Sum of Infinite Geometric Sequence/Proof 4

Theorem

Let $S$ be a standard number field, that is $\Q$, $\R$ or $\C$.

Let $z \in S$.

Let $\size z < 1$, where $\size z$ denotes:

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

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

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

Proof

\(\ds \frac 1 {1 - z}\)

\(=\)

\(\ds \frac 1 {1 + \paren {-z} }\)

\(\ds \)

\(=\)

\(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k \paren {-z}^k\)

Power Series Expansion for $\dfrac 1 {1 + z}$

\(\ds \)

\(=\)

\(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k \paren {-1}^k z^k\)

\(\ds \)

\(=\)

\(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^{2 k} z^k\)

\(\ds \)

\(=\)

\(\ds \sum_{k \mathop = 0}^\infty z^k\)

$\blacksquare$

Sources

• 1953: L. Harwood Clarke: A Note Book in Pure Mathematics ... (previous) ... (next): $\text I$. Algebra: The Binomial Theorem: Negative and fractional indices