Geometric Series of Real Numbers Examples 1

# Geometric Series of Real Numbers Examples 1

Recall from Geometric Series of Real Numbers page the following test for convergence/divergence of a geometric series:

Test for Convergence/Divergence of Geometric Series

Consider the geometric series $\displaystyle{\sum_{n=1}^{\infty} ax^{n-1} = a + ax + ax^2 + ...}$ where $a \in \mathbb{R}$.

a) If we have that $\mid x \mid < 1$, then we conclude that:

• The geometric series $\displaystyle{\sum_{n=1}^{\infty} ax^{n-1}}$ converges to the sum $\displaystyle{\frac{a}{1 - x}}$.

b) If instead we have that $\mid x \mid \geq 1$, then we conclude that:

• The geometric series $\displaystyle{\sum_{n=1}^{\infty} ax^{n-1}}$ diverges.

We will now look at some examples of applying the test to geometric series.

## Example 1

Determine whether the series $\displaystyle{\sum_{n=1}^{\infty} 2 \cdot 3^n}$ converges or diverges, and if it converges, find the sum of the series.

Notice that $x = 3$ in this example, and so $\mid x \mid = \mid 3 \mid \geq 1$. Therefore $\displaystyle{\sum_{n=1}^{\infty} 2 \cdot 3^n}$ diverges.

## Example 2

Determine whether the series $\displaystyle{\sum_{n=1}^{\infty} \frac{2e^2}{\pi^n}}$ converges or diverges, and if it converges, find the sum of the series.

In this example, notice that $a = 2e^2$ and $\displaystyle{x = \frac{1}{\pi}}$. Since $\pi > 1$, $0 < \frac{1}{\pi} < 1$ and hence $\biggr \lvert \frac{1}{\pi} \biggr \rvert < 1$. So the series $\displaystyle{\sum_{n=1}^{\infty} \frac{2e^2}{\pi^n}}$ converges, and:

(1)
\begin{align} \quad \sum_{n=1}^{\infty} \frac{2e^2}{\pi^n} = \frac{2e^2}{1 - \frac{1}{\pi}} = \frac{2e^2 \pi}{\pi - 1} \end{align}

## Example 3

Determine whether the series $\displaystyle{\sum_{n=1}^{\infty} \frac{e 5^n}{e^{2n}}}$ converges or diverges, and if it converges, find the sum of the series.

In this example, $a = e$, and $x = \frac{5}{e^2}$. Notice that $e^2 > 5$ and so $\displaystyle{\biggr \lvert \frac{5}{e^2} \biggr \rvert < 1}$. Therefore the series $\displaystyle{\sum_{n=1}^{\infty} \frac{e 5^n}{e^{2n}}}$ converges and:

(2)
\begin{align} \quad \sum_{n=1}^{\infty} \frac{e 5^n}{e^{2n}} = \frac{e}{1 - \frac{5}{e^2}} = \frac{e^3}{e^2 - 5} \end{align}

## Example 4

Let $S = \{ x_1, x_2, ... \}$ be the collection of all natural numbers that do not contain the digit $0$, i.e. $6 \in S$ but $60 \not \in S$. Prove that $\sum_{k=1}^{\infty} \frac{1}{x_k} < 90$.

Partition $S$ into an infinite collection of sets $\{ S_1, S_2, ..., S_j, ... \}$ where $S_k \subset S$ is the set of all natural numbers that do not contain the digit and have $j$ digits total. If $k = 1$, notice that:

(3)
\begin{align} \quad \frac{1}{9} \leq \frac{1}{x_k} \leq 1 \end{align}

Furthermore, we see that:

(4)
\begin{align} \quad \sum_{x_k \in S_1} \frac{1}{x_k} < 9 \cdot 1 = \frac{9^1}{10^0} \end{align}

If $k = 2$, then notice that:

(5)
\begin{align} \quad \frac{1}{99} \leq \frac{1}{x_k} < \frac{1}{10} \end{align}

There are $90$ numbers between $10$ and $99$ (inclusive), and $9$ of these numbers contain a $0$. So there are $81$ valid reciprocals to sum and:

(6)
\begin{align} \quad \sum_{x_k \in S_2} \frac{1}{x_k} < \frac{81}{10} = \frac{9^2}{10^1} \end{align}

If $k = 3$, then notice that:

(7)
\begin{align} \quad \frac{1}{999} \leq \frac{1}{x_k} < \frac{1}{100} \end{align}

There are $900$ numbers between $100$ and $999$ (inclusive). There are $729$ valid reciprocals to sum and:

(8)
\begin{align} \quad \sum_{x_k \in S_3} \frac{1}{x_k} < \frac{729}{100} = \frac{9^3}{10^2} \end{align}

In general, for $j \in \mathbb{N}$ we see that:

(9)
\begin{align} \quad \sum_{x_k \in S_j} \frac{1}{x_k} < \frac{9^j}{10^{j-1}} \end{align}

So we see that the sum of the reciprocals of numbers in $S$ is bounded since:

(10)
\begin{align} \quad \sum_{k=1}^{\infty} \frac{1}{x_k} = \sum_{j=1}^{\infty} \left ( \sum_{x_k \in S_j} \frac{1}{x_k} \right ) < \sum_{j=1}^{\infty} \frac{9^j}{10^{j-1}} = 9 \sum_{j=1}^{\infty} \left( \frac{9}{10} \right )^{j-1} = 9 \sum_{j=0}^{\infty} \left ( \frac{9}{10} \right )^j = 9 \frac{1}{1 - \frac{9}{10}} = 9 \frac{1}{\frac{1}{10}} = 90 \end{align}