Infinite Series

Section 1.4 Infinite Series

Objectives

Understand the definition of an infinite series.
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Understand the definition of the convergence and divergence of a series.
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Understand if a series is geometric and under what conditions it converges or diverges.
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Learn how to apply convergence tests to other series.
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As we will see below, a power series is a function of $x\text{,}$ however, recall the definition of the series as well as its convergence.

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Definition 1.4.1.

The sum of the terms of a infinite sequence $(a_{1},a_{2},a_{3},\ldots)$ or

\begin{equation*} a_{1} + a_{2} + a_{3} + \cdots + a_{n} + \cdots \end{equation*}

is called an infinite series (or just a series) and is denoted

\begin{equation*} \sum_{n=1}^{\infty}a_{n} \qquad \text{or}\qquad \sum a_{n} \end{equation*}

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Why would we want to add up an infinite number of terms? Let’s answer this with the following picture.

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Figure 1.4.2.

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If we look at this as area, the total area is the area of the square or 1. If we look at this as a sum of the areas of rectangles, then the area is

\begin{equation*} \begin{aligned} s\amp= a_{1} + a_{2} + a_{3} + \cdots \\ \amp= \frac{1}{2}+ \frac{1}{4}+ \frac{1}{8}+ \cdots \end{aligned} \end{equation*}

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And therefore this infinite sum is 1. This is an example of a geometric series which we will generalize below.

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Subsection 1.4.1 Convergence of Series

The example above converged because the infinite sum was a value. We more formally define convergence by looking at a partial sum.

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Definition 1.4.3.

For the series $\sum a_{n}\text{,}$ define the $n$th partial sum, $s_{n}$ as

\begin{equation*} s_{n} = \sum_{k=1}^{n} a_{n} = a_{1} + a_{2} + \cdots + a_{n}. \end{equation*}

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If the sequence $\{ s_{n}\}$ is convergent and $s= \lim_{n\rightarrow \infty}s_{n}\text{,}$ then we write

\begin{equation*} \sum a_{n} = a_{1} + a_{2} + a_{3} + \cdots + a_{n} + \cdots = s \end{equation*}

and the number $s$ is called the sum of the series.

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If $\lim_{n \rightarrow \infty}s_{n}$ does not converge, then we say the series diverges.

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In summary, to determine if an infinite series converges, then

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Find the partial sum of the first $n$ terms. Call this $s_{n}\text{.}$
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Determine if the sequence $\{ s_{n} \}$ converges. Use the standard limit theorems to find this.
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Example 1.4.4.

Show that

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

converges and find the sum.

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The way that you need to find this is to recall that you can use partial fractions to write:

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

therefore we can write:

\begin{equation*} \begin{aligned} \sum_{n=0}^{\infty}\frac{1}{n(n+1)}\amp= \sum_{n=0}^{\infty}\biggl( \frac{1}{n}- \frac{1}{n+1}\biggr) \\\amp= \biggl(1- \frac{1}{2}\biggr) + \biggl(\frac{1}{2}- \frac{1}{3}\biggr) + \biggl(\frac{1}{3}- \frac{1}{4}\biggr) + \cdots \end{aligned} \end{equation*}

and note that the partial sums are

\begin{equation*} \begin{aligned} s_{1}\amp= 1- \frac{1}{2}\\ s_{2}\amp= \biggl(1-\frac{1}{2}\biggr) + \biggl(\frac{1}{2}- \frac{1}{3}\biggr) = 1-\frac{1}{3}\\ s_{3}\amp= \biggl(1-\frac{1}{2}\biggr) + \biggl(\frac{1}{2}- \frac{1}{3}\biggr) + \biggl(\frac{1}{3}- \frac{1}{4}\biggr) = 1- \frac{1}{4}\\ \vdots\amp= \vdots \\ s_{k}\amp= 1- \frac{1}{k+1} \end{aligned} \end{equation*}

so the sum is

\begin{equation*} s = \lim_{n \rightarrow \infty}s_{k} = \lim_{n \rightarrow \infty}\biggl(1- \frac{1}{k+1}\biggr) = 1 \end{equation*}

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Subsection 1.4.2 Geometric Series

Above we saw an example of a geometric series. We now generalize this.

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Definition 1.4.5.

A geometric series is a series in the form:

\begin{equation*} s = a + ar + ar^{2} + \cdots ar^{k} + \cdots = \sum_{n=0}^{\infty}a r^{n-1} \end{equation*}

for real numbers $a$ and $r\text{.}$

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One nice aspect of a geometric series is that ratios of consecutive terms are constant. This means that:

\begin{equation*} \frac{a_{n+1}}{a_{n}} = \frac{a r^{n}}{ar^{n-1}}= r \end{equation*}

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Below is an example that uses this fact to determine if a series is a geometric series. However, we have the following theorem to determine the convergence of a geometric series.

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Theorem 1.4.6. Convergence of Geometric Series.

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Let

\begin{equation*} \sum_{k=0}^{\infty}a_{k} = \sum_{k=0}^{\infty}a r^{k} \end{equation*}

be a geometric series. If $|r| \lt 1\text{,}$ then the series converges. If $|r| \geq 1\text{,}$ the series diverges.

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Proof.

The partial sums of the geometric series are

\begin{equation*} \begin{aligned} s_{1}\amp= a_{1} = a\\ s_{2}\amp= a_{1} + a_{2} =a + ar \\ s_{3}\amp= a_{1} + a_{2} + a_{3} =a + ar+ar^{2} \end{aligned} \end{equation*}

and the $k$th partial sum is

\begin{equation*} s_{k} = a_{1} + a_{2} + \cdots + a_{k} = c +cr + \cdots + cr^{k-1} \end{equation*}

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If we multiply $s_{k}$ by $r$ we get:

\begin{equation*} \begin{aligned} rs_{k}\amp= r(a + ar + ar^{2} + \cdots ar^{k-1}) \\\amp=ar + ar^{2} + ar^{3} + \cdots ar^{k} \end{aligned} \end{equation*}

and subtracting $rs_{k}$ from $s_{k}$ we get:

\begin{equation*} \begin{aligned} s_{k} - rs_{k}\amp= \bigr(a + ar + ar^{2} + \cdots ar^{k-1}\bigr) - \bigl(ar + ar^{2} + ar^{3} + \cdots ar^{k}\bigr) \\\amp= a- ar^{k}. \end{aligned} \end{equation*}

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Next, we solve for $s_{k}\text{:}$

\begin{equation*} \begin{aligned} s_{k} - rs_{k}\amp= c-cr^{k}\\ s_{k}(1-r)\amp= c-cr^{k}\\ s_{k}\amp= \frac{c-cr^{k}}{1-r} \end{aligned} \end{equation*}

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Assume that $|r| < 1\text{,}$ then taking the limit,

\begin{equation*} \begin{aligned} s\amp= \lim_{n \rightarrow \infty}s_{n} = \lim_{n \rightarrow \infty}\frac{c-cr^{n}}{1-r}\\\amp= \frac{c}{1-r}\lim_{n \rightarrow \infty}(1 - r^{n}) \\\amp= \frac{c}{1-r} \end{aligned} \end{equation*}

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If $|r| \gt 1\text{,}$ then

\begin{equation*} \lim_{n \rightarrow \infty}r^{n} \end{equation*}

does not exist, so this diverges. Lastly, we look at $r=1$ and $r=-1\text{.}$ In each case, we return to the series. If $r=1\text{,}$

\begin{equation*} \sum_{n=0}^{\infty}a r^{n} = a+a+a+a+ \cdots \end{equation*}

and this diverges. If $r=-1\text{,}$ then

\begin{equation*} \sum_{n=0}^{\infty}a r^{n} = a-a+a-a+a-a \end{equation*}

which also diverges, so the theorem holds.

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Example 1.4.7.

For each of the following, determine if you have a geometric series. If so, find $r$ and $c\text{.}$

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$\displaystyle \displaystyle \sum_{n=1}^{\infty}3 \biggl(\frac{2}{3}\biggr)^{n}$
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$\displaystyle \displaystyle \sum_{n=0}^{\infty}\frac{1}{n}$
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$\displaystyle \sum_{n=4}^{\infty}\frac{(-2)^{n}}{32}\text{.}$
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$\displaystyle \displaystyle 0.09090909\ldots = 0.\overline{09}$
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In each case above, we try to determine $c$ and $r\text{.}$ x

Note that this fits the form of the geometric series with $r=2/3$ and $a=2\text{.}$ Since $|r| \lt 1\text{,}$ then the series converges to

\begin{equation*} s= \frac{a}{1-r}= \frac{2}{1-2/3}= 6. \end{equation*}

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In this case

\begin{equation*} \frac{a_{n+1}}{a_{n}}= \frac{1/(n+1)}{1/n}= \frac{n}{n+1} \end{equation*}

which is not a constant, so this is not geometric.

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In this case the term starts at $n=4\text{,}$ however, $a$ is still the first term in the series, or 1/2 and $r=-2\text{,}$ so this is geometric. However, since $|r| \gt 1\text{,}$ the series does not converge.
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This doesn’t appear to be a geometric series. However if we write it as:

\begin{equation*} \begin{aligned} s\amp= 0.09 + 0.0009 + 0.000009 + \cdots \\\amp= \frac{9}{100}+ \frac{9}{(100)^{2}}+ \frac{9}{(100)^{3}}+ \cdots \end{aligned} \end{equation*}

and note that

\begin{equation*} \frac{a_{n+1}}{a_{n}} = \frac{\frac{9}{(100)^{n+1}}}{\frac{9}{(100)^{n}}}= \frac{1}{100} \end{equation*}

and since this is constant it is a geometric series with $r=1/100$ and since this satisfies $|r| \lt 1\text{,}$ the series converges. The first term is $a=9/100$ and thus the series converges to

\begin{equation*} \frac{a}{1-r}= \frac{9/100}{1-1/100}= \frac{9/100}{99/100}= \frac{9}{99}= \frac{1}{11} \end{equation*}

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This technique works to find the rational number version of any repeated decimal, which can always be written as a geometric series.

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Subsection 1.4.3 Tests for Convergence of Series

The convergence of a series is a very important concept. Without going over all of tests for series, we refer again to APEX Calculus, specifically Chapter 8. We cover a few tests here:

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Theorem 1.4.8. $p$-test for Series.

The series

\begin{equation*} \sum_{n=1}^{\infty}\frac{1}{n^{p}} \end{equation*}

converges if $p \gt 1$ and diverges if $p \leq 1\text{.}$

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Theorem 1.4.9. Alternating Series Test.

If the alternating series:

\begin{equation*} \sum_{n=1}^{\infty}(-1)^{n+1}b_{n} = b_{1} -b_{2}+b_{3} -b_{4} + b_{5} + \cdots \end{equation*}

for $b_{n} \gt 0$ satisfies:

$b_{n+1}\leq b_{n}$ for all $n\text{,}$ and
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$\displaystyle \displaystyle \lim_{n \rightarrow \infty}b_{n} =0$
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then the series converges.

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Example 1.4.10.

Determine if the following series converge?

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$\displaystyle \displaystyle \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n}$
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$\displaystyle \displaystyle \sum_{n=1}^{\infty}(-1)^{n} \frac{n}{(n+1)^{2}}$
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In the first case $b_{n} = 1/n\text{.}$ Thus $b_{n}$ is positive and decreasing. Also, since

\begin{equation*} \lim_{n \rightarrow \infty}b_{n} = 0 \end{equation*}

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Then the series converges.
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In this case $b_{n} = n/(n+1)^{2}\text{.}$ Again, this is a positive series. To show that $b_{n+1}\leq b_{n}\text{,}$ it’s easier to show that $b_{n} = f(n)$ and $f'(x) \lt 0\text{.}$

\begin{equation*} \begin{aligned} f(x)\amp= x(x+1)^{-2}\\ f'(x)\amp= (x+1)^{-2}- 2x (x+1)^{-3}= \frac{(x+1) -2x}{(x+1)^{3}}\\\amp= \frac{1-x}{(x+1)^{3}} \end{aligned} \end{equation*}

which is negative for all $x \gt 1\text{,}$ so it is a decreasing function. Also,

\begin{equation*} \lim_{n \rightarrow \infty}\frac{n}{(n+1)^{2}}= 0 \end{equation*}

so the series converges by the AST.

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One of the most robust tests, however, is covered here:

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Theorem 1.4.11. Ratio Test for Absolute Convergence (RATFACE).

For the series, $\sum a_{n}\text{,}$ let

\begin{equation*} L = \lim_{n \rightarrow \infty}\biggl\vert \frac{a_{n+1}}{a_{n}}\biggr\vert \end{equation*}

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if $L \lt 1\text{,}$ then $\sum a_{n}$ converges absolutely.
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if $L \gt 1$ or $L$ does not exist, then $\sum a_{n}$ diverges.
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if $L=1\text{,}$ the test is inconclusive.
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Proof.

Consider the case when $L \lt 1\text{.}$ We will compare the series to a geometric series with $L \lt r \lt 1\text{.}$ Since

\begin{equation*} \lim_{n \rightarrow \infty}\biggl\vert \frac{a_{n+1}}{a_{n}}\biggr\vert =L \end{equation*}

converges, then eventually $|a_{n+1}/a_{n}|$ will be less than $r$ and let’s say that this occurs at $N\text{,}$ specifically:

\begin{equation*} \biggl\vert \frac{a_{n+1}}{a_{n}}\biggr\vert \lt r \qquad \text{for all $n\geq N$.} \end{equation*}

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This results in

\begin{equation*} \begin{aligned} |a_{N+1}|\amp \lt |N_{n}| r \\ |a_{N+2}|\amp \lt |a_{N+1}| r \lt |a_{N}| r^{2} \\ \vdots\amp\\ |a_{N+k}|\amp \lt |a_{N}| r^{k} \end{aligned} \end{equation*}

for all $k\geq 1$ an

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Recall that we want to show that the series $\sum a_{n}$ converges absolutely, which means that $\sum |a_{n}|$ converges.

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\begin{equation*} \begin{aligned} \sum_{n=0}^{\infty}|a_{n}|\amp= \sum_{n=1}^{N-1}|a_{n}| + \sum_{n=N}^{\infty}|a_{n}| \\\amp \lt \sum_{n=1}^{N-1}|a_{n}| +\sum_{n=N}^{\infty}|a_{N}| r^{n} \end{aligned} \end{equation*}

and since the second series is geometric with $r \lt 1\text{,}$ it converges, therefore, $\sum |a_{n}|$ converges and that means that $\sum a_{n}$ converges. Similar techniques can be used to show the results if $L \gt 1\text{.}$

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The next example shows how to apply the Ratio Test.

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