Sum of Infinite Geometric Sequence

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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 $\dfrac 1 {1 - z}$.


Corollary 1

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

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


Corollary 2

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

$\ds \sum_{n \mathop = 0}^\infty a z^n = \frac a {1 - z}$


Corollary 3

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

$\ds \sum_{n \mathop = N}^\infty z^n = \frac {z^N } {1 - z}$


Proof 1

From Sum of Geometric Sequence, we have:

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

We have that $\size z < 1$.

So by Sequence of Powers of 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.

$\Box$


It remains to demonstrate absolute convergence:

The absolute value of $\size z$ is just $\size z$.

By assumption:

$\size z < 1$

So $\size z$ fulfils the same condition for convergence as $z$.

Hence:

$\ds \sum_{n \mathop = 0}^\infty \size z^n = \frac 1 {1 - \size z}$

$\blacksquare$


Proof 2

By the Chain Rule for Derivatives and the corollary to $n$th Derivative of Reciprocal of $m$th Power:

$\dfrac {\d^n} {\d z^n} \dfrac 1 {1 - z} = \dfrac {n!} {\paren {1 - z}^{n + 1} }$

Thus the Maclaurin series expansion of $\dfrac 1 {1 - z}$ is:

$\ds \sum_{n \mathop = 0}^\infty \frac {z^n} {n!} \dfrac {n!} {\paren {1 - 0}^{n + 1} }$

whence the result.

$\blacksquare$


Proof 3

Let $S = \ds \sum_{n \mathop = 0}^\infty z^n$.

Then:

\(\ds z S\) \(=\) \(\ds z \sum_{n \mathop = 0}^\infty z^n\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty z^{n + 1}\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 1}^\infty z^n\) Translation of Index Variable of Summation
\(\ds \) \(=\) \(\ds S - z^0\)
\(\ds \) \(=\) \(\ds S - 1\)
\(\ds \leadsto \ \ \) \(\ds 1\) \(=\) \(\ds \paren {1 - z} S\)

Hence the result.

$\blacksquare$


Proof 4

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

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