Sum of Infinite Geometric Progression

Theorem

Let $S$ be a standard number field, i.e. $\Q$, $\R$ or $\C$.

Let $z \in S$.

Let $\left \vert {z}\right \vert < 1$, where $\left \vert {z}\right \vert$ denotes:

the absolute value of $z$, for real and rational $z$

the complex modulus of $z$ for complex $z$.

Then $\displaystyle \sum_{n=0}^\infty z^n$ converges absolutely to to $\dfrac 1 {1 - z}$.

Corollary

With the same restriction on $z \in S$:

$\displaystyle \sum_{n=1}^\infty z^n = \frac z {1-z}$

Proof

From Sum of Geometric Progression, we have:

$\displaystyle s_N = \sum_{n=0}^N z^n = \frac {1 - z^{N+1}} {1 - z}$

We have that $\left \vert {z}\right \vert < 1$.

So by Power of a Number Less Than One:

$z^{N+1} \to 0$ as $N \to \infty$

Hence $s_N \to \dfrac 1 {1 - z}$ as $N \to \infty$.

The result follows.

$\blacksquare$

To demonstrate absolute convergence we note that

$\displaystyle \sum_{n=0}^\infty \left \vert {z}\right \vert^n = \frac 1 {1-\left \vert {z}\right \vert}$

as $\left \vert {z}\right \vert$ fulfils the same condition for convergence as $z$.

$\blacksquare$

Proof of Corollary

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sum_{n=1}^\infty x^n\)	\(=\)	\(\displaystyle -x^0 + \sum_{n=0}^\infty x^n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle -1 + \frac 1 {1-x}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {x - 1 + 1} {1-x}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac x {1-x}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

Or:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sum_{n=1}^\infty x^n\)	\(=\)	\(\displaystyle x \sum_{n=0}^\infty x^n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle x\frac 1 {1-x}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from above
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac x {1-x}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 6.2$