Geometric Series

analysis

Definition

Geometric Series

A series of form:

$$\sum _{n\ge 0}{q}^n=1+q+q^2+q^3+\dots$$

is called geometric series.

Convergence

For $|q|<1$, the value of the geometric series ${\sum }_{n\ge 0}{q}^n$ is given by:

$$\forall q:|q|<1⟹\sum _{n\ge 0}{q}^n=\frac{1}{1-q}$$

Proof

Let $|q|<1$, thus ${\lim }_{n\to \infty }{q}^n=0$. Given that the sequence $a_n=q^n$ is convergent and its limit is $0$, the series ${\sum }_{n\ge 0}{q}^n$ is convergent.

Let $s_n={\sum }_{k=0}^n{q}^k$ be the series’ partial sum.

$$\begin{array}{rl}{s}_n-q{s}_n& =\sum _{k=0}^n{q}^k-\sum _{k=1}^{n+1}{q}^k=1-q^{n-1}\\ (s_n)(1-q)& =1-q^{n-1}\\ ⟺s_n& =\frac{1-q^{n+1}}{1-q}\end{array}$$

Thus:

$$\underset{n\to \infty }{\lim }{s}_n=\underset{n\to \infty }{\lim }\frac{1-q^{n+1}}{1-q}=\frac{{\lim }_{n\to \infty }1-{\lim }_{n\to \infty }{q}^{n+1}}{{\lim }_{n\to \infty }1-{\lim }_{n\to \infty }q}=\frac{1}{1-q}=\sum _{n\ge 0}{q}^n$$

Divergence

For $|q|\ge 1$, the geometric series is divergent, since the sequence of terms, i.e., $(q^n{)}_{n\in \mathbb{N}}$ is not a null sequence.