Interesting Geometric Series and Applications of Geometric Series

A geometric series is a series that can be written as

(1)   \begin{equation*}a + ar + a{r^2} + a{r^3} +  \cdots  = \sum\limits_{n = 1}^\infty  {a{r^{n - 1}}}\end{equation*}

A geometric series will converge if \left| r \right| < 1. For a proof that that the only way a geometric series converges is if \left| r \right| < 1, click here.

Proving that a geometric series converges if \left| r \right| < 1 is simple. The sum of the first n terms of a geometric series is

(2)   \begin{equation*} {S_n} = \frac{{a\left( {1 - {r^n}} \right)}}{{1 - r}} \end{equation*}

Equation (2) is derived as followers.

    \[{S_n} = a + a{r^1} + ... + a{r^{n - 1}} = \sum\limits_{i = 1}^n {a{r^{i - 1}}}\]

Multiply the left side of the above equation and the far right side by r.

    \[r{S_n} = r\left( {a + ar + a{r^2} + ... + a{r^{n - 1}}} \right) = ar + a{r^2} + a{r^3} + ... + a{r^n}\]

Next, subtract the equation above from S_n and solve for S_n:

    \[{S_n} - r{S_n} = \left( {a + ar + a{r^2} + ... + a{r^{n - 1}}} \right) - \left( {ar + a{r^2} + a{r^3} + ... + a{r^n}} \right)\]

    \[S_n =  a - ar^n\]

    \[{S_n}\left( {1 - r} \right) = a\left( {1 - {r^n}} \right)\]

    \[{S_n} = \frac{{a - a{r^n}}}{{1 - r}}\]

(3)   \begin{equation*} {S_n} = \frac{a}{{1 - r}} - \frac{{a{r^n}}}{{1 - r}} \end{equation*}

The sum of a geometric series, if it exists, is defined as

(4)   \begin{equation*} \boxed{S = \mathop {\lim }\limits_{n \to \infty } {S_n} = \mathop {\lim }\limits_{n \to \infty } \left( {\frac{a}{{1 - r}} - \frac{{a{r^n}}}{{1 - r}}} \right)}  \end{equation*}

Since a and r are constants, equation 4 can be written as

(5)   \begin{equation*} S = \frac{a}{{1 - r}} - \frac{a}{{1 - r}}\mathop {\lim }\limits_{n \to \infty } {r^n} \end{equation*}

If \left|r\right| <1:

    \[\mathop {\lim }\limits_{n \to \infty } {r^n} = 0\]

So if \left|r\right| <1, from equation (5), it follows that the geometric series converges, and is equal to

(6)   \begin{equation*} S = \frac{a}{{1 - r}} \end{equation*}

Here are a few interesting consequences of the definition of the sum of a geometric series (equation (4)).

    \[99999.... = .9 + .9\frac{1}{{10}} + .9{\left( {\frac{1}{{10}}} \right)^2} + .9{\left( {\frac{1}{{10}}} \right)^3} + ...\]

In the example above, a = .9 and r = 1/10. So,

    \[.99999... = \frac{a}{{1 - r}} = \frac{{.9}}{{1 - \frac{1}{{10}}}} = \frac{{.9}}{{.9}}\]

So

    \[\boxed{.99999... = 1}\]

    \[\sqrt {2\sqrt {2\sqrt {2\sqrt {2\sqrt {2\sqrt {2...} } } } } }  = 2\]

Proof:

(7)   \begin{equation*}\sqrt {2\sqrt {2\sqrt 2 }... }  = {2^{\frac{1}{2}}} \cdot {2^{\frac{1}{4}}} \cdot {2^{\frac{1}{8}}} + ... = {2^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ....}} \end{equation*}

Focusing on the exponent on the right side, ^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ....}. This is a geometric series with a = \frac{1}{2} and r = \frac{1}{2}.

(8)   \begin{equation*}S = \frac{a}{{1 - r}} =\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... =  \frac{{\frac{1}{2}}}{{1 - \frac{1}{2}}} = \frac{{1/2}}{{1/2}} = 1 \end{equation*}

From (7) and (8):

    \[\sqrt {2\sqrt {2\sqrt 2 ...} }  = {2^{\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ....}} = {2^1}\]

    \[\boxed{\sqrt {2\sqrt {2\sqrt {2\sqrt {2\sqrt {2\sqrt {2...} } } } } }  = 2}\]

In the example above, I used 2. The choice of 2 was arbitrary. If c is a positive number:

    \[\boxed{\sqrt {c\sqrt {c\sqrt {c\sqrt {c\sqrt c ...} } } }  = c}}\]