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Section 9.3 Infinite Series

Sequence vs Series.

Sequence: \(a_1, \quad a_2, \quad a_3, \cdots\)

Series: \(a_1 + a_2 + a_3 + \cdots\)

A series is when you add up a sequence.

Examples of series:

  • \(\displaystyle 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots = \displaystyle \sum_{n = 1}^{\infty} \frac{1}{n} = \infty\)

  • \(\displaystyle 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots = \displaystyle \sum_{n = 1}^{\infty} \frac{1}{2^{n-1}} = 2\)

  • \(\displaystyle 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \dots = \displaystyle \sum_{n = 1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}\)

Why do some series converge and others not?

Definition 9.3.1. Partial Sums.

We define the partial sum, \(S_n\text{,}\) of the first \(n\) terms of a series to be:

\begin{equation*} S_n = \sum_{i = 1}^n a_i = a_1 + a_2 + a_3 + \dots + a_n \end{equation*}

Consider \(\sum_{n = 1}^{\infty} \frac{1}{n} = a + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots\)

\begin{align*} S_1 \amp = 1\\ S_2 \amp = 1 + \frac{1}{2} = \frac{3}{2}\\ S_3 \amp = 1 + \frac{1}{2} + \frac{1}{3} = \frac{11}{6}\\ S_4 \amp = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{25}{16}\\ \amp \vdots \end{align*}

If we make this a sequence of partial sums, does it converge or diverge:

\begin{equation*} 1, \quad \frac{3}{2}, \quad \frac{11}{6}, \quad \frac{25}{16}, \dots \boxed{\text{diverges}} \end{equation*}
Series Convergence.

A series converges if the sequence of partial sums converges and divereges if the series of partial sums diverges.

  • If it converges we write:

    \begin{equation*} \sum_{a = i}^{\infty} a_i = S \end{equation*}

Subsection 9.3.1 Geometric Series

Definition 9.3.2. Geometric Series.

A geometric series is one in which each term is a constant multiple of the term before it.

The first term is \(a\) and the constant multiple (or common ratio) is \(x\text{.}\)

A finite geometric series has the form:

\begin{equation*} a + ax + ax^2 + ax^3 + \dots + ax^{n-1} = \sum_{i=1}^n ax^{i-1} \end{equation*}

An infinite geometric series has the form:

\begin{equation*} a + ax + ax^2 + ax^3 + \dots + ax^{n-1} + \dots = \sum_{i=1}^{\infty} ax^{i-1} \end{equation*}

The sum of a geometric series:

\(\) \(\) \(S_n\) \(= \) \(a\) \(+\) \(ax\) \(+\) \(ax^2\) \(+\) \(\dots\) \(+\) \(ax^{n-2}\) \(+\) \(ax^{n-1}\)
\(\) \(- \) \((xS_n\) \(= \) \(ax\) \(+\) \(ax^2\) \(+\) \(ax^3\) \(+\) \(\dots\) \(+\) \(ax^{n-1}\) \(+\) \(ax^{n})\)
\(S_n\) \(- \) \(xS_n\) \(= \) \(a\) \(- \) \(1\) \(- \) \(ax^n\)
\(S_n\) \((1\) \(- x)\) \(= \) \(a\) \((1\) \(- \) \(x^n)\)

\begin{align*} S_n \amp = \dfrac{a(1-x^n)}{1-x}\\ \sum_{i = 1}^n ax^{i-1} \amp = \dfrac{a(1-x^n)}{1-x} \end{align*}

The infinite geometric series will converge if the limit of the following exists:

\begin{equation*} \lim_{n\to\infty} S_n = \lim_{n\to\infty} \dfrac{a(1-x^n)}{1-x} \end{equation*}

  • if \(\mid x\mid \gt 1\) then the \(\displaystyle \lim_{n\to\infty} S_n\) doesn't exist so the geometric series diverges

  • if \(x = 1\) the geometric series diverges

  • if \(x = - 1\) the geometric series diverge

  • if \(\mid x \mid \lt 1\) the geometric series coverges to \(\dfrac{a}{1-x}\)

Example 9.3.1.

Does the following series converge and if so what does it converge to:

\begin{equation*} 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots \end{equation*}

  • \(a = 1, \quad x = \dfrac{1}{2}, \quad \mid x \mid = \dfrac{1}{2} \lt 1\)

    \(\boxed{\text{converges}}\)

  • \(\displaystyle \frac{a}{1-x} = \dfrac{1}{1-\dfrac{1}{2}} = \dfrac{1}{\frac{1}{2}} = \boxed{2}\)

Subsection 9.3.2 Telescoping Series

A telescoping series can be thought of a "colapsable". An example of a telescoping series is:

\begin{equation*} \sum_{n=1}^{\infty} \frac{1}{(n + 2)(n + 3)} = \frac{1}{12} + \frac{1}{20} + \frac{1}{30} + \frac{1}{42} + \dots \end{equation*}

If we take the partial sums we get:

\begin{align*} S_1 \amp = \frac{1}{12}\\ S_2 \amp = \frac{2}{15}\\ S_3 \amp = \frac{1}{6} \end{align*}

We can rewrite using partial fractions:

\begin{align*} \frac{1}{(n+2)(n+3)} \amp = \frac{A}{n+2} + \frac{B}{n+3}\\ \amp = \frac{A (n+3) }{(n+2)(n+3)} + \frac{B(n+2)}{(n+3)(n+2)}\\ \amp = \frac{A (n+3) + B(n+2) }{(n+2)(n+3)} \\ 1 \amp = A(n+3) + B(n+2) \end{align*}

Solve for A:

\begin{equation*} n = -2 \end{equation*}
\begin{align*} 1 \amp = A(-2+3) + B(-2+2)\\ \amp = A(1) + 0\\ 1 \amp = A \end{align*}

Solve for B:

\begin{equation*} n = -3 \end{equation*}
\begin{align*} 1 \amp = A(-3+3) + B(-3+2)\\ \amp = B(-1) + 0\\ -1 \amp = B \end{align*}

\begin{equation*} \frac{1}{(n+2)(n+3)} = \frac{1}{n+2} - \frac{1}{n+3} \end{equation*}

Now if we find the sum using this new notation we get:

\begin{align*} S_N \amp = \sum_{n=1}^{N} \frac{1}{(n+2)(n+3)} \\ \amp = \sum_{n=1}^{N} \left( \frac{1}{n+2} - \frac{1}{n+3} \right) \\ \amp = \left( \frac{1}{3} - \frac{1}{4} \right) + \left( \frac{1}{4} - \frac{1}{5} \right) + \left( \frac{1}{5} - \frac{1}{6} \right) + \dots + \left( \frac{1}{N+2} - \frac{1}{N+3} \right)\\ \amp = \frac{1}{3} - \frac{1}{N+3} \end{align*}

So if we find the limit as \(N\) goes to infinity we get:

\begin{align*} \lim_{N\to\infty} S_N \amp = \lim_{N\to\infty} \frac{1}{3} - \frac{1}{N+3} \\ \amp = \frac{1}{3} - 0\\ \amp = \frac{1}{3} \end{align*}

So the original series converges:

\begin{equation*} \sum_{n = 1}^{\infty} \frac{1}{(n+2)(n+3)} = \frac{1}{3} \end{equation*}