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Geometric Series 📂Calculus

Geometric Series

Definition1

The following series is called a geometric series for $a \ne 0$.

$$ a + ar + ar^{2} + ar^{3} + \cdots = \sum_{n=0}^{\infty} ar^{n} $$

Explanation

It is the infinite sum of a geometric sequence with the first term $a$ and the common ratio $r$. The ▷eq04

Definition1

The following series is called a geometric series for $a \ne 0$.

$$ a + ar + ar^{2} + ar^{3} + \cdots = \sum_{n=0}^{\infty} ar^{n} $$

Explanation

It is the infinite sum of a geometric sequence with the first term $a$ and the common ratio $r$. The $n$th term is the geometric mean of the $n-1$th term and the $n+2$th term.

$$ \sqrt{(ar^{n-1})(ar^{n+1})} = ar^{n} $$

Partial Sum

The partial sum $s_{n}$ is as follows.

$$ s_{n} = \dfrac{a(1 - r^{n})}{1 - r} $$

Convergence

The geometric series $\sum ar^{n} = a + ar + ar^{2} + ar^{3} + \cdots$ converges when $|r| \lt 1$, and its value is

$$ \sum\limits_{n = 1}^{\infty} ar^{n} = \dfrac{a}{1 - r} \qquad (|r| \lt 1) $$

When $|r| \ge 1$, it diverges.

Proof

When $ |r| \lt 1$

Since the limit of the geometric sequence is $0$,

$$ \begin{align*} \lim\limits_{n \to \infty} s_{n} = \lim\limits_{n \to \infty} \dfrac{a(1 - r^{n})}{1 - r} & = \lim\limits_{n \to \infty} \left( \dfrac{a}{1 - r} - \dfrac{ar^{n}}{1 - r} \right) \\ & = \dfrac{a}{1 - r} - \dfrac{a}{1 - r}\lim\limits_{n \to \infty} r^{n} \\ & = \dfrac{a}{1 - r} \end{align*} $$

When $|r| \ge 1$

In this case, $\{ ar^{n} \}$ does not converge to $0$, and by the divergence test, it diverges.


  1. James Stewart, Daniel Clegg, and Saleem Watson, Calculus (early transcendentals, 9E), p742 ↩︎ ↩︎