Calculus

Infinite Sequences and Series

Sequences and Series Logo

Geometric Series

A sequence of numbers {an} is called a geometric sequence if the quotient of successive terms is a constant, called the common ratio. Thus an+1/an = q or an+1 = qan for all terms of the sequence. It's supposed that q ≠ 0 and q ≠ 1.

For any geometric sequence:

\[{a_n} = {a_1}{q^{n - 1}}.\]

A geometric series is the indicated sum of the terms of a geometric sequence. For a geometric series with q ≠ 1,

\[{S_n} = {a_1} + {a_2} + \ldots + {a_n} = {a_1}\frac{{1 - {q^n}}}{{1 - q}},\;\; q \ne 1.\]

We say that the geometric series converges if the limit \(\lim\limits_{n \to \infty } {S_n}\) exists and is finite. Otherwise the series is said to diverge.

Let

\[S = \sum\limits_{n = 0}^\infty {{a_n}} = {a_1}\sum\limits_{n = 0}^\infty {{q^n}} \]

be a geometric series. Then the series converges to \(\frac{{{a_1}}}{{1 - q}}\) if \(\left| q \right| \lt 1,\) and the series diverges if \(\left| q \right| \gt 1.\)

Solved Problems

Example 1.

Find the sum of the first \(8\) terms of the geometric sequence

\[3,6,12, \ldots\]

Solution.

Here \({a_1} = 3\) and \(q = 2.\) For \(n = 8\) we have

\[{S_8} = {a_1}\frac{{1 - {q^8}}}{{1 - q}} = 3 \cdot \frac{{1 - {2^8}}}{{1 - 2}} = 3 \cdot \frac{{1 - 256}}{{\left( { - 1} \right)}} = 765.\]

Example 2.

Find the sum of the series

\[1 - 0,37 + 0,37^2 - 0,37^3 + \ldots\]

Solution.

This is an infinite geometric series with ratio \(q = -0,37.\) Hence, the series converges to

\[S = \sum\limits_{n = 0}^\infty {{q^n}} = \frac{1}{{1 - \left( { - 0,37} \right)}} = \frac{1}{{1 + 0,37}} = \frac{1}{{1,37}} = \frac{{100}}{{137}}.\]

See more problems on Page 2.

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