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# 8.2: Infinite Series

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Demonstrate an understanding of series and the sequence of partial sums
• Recognize geometric series and determine when they converge or diverge
• Compute the sum of a convergent geometric series
• Determine convergence or divergence of series using the nth-Term Test

## Infinite Series (series, sequence of partial sums, convergence, divergence)

### Series

Another topic that involves an infinite number of terms is the topic of infinite series. We can represent certain functions and numbers with an infinite series. For example, any real number that can be written as a non-terminating decimal can be represented as an infinite series.

Example 1

The rational number 49\begin{align*}\frac{4}{9}\end{align*} can be written as 0.44444\begin{align*}0.44444\end{align*}…. We can expand the decimal notation as an infinite series:

49=0.4+0.04+0.004+0.0004+=410+4100+41000+410,000+=410+4102+4103+4104+\begin{align*}\frac {4}{9} & = 0.4+0.04+0.004+0.0004+\ldots \\ & = \frac {4}{10}+\frac {4}{100}+\frac {4}{1000}+\frac {4}{10,000}+\ldots \\ & = \frac {4}{10}+\frac {4}{10^2}+\frac {4}{10^3}+\frac {4}{10^4}+\ldots\end{align*}

On the other hand, the number 14\begin{align*}\frac{1}{4}\end{align*} can be written as 0.25\begin{align*}0.25\end{align*}. If we expand the decimal notation, we get a finite series:

14=0.2+0.05=210+5100=210+5102\begin{align*}\frac {1}{4} & = 0.2+0.05 \\ & = \frac {2}{10}+\frac {5}{100} \\ & = \frac {2}{10}+\frac {5}{10^2}\end{align*}

Do you see the difference between an infinite series and a finite series? Let’s define what we mean by an infinite series.

Infinite Series

An infinite series is the sum of an infinite number of terms, u1,u2,u3,u4,,\begin{align*}u_1, u_2, u_3, u_4,\ldots,\end{align*} usually written as.

u1+u2+u3+u4+.\begin{align*}u_1 + u_2 + u_3 + u_4 +\ldots.\end{align*}

A shorthand notation for an infinite series is to use sigma notation:

k=1uk\begin{align*}\sum_{k=1}^\infty u_k\end{align*}, which can be read as “the sum of the terms uk\begin{align*}u_k\end{align*} ’s for k\begin{align*}k\end{align*} equal to 1\begin{align*}1\end{align*} to infinity.”

We can make finite sums from the terms of the infinite series:

s1s2s3=u1=u1+u2=u1+u2+u3\begin{align*}s_1 & = u_1\\ s_2 & = u_1 + u_2\\ s_3 & = u_1 + u_2 + u_3\end{align*}

The first sum is the first term of the sequence. The second sum is the sum of the first two terms. The third term is the sum of the first three terms. Thus, the n\begin{align*}n\end{align*}th finite sum, sn\begin{align*}s_n\end{align*} is the sum of the first n\begin{align*}n\end{align*} terms of the infinite series: sn=u1+u2+u3++un\begin{align*}s_n=u_1 + u_2 + u_3 + \ldots + u_n\end{align*}.

### Sequence of Partial Sums

As you can see, the sums sn=u1+u2+u3++un\begin{align*}s_n = u_1 + u_2 + u_3 +\ldots+ u_n\end{align*} form a sequence. The sequence is very important for the study of the related infinite series for it tells a lot about the infinite series.

Partial Sums

For an infinite series k=1uk\begin{align*}\sum_{k=1}^\infty u_k\end{align*} , the n\begin{align*}n\end{align*}th partial sum, sn\begin{align*}s_n\end{align*} is the sum of the first n\begin{align*}n\end{align*} terms of the infinite series: sn=k=1nuk\begin{align*}s_n=\sum_{k=1}^n u_k\end{align*} .

The sequence {sn}\begin{align*}\left \{s_n \right \}\end{align*} formed from these sums is called the sequence of partial sums.

Example 2

Find the first five partial sums of the infinite series 1+0.1+0.01+0.001+....\begin{align*}1 + 0.1 + 0.01 + 0.001 + ....\end{align*}

Solution

s1s2s3s4s5=u1=1=u1+u2=1+0.1=1.1=1+0.1+0.01=1.11=1+0.1+0.01+0.001+0.0001=1.111=1+0.1+0.01+0.001+0.0001=1.1111\begin{align*}s_1 & = u_1 = 1 \\ s_2 & = u_1 + u_2 = 1 + 0.1 = 1.1 \\ s_3 & = 1 + 0.1 + 0.01 = 1.11 \\ s_4 & = 1 + 0.1 + 0.01 + 0.001 + 0.0001 = 1.111 \\ s_5 & = 1 + 0.1 + 0.01 + 0.001 + 0.0001 = 1.1111\end{align*}

To further explore series, try experimenting with this applet. The applet shows the terms of a series as well as selected partial sums of the series. Series Applet. As you see from this applet, for some series the partial sums appear to approach a fixed number, while for other series the partial sums do not. Exploring this phenomenon is the topic of the next sections.

## Convergence and Divergence

Just as with sequences, we can talk about convergence and divergence of infinite series. It turns out that the convergence or divergence of an infinite series depends on the convergence or divergence of the sequence of partial sums.

Convergence/Divergence of Series

Let k=1uk\begin{align*}\sum_{k=1}^\infty u_k\end{align*} be an infinite series and let {sn}\begin{align*}\left \{s_n \right \}\end{align*} be the sequence of partial sums for the series. If {sn}\begin{align*}\left \{s_n \right \}\end{align*} has a finite limit l\begin{align*}l\end{align*} , then the infinite series converges and k=1uk=l\begin{align*}\sum_{k=1}^\infty u_k = l\end{align*}.

If {sn}\begin{align*}\left \{s_n \right \}\end{align*} does not have a finite limit, then the infinite series diverges. The infinite series does not have a sum.

Example 3

Does the infinite series 1+0.1+0.01+0.001+\begin{align*}1 + 0.1 + 0.01 + 0.001 + \ldots\end{align*} converge or diverge?

Solution

To make our work easier, write the infinite series 1+0.1+0.01+0.001+\begin{align*}1 + 0.1 + 0.01 + 0.001 + \ldots\end{align*} as an infinite series of fractions:

1+110+1102+1103+\begin{align*}1 +\frac {1}{10}+\frac {1}{10^2}+\frac {1}{10^3}+ \ldots \end{align*}

To solve for convergence or divergence of the infinite series, write the formula for the n\begin{align*}n\end{align*}th partial sum sn=k=1nuk:sn=1+110+1102+1103++110n1\begin{align*}s_n=\sum_{k=1}^n u_k : s_n = 1 +\frac {1}{10}+\frac {1}{10^2}+\frac {1}{10^3}+ \ldots+ \frac {1}{10^{n-1}}\end{align*} . Note that the n\begin{align*}n\end{align*}th partial sum ends with a power of n1\begin{align*}n - 1\end{align*} in the denominator because 1\begin{align*}1\end{align*} is the first term of the infinite series.

It is rather difficult to find limn+sn=limn+1+110+1102+1103++110n1\begin{align*}\lim_{n \to +\infty} s_n =\lim_{n \to +\infty}1 +\frac {1}{10}+\frac {1}{10^2}+\frac {1}{10^3}+\ldots+\frac {1}{10^{n-1}}\end{align*} as it is written. We will “work” the sum into a different form so that we can find the limit of the sequence of partial sums.

First, multiply both sides of the equation sn=1+110+1102+1103++110n1\begin{align*}s_n = 1 +\frac {1}{10}+\frac {1}{10^2}+\frac {1}{10^3}+\ldots+\frac {1}{10^{n-1}}\end{align*} by 110\begin{align*}\frac {1}{10}\end{align*}:

110sn110sn=110(1+110+1102+1103++110n1)=110+1102+1103+1104++110n\begin{align*}\frac {1}{10}s_n & = \frac {1}{10}\left(1 +\frac {1}{10}+\frac {1}{10^2}+\frac {1}{10^3}+\ldots+\frac {1}{10^{n-1}}\right)\\ \frac {1}{10}s_n & = \frac {1}{10} +\frac {1}{10^2}+\frac {1}{10^3}+\frac {1}{10^4}+\ldots+\frac {1}{10^n}\end{align*}

Now we have two equations:

sn110sn=1+110+1102+1103++110n1=110+1102+1103+1104++110n\begin{align*}s_n & = 1 +\frac {1}{10}+\frac {1}{10^2}+\frac {1}{10^3}+\ldots+\frac {1}{10^{n-1}}\\ \frac {1}{10}s_n & = \frac {1}{10} +\frac {1}{10^2}+\frac {1}{10^3}+\frac {1}{10^4}+ \ldots + \frac {1}{10^n}\end{align*}

Subtract the bottom equation from the top equation to cancel terms and simplifying:

sn=1+110+1102+1103+1104++110n1110sn=(110+1102+1103+1104++110n)sn110sn=1110n910sn=1110n\begin{align*}& s_n =1+ \frac {1}{10} +\frac {1}{10^2}+\frac {1}{10^3}+\frac {1}{10^4} + \ldots+ \frac {1}{10^{n-1}} \\ & -\frac {1}{10}s_n = - \left(\frac {1}{10} +\frac {1}{10^2}+\frac {1}{10^3}+\frac {1}{10^4} + \ldots + \frac {1}{10^n}\right)\\ & \underline{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\\ & s_n-\frac {1}{10}s_n = 1- \frac {1}{10^n} \\ & \frac {9}{10}s_n = 1- \frac {1}{10^n}\end{align*}

Solve for sn\begin{align*}s_n\end{align*} by multiplying both sides of the last equation by 109\begin{align*}\frac {10}{9}\end{align*} :

sn=109(1110n)\begin{align*}s_n=\frac {10}{9}(1- \frac {1}{10^n})\end{align*}

Now we find the limit of both sides:

limn+snlimn+109(1110n)=limn109(1110n)=limn109limn+109(110n)=1090=109\begin{align*}\lim_{n \to +\infty} s_n & = \lim_{n \to \infty}\frac {10}{9} \left (1- \frac {1}{10^n}\right ) \\ \lim_{n \to +\infty}\frac {10}{9} \left (1- \frac {1}{10^n}\right ) & = \lim_{n \to \infty}\frac {10}{9} - \lim_{n \to +\infty}\frac {10}{9}\left(\frac {1}{10^n}\right)\\ & = \frac {10}{9}-0= \frac {10}{9}\end{align*}

The sum of the infinite series is 109\begin{align*}\frac {10}{9}\end{align*} and so the series converges.

## Geometric Series

The geometric series is a special kind of infinite series whose convergence or divergence is based on a certain number associated with the series.

Geometric Series

A geometric series is an infinite series written as

a+ar+ar2+ar3++ar{i1}+.\begin{align*}a + ar + ar^2 + ar^3 + \ldots+ ar^{\left \{i-1 \right \} }+ \ldots . \end{align*}

In sigma notation, a geometric series is written as k=1ark1\begin{align*}\sum_{k=1}^\infty ar^{k-1}\end{align*}.

The number r\begin{align*}r\end{align*} is the ratio of the series.

Example 4

Here are some examples of geometric series.

Geometric Series a\begin{align*}a\end{align*} r\begin{align*}r\end{align*}
1+14+142+143++14k1+\begin{align*}1+\frac {1}{4}+\frac {1}{4^2} +\frac {1}{4^3}+\ldots+\frac {1}{4^{k-1}}+\ldots\end{align*} 1\begin{align*}1\end{align*} 14\begin{align*}\frac {1}{4}\end{align*}
56+562563++(56)(16)+\begin{align*}-\frac {5}{6}+\frac {5}{6^2} -\frac {5}{6^3}+\ldots+\left(-\frac {5}{6}\right) \left(-\frac{1}{6}\right)+\ldots\end{align*} 56\begin{align*}-\frac{5}{6}\end{align*} 16\begin{align*}-\frac {1}{6}\end{align*}
1+3+32+33++3k1+\begin{align*}1 + 3 + 3^2 + 3^3 + \ldots+ 3^{k-1} + \ldots\end{align*} 1\begin{align*}1\end{align*} 3\begin{align*}3\end{align*}

The convergence or divergence of a geometric series depends on r\begin{align*}r\end{align*}.

Theorem

Suppose that the geometric series k=1ark1\begin{align*}\sum_{k=1}^\infty ar^{k-1}\end{align*} has ratio r\begin{align*}r\end{align*}.

1. The geometric series converges if |r|<1\begin{align*}|r| < 1\end{align*} and the sum of the series is a1r\begin{align*}\frac{a}{1-r}\end{align*}.
2. The geometric series diverges if |r|1\begin{align*}|r| \ge 1\end{align*}.

Example 5

Determine if the series 7+78+782+783++78i1+\begin{align*}7+\frac {7}{8}+\frac {7}{8^2} +\frac {7}{8^3}+\ldots+\frac {7}{8^{i-1}}+\ldots\end{align*} converges or diverges. If it converges, find the sum of the series.

Solution

The series is a geometric series that can be written as k=17(18)k1\begin{align*}\sum_{k=1}^\infty 7\left(\frac{1}{8}\right)^{k-1}\end{align*}. Then a=7\begin{align*}a = 7\end{align*} and the ratio r=18\begin{align*}r=\frac {1}{8}\end{align*}. Because 18<1\begin{align*}\left | \frac {1}{8} \right |< 1\end{align*} , the series converges. The sum of the series is a1r=7118=778=8\begin{align*}\frac{a}{1-r}=\frac{7}{1-\frac{1}{8}}=\frac{7}{\frac{7}{8}}=8\end{align*}.

Example 6

Determine if the series k=1+9k1\begin{align*}\sum_{k=1}^{+\infty} 9^{k-1}\end{align*} converges or diverges. If it converges, find the sum of the series.

Solution

The series is a geometric series with a=1\begin{align*}a = 1\end{align*} and the ratio r=9\begin{align*}r = 9\end{align*}. Because |9|>1\begin{align*}|9|>1\end{align*}, the series diverges.

Example 7

Determine if 34+342+343++3(1)k4k+\begin{align*}\frac {3}{4}+\frac {3}{4^2} +\frac {3}{4^3}+\ldots+\frac {3(-1)^k}{4^k}+ \ldots\end{align*} converges or diverges. If it converges, find the sum of the series.

Solution

If we rewrite the series in terms of powers of k\begin{align*}k\end{align*}, the series looks like this:

3(1)141+3(1)242+3(1)343++3(1)k4k+=3(14)1+3(14)2++3(14)k+.\begin{align*}\frac {3(-1)^1}{4^1}+\frac {3(-1)^2}{4^2}+\frac {3(-1)^3}{4^3}+ \ldots +\frac {3(-1)^k }{4^k}+ \ldots=3\left(-\frac{1}{4}\right)^1+3\left(-\frac{1}{4}\right)^2+ \ldots +3\left(-\frac{1}{4}\right)^k+ \ldots.\end{align*}

It looks like a geometric series with a=3\begin{align*}a = 3\end{align*} and r=14\begin{align*}r=-\frac{1}{4}\end{align*}.Since 14=14<1\begin{align*}\left | -\frac {1}{4} \right |= \frac {1}{4} < 1\end{align*}, the series converges.

However, if we write the definition of a geometric series for a=3\begin{align*}a = 3\end{align*} and r=14\begin{align*}r=-\frac{1}{4}\end{align*} , the series looks like this:

k=1+3(14k1)=3(14)0+3(14)1+3(14)2+=3+3(14)1+3(14)2+\begin{align*}\sum_{k=1}^{+\infty}3 \left (-\frac{1}{4}^{k-1}\right )& =3\left(-\frac{1}{4}\right)^0+3\left(-\frac{1}{4}\right)^1+3\left(-\frac{1}{4}\right)^2+\ldots \\ & = 3+3\left(-\frac{1}{4}\right)^{1}+3\left(-\frac{1}{4}\right)^2+\ldots\end{align*}

The original problem, 3(1)141+3(1)242+3(1)343++3(1)k4k+\begin{align*}\frac {3(-1)^1}{4^1}+\frac {3(-1)^2}{4^2}+\frac {3(-1)^3}{4^3}+ \ldots +\frac {3(-1)^k }{4^k}+\ldots\end{align*}, does not have the leading term of 3\begin{align*}3\end{align*}. This does not affect the convergence but will affect the sum of the series. We need to subtract 3\begin{align*}3\end{align*} from the sum of the series 3+3(14)1+3(14)2+\begin{align*}3+3\left(-\frac{1}{4}\right)^1+3\left(-\frac{1}{4}\right)^2+ \ldots\end{align*} to get the sum of 3(1)141+3(1)242+3(1)343++3(1)k4k+\begin{align*}\frac {3(-1)^1}{4^1}+\frac {3(-1)^2}{4^2}+\frac {3(-1)^3}{4^3}+ \ldots +\frac {3(-1)^k }{4^k}+ \ldots\end{align*}.

The sum of the series is: a1r3=31(14)3=3543=1253=125155=35\begin{align*}\frac{a}{1-r}-3=\frac{3}{1-\left(-\frac{1}{4}\right)}-3=\frac{3}{\frac{5}{4}}-3=\frac{12}{5}-3=\frac{12}{5}-\frac{15}{5}=-\frac{3}{5}\end{align*}.

## Other Convergent Series

There are other infinite series that will converge.

Example 8

Determine if k=1+(2k2k+1)\begin{align*}\sum_{k=1}^{+\infty} \left(\frac{2}{k}-\frac{2}{k+1}\right)\end{align*} converges or diverges. If it converges, find the sum.

Solution

The n\begin{align*}n\end{align*}th partial sum sn\begin{align*}s_n\end{align*} is:

Sn=k=1n(2k2k+1)=(2121+1)+(2222+1)+(2323+1)++(2n2n+1)=(2122)+(2223)+(2324)++(2n2n+1)\begin{align*}S_n & =\sum_{k=1}^n\left(\frac{2}{k}-\frac{2}{k+1}\right) \\ & = \left(\frac{2}{1}-\frac{2}{1+1}\right) +\left(\frac{2}{2}-\frac{2}{2+1}\right)+\left(\frac{2}{3}-\frac{2}{3+1}\right)+\ldots+\left(\frac{2}{n}-\frac{2}{n+1}\right)\\ & = \left(\frac{2}{1}-\frac{2}{2}\right)+\left(\frac{2}{2}-\frac{2}{3}\right)+\left(\frac{2}{3}-\frac{2}{4}\right)+\ldots+\left(\frac{2}{n}-\frac{2}{n+1}\right)\end{align*}

We can simplify sn\begin{align*}s_n\end{align*} further. Notice that the first parentheses has 22\begin{align*}-\frac{2}{2}\end{align*} while the second parentheses has 22\begin{align*}\frac{2}{2}\end{align*} . These will add up to 0\begin{align*}0\end{align*} and cancel out. Likewise, the 23\begin{align*}-\frac{2}{3}\end{align*} and 23\begin{align*}\frac{2}{3}\end{align*}

will cancel out. Continue in this way to cancel opposite terms. This sum is a telescoping sum, which is a sum of terms that cancel each other out so that the sum will fold neatly like a folding telescope. Thus, we can write the partial sum as

sn=212n+1=22n+1.\begin{align*}s_n=\frac{2}{1}-\frac{2}{n+1}=2-\frac{2}{n+1}.\end{align*}

Then limn+ sn=limn+ (22n+1)=2\begin{align*}\lim_{n \to +\infty}\ s_n= \lim_{n \to +\infty}\ \left (2-\frac{2}{n+1} \right )= 2\end{align*} and k=1+(2k2k+1)=2\begin{align*}\sum_{k=1}^{+\infty} \left (\frac{2}{k}-\frac{2}{k+1} \right )=2\end{align*}.

Other Divergent Series (n\begin{align*}n\end{align*}th-Term Test)

Determining convergence by using the limit of the sequence of partial sums is not always feasible or practical. For other series, it is more useful to apply tests to determine if an infinite series converges or diverges. Here are two theorems that help us determine convergence or divergence.

Theorem (The n\begin{align*}n\end{align*}th-Term Test)

If the infinite series k=1uk\begin{align*}\sum_{k=1}^\infty u_k\end{align*} converges, then limk+uk=0\begin{align*}\lim_{k \to +\infty}u_{k=0}\end{align*}

Theorem

If limk+uk0\begin{align*}\lim_{k \to +\infty} u_k \neq 0\end{align*} or limk+uk\begin{align*}\lim_{k \to +\infty} u_k\end{align*} does not exist, then the infinite series k=1uk\begin{align*}\sum_{k=1}^\infty u_k\end{align*} diverges.

The first theorem tells us that if an infinite series converges, then the limit of the sequence of terms is 0\begin{align*}0\end{align*}. The converse is not true: If the limit of the sequence of terms is 0\begin{align*}0\end{align*}, then the series converges. So, we cannot use this theorem as a test of convergence.

The second theorem tells us that if limit of the sequence of terms is not zero, then the infinites series diverges. This gives us the first test of divergence: the n\begin{align*}n\end{align*}th-Term Test or Divergence Test. Note that if the test is applied and the limit of the sequence of terms is 0\begin{align*}0\end{align*}, we cannot conclude anything and must use another test.

Example 9

Determine if k=1kk+5\begin{align*}\sum_{k=1}^\infty \frac{k}{k+5}\end{align*} converges or diverges.

Solution

We can use the n\begin{align*}n\end{align*}th-Term Test to determine if the series diverges. Then we do not have to check for convergence.

limk+ kk+5=limk+ kkk+5k=limk+ 11+5k=limk+1limk+1+5k=1\begin{align*}\lim_{k \to +\infty}\ \frac{k}{k+5}=\lim_{k \to +\infty}\ \frac {\frac{k}{k}}{\frac{k+5}{k}}= \lim_{k \to +\infty}\ \frac{1}{1+\frac{5}{k}}=\frac {\lim_{k \to +\infty}1}{\lim_{k \to +\infty}1+\frac{5}{k}}=1\end{align*}

Because limk+ kk+50\begin{align*}\lim_{k \to +\infty}\ \frac{k}{k+5} \neq 0\end{align*}, the series k=1kk+5\begin{align*}\sum_{k=1}^\infty\frac{k}{k+5}\end{align*} diverges.

Example 10

Determine if k=18k3\begin{align*}\sum_{k=1}^\infty \frac{8}{k-3}\end{align*} converges or diverges.

Solution

Using the n\begin{align*}n\end{align*}th-Term Test, limk+ 8k3\begin{align*}\lim_{k \to +\infty}\ \frac{8}{k-3} \end{align*} . Since the limit is 0\begin{align*}0\end{align*}, we cannot make a conclusion about convergence or divergence.

## Rules for Convergent Series, Reindexing

### Rules

As with sequences, there are rules for convergent infinite series that help make it easier to determine convergence.

Theorem (Rules for Convergent Series)

1. Suppose k=1uk\begin{align*}\sum_{k=1}^\infty u_k\end{align*} and n=1vk\begin{align*}\sum_{n=1}^\infty v_k\end{align*} are convergent series with k=1uk=S1\begin{align*}\sum_{k=1}^\infty u_k =S_1\end{align*} and k=1vk=S2\begin{align*}\sum_{k=1}^\infty v_k=S_2\end{align*}.

Then k=1(uk+vk)\begin{align*}\sum_{k=1}^\infty (u_k+v_k)\end{align*} and k=1(ukvk)\begin{align*}\sum_{k=1}^\infty (u_k-v_k)\end{align*} are also convergent where

k=1(uk+vk)=k=1(uk)+k=1(vk)=S1+S2\begin{align*}\sum_{k=1}^\infty (u_k+v_k) =\sum_{k=1}^\infty (u_k)+\sum_{k=1}^\infty (v_k)=S_1+S_2\end{align*} and k=1(ukvk)=k=1(uk)k=1(vk)=S1S2\begin{align*}\sum_{k=1}^\infty (u_k-v_k) =\sum_{k=1}^\infty (u_k)-\sum_{k=1}^\infty (v_k)=S_1-S_2\end{align*}

(The sum or difference of convergent series is also convergent.)

2. Let c0\begin{align*}c \neq 0\end{align*} be a constant.

Suppose k=1uk\begin{align*}\sum_{k=1}^\infty u_k\end{align*} converges and k=1uk=S\begin{align*}\sum_{k=1}^\infty u_k = S\end{align*} Then k=1cuk\begin{align*}\sum_{k=1}^\infty cu_k\end{align*} also converges where.

k=1cui=ck=1uk=cS\begin{align*}\sum_{k=1}^\infty cu_i = c\sum_{k=1}^\infty u_k = cS\end{align*}

If k=1uk\begin{align*}\sum_{k=1}^\infty u_k\end{align*} diverges, then k=1cuk\begin{align*}\sum_{k=1}^\infty cu_k\end{align*} also diverges.

(Multiplying by a nonzero constant does not affect convergence or divergence.)

Example 10

Find the sum of k=1(23k1+18k1)\begin{align*}\sum_{k=1}^\infty \left(\frac{2}{3^{k-1}}+\frac{1}{8^{k-1}}\right)\end{align*}.

Solution

Using the Rules Theorem, k=1(23k1+18k1)=k=123k1+k=118k1\begin{align*}\sum_{k=1}^\infty \left(\frac{2}{3^{k-1}}+\frac{1}{8^{k-1}}\right) = \sum_{k=1}^\infty \frac{2}{3^{k-1}}+ \sum_{k=1}^\infty \frac{1}{8^{k-1}}\end{align*} .

k=123k1\begin{align*}\sum_{k=1}^\infty \frac{2}{3^{k-1}}\end{align*} is a convergent geometric series with a=2\begin{align*}a = 2\end{align*} and r=13\begin{align*}r = \frac{1}{3}\end{align*} . Its sum is 2113=223=3\begin{align*}\frac{2}{1-\frac{1}{3}} = \frac{2}{\frac{2}{3}} = 3\end{align*} .

k=118k1\begin{align*}\sum_{k=1}^\infty \frac{1}{8^{k-1}}\end{align*} is a convergent geometric series with a=2\begin{align*}a = 2\end{align*} and r=18\begin{align*}r = \frac{1}{8}\end{align*} . Its sum is 2118=178=87\begin{align*}\frac{2}{1-\frac{1}{8}} = \frac{1}{\frac{7}{8}} = \frac{8}{7}\end{align*} .

Then k=1(23k1+18k1)=k=123k1+i=118k1=3+87=297\begin{align*}\sum_{k=1}^\infty \left(\frac{2}{3^{k-1}} + \frac{1}{8^{k-1}}\right) = \sum_{k=1}^\infty \frac{2}{3^{k-1}}+ \sum_{i=1}^\infty \frac{1}{8^{k-1}} = 3 + \frac{8}{7} = \frac{29}{7}\end{align*}

Example 11

Find the sum of k=12(56k1)\begin{align*}\sum_{k=1}^\infty 2\left(\frac{5}{6^{k-1}}\right)\end{align*} .

Solution

By the rules for constant in infinite series, k=12(56k1)=2k=156k1\begin{align*}\sum_{k=1}^\infty 2\left(\frac{5}{6^{k-1}}\right) = 2 \sum_{k=1}^\infty \frac{5}{6^{k-1}}\end{align*}. The series k=156k1\begin{align*}\sum_{k=1}^\infty \frac{5}{6^{k-1}}\end{align*} is a geometric series with a=5\begin{align*}a = 5\end{align*} and r=16\begin{align*}r=\frac{1}{6}\end{align*}. Note that, by the Theorem on convergence of geometric series, this series converges to 6, that is k=156k1=5116=556=6\begin{align*}\sum_{k=1}^\infty \frac{5}{6^{k-1}} = \frac{5}{1-\frac{1}{6}} = \frac{5}{\frac{5}{6}} = 6\end{align*}.

Then k=12(56k1)=2×6=12\begin{align*}\sum_{k=1}^\infty 2\left(\frac{5}{6^{k-1}}\right) = 2 \times 6 = 12\end{align*}.

Adding or subtracting a finite number of terms from an infinite series does not affect convergence or divergence.

Theorem

If k=1uk\begin{align*}\sum_{k=1}^\infty u_k\end{align*} converges, then k=1uk+(u1+u2++um)\begin{align*}\sum_{k=1}^\infty u_k + (u_1 +u_2 + \ldots +u_m)\end{align*} is also convergent.

If k=1uk\begin{align*}\sum_{k=1}^\infty u_k\end{align*} converges, then k=1uk(u1+u2++um)\begin{align*}\sum_{k=1}^\infty u_k - (u_1 +u_2 + \ldots +u_m)\end{align*} is also convergent.

Likewise, if k=1uk\begin{align*}\sum_{k=1}^\infty u_k\end{align*} diverges, then k=1uk+(u1+u2++um)\begin{align*}\sum_{k=1}^\infty u_k + (u_1 +u_2 + \ldots +u_m)\end{align*} and