a}{1 - r}\).

Consider the GP: \(24,\ 12,\ 6,\ 3,\ \ldots\), \(S_{4} = 45\), \(S_{10} = 47.953125\), \(S_{20} = 47.999954\). It is obvious that as \(n \rightarrow \infty\), \(S_{n}\ \rightarrow 48\) as \(r = \frac{12}{24} = \frac{1}{2}\), which is less than \(1\).

From this definition, we can examine what a convergent and divergent series is. A convergent series is a series whose sum to infinite tends to a definite value. This is true if and only if \(- 1 < r < 1\) or \(|r| < 1\).

Example 5 Which of the following GPs are convergent? If there are convergent, find the sum to infinite in each case

1. \(12 + 18 + 27 + \ldots\)

\(18 + 12 + 8 + \ldots\)

\(64 - 48 + 36 - 27 + \ldots\)

\(16 - 40 + 100 - 250 + \ldots\)

Solution

1. Given \(12 + 18 + 27 + \ldots\), \(r = \frac{18}{12} = \frac{3}{2}\) and so \(r \nless 1\); \(r > 1\)

This series is not a convergent series

1. Given \(18 + 12 + 8 + \ldots\), \(r = \frac{12}{18} = \frac{2}{3}\), and since \(r < 1\), this series is a convergent series

Hence, \(S_{\infty} = \frac{a}{1 - r} = \frac{18}{1 - \frac{2}{3}} = \frac{18}{\frac{1}{3}} = 18 \times 3 = 54\)

1. Given \(64 - 48 + 36 - 27 + \ldots\), \(r = \frac{- 48}{64} = - \frac{3}{4}\), \(|r| = \left| \frac{- 3}{4} \right| = \frac{3}{4} < 1\), this is a convergent series

\[S_{\infty} = \frac{a}{1 - r} = \frac{64}{1 + \frac{3}{4}} = \frac{64}{\frac{7}{4}} = 36\frac{4}{7}\]

1. Given \(16 - 40 + 100 - 250 + \ldots\), \(r = \frac{- 250}{100} = \frac{- 5}{2}\) and so \(|r| \nless 1\); this series in not convergent.