Section 8.2 Infinite Series
Subsection 8.2.1 Convergence of sequences
Definition 8.2.2. Infinite Series, nth Partial Sums, Convergence, Divergence.
Let {an} be a sequence.
The sum ∞∑n=1an is an infinite series (or, simply series).
Let Sn=n∑i=1ai ; the sequence {Sn} is the sequence of nth partial sums of {an}.
If the sequence {Sn} converges to L, we say the series ∞∑n=1an converges to L, and we write ∞∑n=1an=L.
If the sequence {Sn} diverges, the series ∞∑n=1an diverges.
Example 8.2.3. Showing series diverge.
Let {an}={n2}. Show ∞∑n=1an diverges.
Let {bn}={(−1)n+1}. Show ∞∑n=1bn diverges.
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Consider \(S_n\text{,}\) the \(n\)th partial sum.
\begin{align*} S_n \amp = a_1+a_2+a_3+\cdots+a_n\\ \amp = 1^2+2^2+3^2\cdots + n^2.\\ \end{align*}By Theorem 5.3.9, this is
\begin{align*} \amp = \frac{n(n+1)(2n+1)}{6}\text{.} \end{align*}Since \(\lim\limits_{n\to\infty}S_n = \infty\text{,}\) we conclude that the series \(\ds \infser n^2\) diverges. It is instructive to write \(\ds \infser n^2=\infty\) for this tells us how the series diverges: it grows without bound. A scatter plot of the sequences \(\{a_n\}\) and \(\{S_n\}\) is given in Figure 8.2.4.(a). The terms of \(\{a_n\}\) are growing, so the terms of the partial sums \(\{S_n\}\) are growing even faster, illustrating that the series diverges.
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The sequence \(\{b_n\}\) starts with 1, \(-1\text{,}\) 1, \(-1\text{,}\) \(\ldots\text{.}\) Consider some of the partial sums \(S_n\) of \(\{b_n\}\text{:}\)
\begin{align*} S_1 \amp = 1\\ S_2 \amp = 0\\ S_3 \amp = 1\\ S_4 \amp = 0 \end{align*}This pattern repeats; we find that \(S_n = \begin{cases} 1 \amp n\, \text{ is odd } \\, 0 \amp n\, \text{ is even } \end{cases}\text{.}\) As \(\{S_n\}\) oscillates, repeating 1, 0, 1, 0, \(\ldots\text{,}\) we conclude that \(\lim\limits_{n\to\infty}S_n\) does not exist, hence \(\ds\infser (-1)^{n+1}\) diverges. A scatter plot of the sequence \(\{b_n\}\) and the partial sums \(\{S_n\}\) is given in Figure 8.2.4.(b). When \(n\) is odd, \(b_n = S_n\) so the marks for \(b_n\) are drawn oversized to show they coincide.
(a) (b) Figure 8.2.4. Scatter plots relating to Example 8.2.3
Subsection 8.2.2 Geometric Series
One important type of series is a geometric series.Definition 8.2.5. Geometric Series.
A geometric series is a series of the form
Note that the index starts at n=0, not n=1.
Theorem 8.2.6. Geometric Series Test.
Consider the geometric series ∞∑n=0rn.
The nth partial sum is: Sn=1−rn+11−r.
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The series converges if, and only if, |r|<1. When |r|<1,
∞∑n=0rn=11−r.
Proof.
We begin by proving the formula for the simplied form for the partial sums. Consider the \(n\)th partial sum of the geometric series, \(S_n=\sum_{i=0}^n r^i\text{:}\)
Multiply both sides by \(r\text{:}\)
\begin{align*} r\cdot S_n \amp = r+r^2+r^3+\dots+r^{n}+r^{n+1}\\ \end{align*}Now subtract the second line from the first and solve for \(S_n\text{:}\)
\begin{align*} S_n-r\cdot S_n \amp = 1-r^{n+1}\\ S_n(1-r) \amp = 1-r^{n+1}\\ S_n \amp = \frac{1-r^{n+1}}{1-r}\text{.} \end{align*}We have shown Part 1 of Geometric Series Test.
Now, examining the partial sums, we consider four cases to determine when \(S_n\) converges:
If \(\abs{r}\lt 1\text{,}\) then \(r^n \to 0\) as \(n \to \infty\text{,}\) so we have \(\inflim S_n=\frac{1-0}{1-r}=\frac{1}{1-r}\text{,}\) a convergent sequence of partial sums.
If \(\abs{r}\gt 1\text{,}\) then \(r^n \to \pm \infty\) (depending on the sign of \(r\)) as \(n \to \infty\text{.}\) We also have \(1-r\) a constant, so \(\inflim S_n\) does not exist (if \(r \gt 0\text{,}\) the partial sums diverge to \(\infty\) and if \(r \lt 0\text{,}\) the partial sums diverge to \(-\infty\)).
- If \(r=1\text{,}\) then \(S_n = \frac{1-1^{n+1}}{1-1}\) is undefined. However, examining \(S_n = 1+r+r^2+\dots+r^{n-1}+r^n\) for \(r=1\text{,}\) we can see that the partial sums simplify to \(S_n=n+1\text{.}\) Clearly this sequence diverges to \(\infty\text{.}\)
- If \(r=-1\text{,}\) then \(S_n = \frac{1-(-1)^{n+1}}{2}\text{.}\) For even values of \(n\text{,}\) the partial sums are always \(1\text{.}\) For odd values of \(n\text{,}\) the partial sums are always \(0\text{.}\) So the sequence of partial sums diverges.
Therefore, a geometric series converges if and only if \(\abs{r} \lt 1\text{.}\)
Example 8.2.8. Exploring geometric series.
Check the convergence of the following series. If the series converges, find its sum.
∞∑n=2(34)n
∞∑n=0(−12)n
∞∑n=03n
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Since \(r=3/4\lt 1\text{,}\) this series converges. By Theorem 8.2.6, we have that
\begin{equation*} \infser[0] \left(\frac34\right)^n = \frac{1}{1-3/4} = 4\text{.} \end{equation*}However, note the subscript of the summation in the given series: we are to start with \(n=2\text{.}\) Therefore we subtract off the first two terms, giving:
\begin{equation*} \sum_{n=2}^\infty \left(\frac34\right)^n = 4 - 1 - \frac34 = \frac94\text{.} \end{equation*}This is illustrated in Figure 8.2.9.
Figure 8.2.9. Scatter plots for the series in Item 1 -
Since \(\abs{r} = 1/2 \lt 1\text{,}\) this series converges, and by Theorem 8.2.6,
\begin{equation*} \infser[0] \left(\frac{-1}{2}\right)^n = \frac{1}{1-(-1/2)} = \frac23\text{.} \end{equation*}The partial sums of this series are plotted in Figure 8.2.10. Note how the partial sums are not purely increasing as some of the terms of the sequence \(\{(-1/2)^n\}\) are negative.
Figure 8.2.10. Scatter plots for the series in Item 2 -
Since \(r \gt 1\text{,}\) the series diverges. (This makes “common sense”; we expect the sum
\begin{equation*} 1+3+9+27 + 81+243+\cdots \end{equation*}to diverge.) This is illustrated in Figure 8.2.11.
Figure 8.2.11. Scatter plots for the series in Item 3
Subsection 8.2.3 p-Series
Another important type of series is the p-series.Definition 8.2.12. p-Series, General p-Series.
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A p-series is a series of the form
∞∑n=11np, where p>0. -
A general p-series is a series of the form
∞∑n=11(an+b)p,where p>0 and a,b are real numbers such that a≠0 and an+b>0 for all n≥1.
Theorem 8.2.13. p-Series Test.
A general p-series ∞∑n=11(an+b)p will converge if, and only if, p>1.
Example 8.2.14. Determining convergence of series.
Determine the convergence of the following series.
∞∑n=11n
∞∑n=11n2
∞∑n=11√n
∞∑n=1(−1)nn
∞∑n=111(12n−5)3
∞∑n=112n
This is a \(p\)-series with \(p=1\text{.}\) By Theorem 8.2.13, this series diverges. This series is a famous series, called the Harmonic Series, so named because of its relationship to harmonics in the study of music and sound.
This is a \(p\)-series with \(p=2\text{.}\) By Theorem 8.2.13, it converges. Note that the theorem does not give a formula by which we can determine what the series converges to; we just know it converges. A famous, unexpected result is that this series converges to \(\ds{\pi^2}/{6}\text{.}\)
This is a \(p\)-series with \(p=1/2\text{;}\) the theorem states that it diverges.
This is not a \(p\)-series; the definition does not allow for alternating signs. Therefore we cannot apply Theorem 8.2.13. (Another famous result states that this series, the Alternating Harmonic Series, converges to \(\ln(2)\text{.}\))
This is a general \(p\)-series with \(p=3\text{,}\) therefore it converges.
This is not a \(p\)-series, but a geometric series with \(r=1/2\text{.}\) It converges.
Example 8.2.15. Telescoping series.
Evaluate the sum ∞∑n=1(1n−1n+1).
It will help to write down some of the first few partial sums of this series.
Note how most of the terms in each partial sum are canceled out! In general, we see that \(\ds S_n = 1-\frac{1}{n+1}\text{.}\) The sequence \(\{S_n\}\) converges, as \(\lim\limits_{n\to\infty}S_n = \lim_{n\to\infty}\left(1-\frac1{n+1}\right) = 1\text{,}\) and so we conclude that \(\ds \infser \left(\frac1n-\frac1{n+1}\right) = 1\text{.}\) Partial sums of the series are plotted in Figure 8.2.16.
Example 8.2.17. Evaluating series.
Evaluate each of the following infinite series.
∞∑n=12n2+2n
∞∑n=1ln(n+1n)
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We can decompose the fraction \(2/(n^2+2n)\) as
\begin{equation*} \frac2{n^2+2n} = \frac1n-\frac1{n+2}\text{.} \end{equation*}(See Section 6.5, Partial Fraction Decomposition, to recall how this is done, if necessary.) Expressing the terms of \(\{S_n\}\) is now more instructive:
\begin{align*} S_1 \amp = 1-\frac13 \\ S_2 \amp = \left(1-\frac13\right) + \left(\frac12-\frac14\right)\\ \amp = 1+\frac12-\frac13-\frac14\\ S_3 \amp = \left(1-\frac13\right) + \left(\frac12-\frac14\right)+\left(\frac13-\frac15\right)\\ \amp = 1+\frac12-\frac14-\frac15\\ S_4 \amp = \left(1-\frac13\right) + \left(\frac12-\frac14\right)+\left(\frac13-\frac15\right)+\left(\frac14-\frac16\right)\\ \amp = 1+\frac12-\frac15-\frac16\\ S_5 \amp = \left(1-\frac13\right) + \left(\frac12-\frac14\right)+\left(\frac13-\frac15\right)+\left(\frac14-\frac16\right)+\left(\frac15-\frac17\right)\\ \amp = 1+\frac12-\frac16-\frac17 \end{align*}We again have a telescoping series. In each partial sum, most of the terms cancel and we obtain the formula \(\ds S_n = 1+\frac12-\frac1{n+1}-\frac1{n+2}\text{.}\) Taking limits allows us to determine the convergence of the series:
\begin{equation*} \lim_{n\to\infty}S_n = \lim_{n\to\infty} \left(1+\frac12-\frac1{n+1}-\frac1{n+2}\right) = \frac32\text{,} \end{equation*}so \(\infser \frac1{n^2+2n} = \frac32\text{.}\) This is illustrated in Figure 8.2.18.(a).
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We begin by writing the first few partial sums of the series:
\begin{align*} S_1 \amp = \ln\left(2\right)\\ S_2 \amp = \ln\left(2\right)+\ln\left(\frac32\right)\\ S_3 \amp = \ln\left(2\right)+\ln\left(\frac32\right)+\ln\left(\frac43\right)\\ S_4 \amp = \ln\left(2\right)+\ln\left(\frac32\right)+\ln\left(\frac43\right)+\ln\left(\frac54\right) \end{align*}At first, this does not seem helpful, but recall the logarithmic identity: \(\ln(x) +\ln(y) = \ln(xy)\text{.}\) Applying this to \(S_4\) gives:
\begin{align*} S_4 \amp = \ln\left(2\right)+\ln\left(\frac32\right)+\ln\left(\frac43\right)+\ln\left(\frac54\right)\\ \amp = \ln\left(\frac21\cdot\frac32\cdot\frac43\cdot\frac54\right) = \ln\left(5\right)\text{.} \end{align*}We can conclude that \(\{S_n\} = \big\{\ln(n+1)\big\}\text{.}\) This sequence does not converge, as \(\lim\limits_{n\to\infty}S_n=\infty\text{.}\) Therefore \(\ds\infser \ln\left(\frac{n+1}{n}\right)=\infty\text{;}\) the series diverges. Note in Figure 8.2.18.(b) how the sequence of partial sums grows slowly; after 100 terms, it is not yet over 5. Graphically we may be fooled into thinking the series converges, but our analysis above shows that it does not.
Theorem 8.2.19. Properties of Infinite Series.
Let ∞∑n=1an=L,∞∑n=1bn=K, and let c be a constant.
Constant Multiple Rule: ∞∑n=1c⋅an=c⋅∞∑n=1an=c⋅L.
Sum/Difference Rule: ∞∑n=1(an±bn)=∞∑n=1an±∞∑n=1bn=L±K.
Key Idea 8.2.20. Important Series.
∞∑n=01n!=e. (Note that the index starts with n=0.)
∞∑n=11n2=π26.
∞∑n=1(−1)n+1n2=π212.
∞∑n=0(−1)n2n+1=π4.
∞∑n=11n diverges. (This is called the Harmonic Series.)
∞∑n=1(−1)n+1n=ln(2). (This is called the Alternating Harmonic Series.)
Example 8.2.21. Evaluating series.
Evaluate the given series.
∞∑n=1(−1)n+1(n2−n)n3
∞∑n=11000n!
116+125+136+149+⋯
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We start by using algebra to break the series apart:
\begin{align*} \infser \frac{(-1)^{n+1}\big(n^2-n\big)}{n^3} \amp = \infser\left(\frac{(-1)^{n+1}n^2}{n^3}-\frac{(-1)^{n+1}n}{n^3}\right)\\ \amp = \infser\frac{(-1)^{n+1}}{n}-\infser\frac{(-1)^{n+1}}{n^2}\\ \amp = \ln(2) - \frac{\pi^2}{12} \approx -0.1293\text{.} \end{align*}This is illustrated in Figure 8.2.22.(a).
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This looks very similar to the series that involves \(e\) in Key Idea 8.2.20. Note, however, that the series given in this example starts with \(n=1\) and not \(n=0\text{.}\) The first term of the series in the Key Idea is \(1/0! = 1\text{,}\) so we will subtract this from our result below:
\begin{align*} \infser \frac{1000}{n!} \amp = 1000\cdot\infser \frac{1}{n!}\\ \amp = 1000\cdot (e-1) \approx 1718.28\text{.} \end{align*}This is illustrated in Figure 8.2.22.(b). The graph shows how this particular series converges very rapidly.
(a) (b) Figure 8.2.22. Scatter plots relating to the series in Example 8.2.21 -
The denominators in each term are perfect squares; we are adding \(\ds \sum_{n=4}^\infty \frac{1}{n^2}\) (note we start with \(n=4\text{,}\) not \(n=1\)). This series will converge. Using the formula from Key Idea 8.2.20, we have the following:
\begin{align*} \infser \frac1{n^2} \amp = \sum_{n=1}^3 \frac1{n^2} +\sum_{n=4}^\infty \frac1{n^2}\\ \infser \frac1{n^2} - \sum_{n=1}^3 \frac1{n^2} \amp =\sum_{n=4}^\infty \frac1{n^2}\\ \frac{\pi^2}{6} - \left(\frac11+\frac14+\frac19\right) \amp = \sum_{n=4}^\infty \frac1{n^2}\\ \frac{\pi^2}{6} - \frac{49}{36} \amp = \sum_{n=4}^\infty \frac1{n^2}\\ 0.2838\amp \approx \sum_{n=4}^\infty \frac1{n^2} \end{align*}
In order to add an infinite list of nonzero numbers and get a finite result, “most” of those numbers must be “very near” 0.
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If a series diverges, it means that the sum of an infinite list of numbers is not finite (it may approach ±∞ or it may oscillate), and:
The series will still diverge if the first term is removed.
The series will still diverge if the first 10 terms are removed.
The series will still diverge if the first 1,000,000 terms are removed.
The series will still diverge if any finite number of terms from anywhere in the series are removed.
Theorem 8.2.23. nth-Term Test for Divergence.
Consider the series ∞∑n=1an. If limn→∞an≠0, then ∞∑n=1an diverges.
Theorem 8.2.24. Infinite Nature of Series.
The convergence or divergence of an infinite series remains unchanged by the addition or subtraction of any finite number of terms. That is:
A divergent series will remain divergent with the addition or subtraction of any finite number of terms.
A convergent series will remain convergent with the addition or subtraction of any finite number of terms. (Of course, the sum will likely change.)
Divergence of the harmonic series.
If you just consider the partial sums
it is not apparent that the partial sums diverge. Indeed they do diverge, but very, very slowly. (If you graph them on a logarithmic scale however, you can clearly see the divergence of the partial sums.) Instead, we will consider the partial sums, indexed by powers of \(2\text{.}\) That is, we will consider \(S_2,S_4, S_8, S_{16}, \dots\text{.}\)
Next, we consider grouping together terms in each partial sum. We will use these groupings to set up inequalities.
In the partial sum \(S_4\text{,}\) we note that since \(1/3\gt 1/4\text{,}\) we can say
Do the same in \(S_8\) and also note that every term in the group \(\left(\frac15+\frac16+\frac17+\frac18\right)\) is larger than \(1/8\text{.}\) So
Generally, we can see that \(S_{2^n} \gt 1+\frac{n}2\text{.}\) (In order to really show this, we should employ proof by induction.) Since the sequence of partial sums clearly diverges, so does the series \(\infser 1/n\text{.}\)
Exercises 8.2.4 Exercises
Terms and Concepts
1.
Use your own words to describe how sequences and series are related.
2.
Use your own words to define a partial sum.
3.
Given a series ∞∑n=1an, describe the two sequences related to the series that are important.
4.
Use your own words to explain what a geometric series is.
5.
T/F: If {an} is convergent, then ∞∑n=1an is also convergent.
6.
T/F: If {an} converges to 0, then ∞∑n=0an converges.
In the following exercises, a series ∞∑n=1an is given.
Give the first 5 partial sums of the series.
Give a graph of the first 5 terms of an and Sn on the same axes.
In the following exercises, use Theorem 8.2.23 to show the given series diverges.
In the following exercises, state whether the given series converges or diverges.
In the following exercises, a series is given.
Find a formula for Sn, the nth partial sum of the series.
Determine whether the series converges or diverges. If it converges, state what it converges to.
31.
∞∑n=014n
32.
∞∑n=12
33.
13+23+33+43+⋯
34.
∞∑n=1(−1)nn
35.
∞∑n=052n
36.
∞∑n=0e−n
37.
1−13+19−127+181+⋯
38.
∞∑n=11n(n+1)
39.
∞∑n=13n(n+2)
40.
∞∑n=11(2n−1)(2n+1)
41.
∞∑n=1ln(nn+1)
42.
∞∑n=12n+1n2(n+1)2
43.
11⋅4+12⋅5+13⋅6+14⋅7+⋯
44.
2+(12+13)+(14+19)+(18+127)+⋯
45.
∞∑n=21n2−1
46.
∞∑n=0(sin(1))n
47.
Break the Harmonic Series into the sum of the odd and even terms:
The goal is to show that each of the series on the right diverge.
Show why ∞∑n=112n−1>∞∑n=112n. (Compare each nth partial sum.)
Show why ∞∑n=112n−1<1+∞∑n=112n
Explain why (a) and (b) demonstrate that the series of odd terms is convergent, if, and only if, the series of even terms is also convergent. (That is, show both converge or both diverge.)
Explain why knowing the Harmonic Series is divergent determines that the even and odd series are also divergent.
48.
Show the series ∞∑n=1n(2n−1)(2n+1) diverges.