# 8.3: Series of Real Numbers

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Learning Objectives

In this section, we strive to understand the ideas generated by the following important questions:

- What is an infinite series?
- What is the nth partial sum of an infinite series?
- How do we add up an infinite number of numbers? In other words, what does it mean for an infinite series of real numbers to converge?
- What does is mean for an infinite series of real numbers to diverge?

In Section 8.2, we encountered several situations where we naturally considered an infinite sum of numbers called a geometric series. For example, by writing

\[N = 0.1212121212 · · · = \dfrac{12}{100} + \dfrac{12}{100} · \dfrac{1}{100} + \dfrac{12}{100} · \dfrac{1}{1002} + · · · \nonumber\]

as a geometric series, we found a way to write the repeating decimal expansion of \9N\) as a single fraction: \(N = \dfrac{4}{33}\). There are many other situations in mathematics where infinite sums of numbers arise, but often these are not geometric. In this section, we begin exploring these other types of infinite sums. Preview Activity \(\PageIndex{1}\) provides a context in which we see how one such sum is related to the famous number e.

Preview Activity \(\PageIndex{1}\)

Have you ever wondered how your calculator can produce a numeric approximation for complicated numbers like \(e\), \(π\) or \(\ln(2)\)? After all, the only operations a calculator can really perform are addition, subtraction, multiplication, and division, the operations that make up polynomials. This activity provides the first steps in understanding how this process works. Throughout the activity, let \(f (x) = e^x \).

- Find the tangent line to \(f\) at \(x = 0\) and use this linearization to approximate \(e\). That is, find a formula \(L(x)\) for the tangent line, and compute \(L(1)\), since \( L(1) \approx f (1) = e\).
- The linearization of \(e^x\) does not provide a good approximation to \(e\) since 1 is not very close to 0. To obtain a better approximation, we alter our approach a bit. Instead of using a straight line to approximate \(e\), we put an appropriate bend in our estimating function to make it better fit the graph of \(e^x\) for \(x\) close to 0. With the linearization, we had both \(f (x)\) and \(f' (x)\) share the same value as the linearization at \(x = 0\). We will now use a quadratic approximation \(P_2(x)\) to \(f (x) = e^x\) centered at \(x = 0\) which has the property that \(P_2(0) = f (0)\), \(P'_2 (0) = f 0 (0)\), and \(P''_2 (0) = f''(0)\).
- Let \(P_2(x) = 1 + x + \dfrac{x^2}{2}\). Show that \(P_2(0) = f (0)\), \(P'_2 (0) = f'(0)\), and \(P''_2 (0) = f''(0)\). Then, use \(P_2(x)\) to approximate \(e\) by observing that \(P_2(1) \approx f (1)\).
- We can continue approximating e with polynomials of larger degree whose higher derivatives agree with those of \(f\) at 0. This turns out to make the polynomials fit the graph of \(f\) better for more values of \(x\) around 0. For example, let \(P_3(x) = 1 + x + \dfrac{x^2}{2} + \frac{x^3}{6}\). Show that \(P_3(0) = f (0), P'_3(0) = f'(0), P''_3 (0) = f''(0)\), and \(P'''_3 (0) = f'''(0)\). Use \(P_3(x)\) to approximate e in a way similar to how you did so with \(P_2(x)\) above.

Preview Activity \(\PageIndex{1}\) shows that an approximation to e using a linear polynomial is 2, an approximation to e using a quadratic polynomial is 2.5, and an approximation using a cubic polynomial is 2.6667. As we will see later, if we continue this process we can obtain approximations from quartic (degree 4), quintic (degree 5), and higher degree polynomials giving us the following approximations to \(e\):

- linear: \(e \approx 1 + 1 = 2\)
- quadratic: \(e \approx 1 + 1 + \dfrac{1}{2} =2.5\)
- cubic: \(e \approx 1 + 1 + \dfrac{1}{2} + \dfrac{1}{6} = 2.6\)
- quartic: \(e \approx 1 + 1 + \dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{24} = 2.7083\)
- quintic: \(e \approx 1 + 1 + \dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{24} + \dfrac{1}{120} = 2.716\)

We see an interesting pattern here. The number \(e\) is being approximated by the sum

\[1 + 1 + \dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{24} + \dfrac{1}{120} + · · · + \dfrac{1}{n!} \tag{8.11}\label{8.11} \]

for increasing values of \(n\). And just as we did with Riemann sums, we can use the summation notation as a shorthand^{4} for writing the sum in Equation \(\ref{8.11}\) so that

\[e \approx 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + \frac{1}{120} + · · · + \dfrac{1}{ n!} \sum_n^{k=0} \dfrac{1}{k!}. \label{8.12} \]

We can calculate this sum using as large an \(n\) as we want, and the larger \(n\) is the more accurate the approximation (Equation \ref{8.12}) is. Ultimately, this argument shows that we can write the number e as the infinite sum:

\[ e = \sum_{k=0}^\infty \dfrac{1}{k!}.\tag{8.13} \label{8.13} \]

Note that 0! appears in Equation \(\ref{8.12}\) and by definition, 0! = 1. This sum is an example of a series (or an infinite series). Note that the series in Equation \ref{8.13} is the sum of the terms of the (infinite) sequence {\(\dfrac{1}{n!}\)}. In general, we use the following notation and terminology. Definition 8.3. An infinite series of real numbers is the sum of the entries in an infinite sequence of real numbers. In other words, an infinite series is sum of the form

\[a_1 + a_2 + · · · + a_n + · · · = \sum_{k=1}^\infty ak, \nonumber\]

where \(a_1, a_2, . . .,\) are real numbers.

We will normally use summation notation to identify a series. If the series adds the entries of a sequence {\(a_n\)}\(n\geq1\), then we will write the series as

\[\sum_{k \geq1} a_k \nonumber\]

or

\[\sum_{k =1} a_k \nonumber\]

Note well: each of these notations is simply shorthand for the infinite sum \(a_1 + a_2 + · · · + a_n + · · · \).

Is it even possible to sum an infinite list of numbers? This question is one whose answer shouldn’t come as a surprise. After all, we have used the definite integral to add up continuous (infinite) collections of numbers, so summing the entries of a sequence might be even easier. Moreover, we have already examined the special case of geometric series in the previous section. The next activity provides some more insight into how we make sense of the process of summing an infinite list of numbers.

Activity \(\PageIndex{2}\):

Consider the series

\[\sum_{k = 1} \dfrac{1}{k^2}. \nonumber\]

While it is physically impossible to add an infinite collection of numbers, we can, of course, add any finite collection of them. In what follows, we investigate how understanding how to find the nth partial sum (that is, the sum of the first n terms) enables us to make sense of the infinite sum.

a. Sum the first two numbers in this series. That is, find a numeric value for

\[\sum_{k=1}^a \dfrac{1}{k^2} \nonumber\]

b. (b) Next, add the first three numbers in the series.

c. Continue adding terms in this series to complete Table 8.4. Carry each sum to at least 8 decimal places.

\(\sum_{k=1}^1 \dfrac{1}{k^2} = 1\) | \(\sum_{k=1}^6 \dfrac{1}{k^2} = \) |

\(\sum_{k=1}^2 \dfrac{1}{k^2} = \) | \(\sum_{k=1}^7 \dfrac{1}{k^2} =\) |

\(\sum_{k=1}^3 \dfrac{1}{k^2} = \) | \(\sum_{k=1}^8 \dfrac{1}{k^2} =\) |

\(\sum_{k=1}^4 \dfrac{1}{k^2} =\) | \(\sum_{k=1}^9 \dfrac{1}{k^2} =\) |

\(\sum_{k=1}^5 \dfrac{1}{k^2} = \) | \(\sum_{k=1}^10 \dfrac{1}{k^2} =\) |

Table 8.4: Sums of some of the first terms of the series \(\sum_{k=1}^\infty \dfrac{1}{k^2}\)

(d) The sums in the table in (c) form a sequence whose \(n\)th term is \(S_n = \sum_{k=1}^\infty \dfrac{1}{k^2}\). Based on your calculations in the table, do you think the sequence {\(S_n\)} converges or diverges? Explain. How do you think this sequence {\(S_n\)} is related to the series \( \sum_{k=1}^\infty \dfrac{1}{k^2}\)?

The example in Activity \(\PageIndex{2}\) illustrates how we define the sum of an infinite series. We can add up the first n terms of the series to obtain a new sequence of numbers (called the sequence of partial sums). Provided that sequence converges, the corresponding infinite series is said to converge, and we say that we can find the sum of the series.

Definition

The \(n\)th partial sum of the series \(\sum_{k=1}^\infty a_k\) is the finite sum \(S_n = \sum_{k=1}^\infty a_k\). In other words, the nth partial sum Sn of a series is the sum of the first n terms in the series, or

\[S_n = a_1 + a_2 + · · · + a_n. \nonumber\]

We then investigate the behavior of a given series by examining the sequence

\[S_1, S_2, . . ., S_n, . . . \nonumber\]

of its partial sums. If the sequence of partial sums converges to some finite number, then we say that the corresponding series converges. Otherwise, we say the series diverges. From our work in Activity \(\PageIndex{2}\), the series

\[ \sum_{k=1}^\infty \dfrac{1}{k^2} \nonumber\]

appears to converge to some number near 1.54977. We formalize the concept of convergence and divergence of an infinite series in the following definition.

Definition

The infinite series

\[\sum_{k=1}^\infty a_k \nonumber\]

converges (or is convergent) if the sequence {\(S_n\)} of partial sums converges, where

\[S_n = \sum_{k=1}^\infty a_k. \nonumber\]

If \(\lim_{n \rightarrow \infty} S_n = S\), then we call \(S\) the sum of the series \(\sum_{k=1}^\infty a_k\). That is,

\[\sum_{k=1}^\infty a_k = \lim_{n \rightarrow \infty} S_n = S. \nonumber\]

If the sequence of partial sums does not converge, then the series

\[\sum_{k=1}^\infty a_k \nonumber\]

*diverges* (or is *divergent*).

The early terms in a series do not contribute to whether or not the series converges or diverges. Rather, the convergence or divergence of a series

\[\sum_{k=1}^\infty a_k \nonumber\]

is determined by what happens to the terms \(a_k\) for very large values of \(k\). To see why, suppose that \(m\) is some constant larger than 1. Then

\[\sum_{k=1}^\infty a_k = (a_1+a_2+...a_m) + \sum_{k=m+1}^\infty a_k. \nonumber\]

Since \(a_1 + a_2 + · · · + a_m\) is a finite number, the series \(\sum_{k=1}^\infty a_k\) will converge if and only if the series \(\sum_{k=m+1}^\infty a_k\) converges. Because the starting index of the series doesn’t affect whether the series converges or diverges, we will often just write

\[ \sum a_k \nonumber\]

when we are interested in questions of convergence/divergence and not necessarily the exact sum of a series.

In Section 8.2 we encountered the special family of infinite geometric series whose convergence or divergence we completely determined. Recall that a geometric series is a special series of the form \(\sum_{k=1}^\infty a_k\) where \(a\) and \(r\) are real numbers (and \(r \neq 1\) ). We found that the \(n\)th partial sum \(S_n\) of a geometric series is given by the convenient formula

\[S_n = \dfrac{1 − r^n}{1 − r}, \nonumber\]

and thus a geometric series converges if |\(r\)| < 1. Geometric series diverge for all other values of \(r\). While we have completely determined the convergence or divergence of geometric series, it is generally a difficult question to determine if a given nongeometric series converges or diverges. There are several tests we can use that we will consider in the following sections.

### The Divergence Test

The first question we ask about any infinite series is usually “Does the series converge or diverge?” There is a straightforward way to check that certain series diverge; we explore this test in the next activity.

Activity \(\PageIndex{3}\):

If the series \( \sum a_k\) converges, then an important result necessarily follows regarding the sequence {\(a_n\)}. This activity explores this result. Assume that the series \(\sum_{k=1}^\infty a_k\) converges and has sum equal to \(L\).

- What is the \(n\)th partial sum \(S_n\) of the series \(\sum_{k=1}^\infty a_k\) ?
- What is the \((n − 1)\)st partial sum \(S_n−1\) of the series \(\sum_{k=1}^\infty a_k\) ?
- What is the difference between the nth partial sum and the \((n − 1)\)st partial sum of the series \(\sum_{k=1}^\infty a_k\) ?
- Since we are assuming that \(\sum_{k=1}^\infty a_k = L \), what does that tell us about \(\lim_{n \rightarrow \infty} S_n\)? Why? What does that tell us about \( \lim_{n \rightarrow \infty} S_{n-1}\)? Why?
- Combine the results of the previous two parts of this activity to determine \( \lim_{n \rightarrow \infty} a_n = \lim_{n \rightarrow \infty} (S_n - S_{n-1}\)

The result of Activity \(\PageIndex{3}\) is the following important conditional statement:

If the series \(\sum_{k=1}^\infty a_k\) converges, then the sequence {\(a_k\)} of \(k\)th terms converges to 0. It is logically equivalent to say that if the sequence {ak } of n terms does not converge to 0, then the series \( \sum_{k=1}^\infty a_k\) cannot converge. This statement is called the Divergence Test.

THe divergence Test

If \(\lim_{k \rightarrow \infty}a_k \neq 0, \), then the series \( \sum ak\) diverges.

Activity \(\PageIndex{4}\):

Determine if the Divergence Test applies to the following series. If the test does not apply, explain why. If the test does apply, what does it tell us about the series?

a. \( \sum \dfrac{k}{k+1} \)

b. \( \sum (-1)^k \)

c. \( \sum \frac{1}{k} \)

Note well: be very careful with the Divergence Test. This test only tells us what happens to a series if the terms of the corresponding sequence do not converge to 0. If the sequence of the terms of the series does converge to 0, the Divergence Test does not apply: indeed, as we will soon see, a series whose terms go to zero may either converge or diverge.

## The Integral Test

The Divergence Test settles the questions of divergence or convergence of series \( \sum a_k\) in which \(\lim_{k \rightarrow \infty}a_k \neq 0\). Determining the convergence or divergence of series \( \sum a_k \) in which \(\lim_{k \rightarrow \infty}a_k = 0\) turns out to be more complicated. Often, we have to investigate the sequence of partial sums or apply some other technique.

As an example, consider the **harmonic ****series**

\[\sum_{k=1}^\infty \dfrac{1}{k}. \nonumber\] ^{5}

^{5} This series is called harmonic because each term in the series after the first is the harmonic mean of the term before it and the term after it. The harmonic mean of two numbers \(a\) and \(b\) is \(\dfrac{2ab}{a+b}\). See “What’s Harmonic about the Harmonic Series", by David E. Kullman (in the College Mathematics Journal, Vol. 32, No. 3 (May, 2001), 201-203) for an interesting discussion of the harmonic mean.

Table 8.3 shows some partial sums of this series.

\(\sum_{k=1}^1 \dfrac{1}{k}\). | 1 | \(\sum_{k=1}^6 \dfrac{1}{k}\). | 2.450000000 |

\(\sum_{k=1}^2 \dfrac{1}{k}\). | 1.5 | \(\sum_{k=1}^7 \dfrac{1}{k}\). | 2.592857143 |

\(\sum_{k=1}^3 \dfrac{1}{k}\). | 1.833333333 | \(\sum_{k=1}^8 \dfrac{1}{k}\). | 2.717857143 |

\(\sum_{k=1}^4 \dfrac{1}{k}\). | 2.083333333 | \(\sum_{k=1}^9 \dfrac{1}{k}\). | 2.828968254 |

\(\sum_{k=1}^5 \dfrac{1}{k}\). | 2.283333333 | \(\sum_{k=1}^10 \dfrac{1}{k}\). | 2.928968254 |

This information doesn’t seem to be enough to tell us if the series\(\sum_{k=1}^\infty \dfrac{1}{k}\) converges or diverges. The partial sums could eventually level off to some fixed number or continue to grow without bound. Even if we look at larger partial sums, such as \(\sum_{n=1}^{1000} \approx 7.485470861\), the result doesn’t particularly sway us one way or another. The Integral Test is one way to determine whether or not the harmonic series converges, and we explore this further in the next activity.

Activity \(\PageIndex{5}\):

Consider the harmonic series \(\sum_{k=1}^\infty \dfrac{1}{k}\). Recall that the harmonic series will converge provided that its sequence of partial sums converges. The \(n\)th partial sum \(S_n\) of the series \(\sum_{k=1}^\infty \dfrac{1}{k}\) is

\[S_n = \sum_{k=1}^n \dfrac{1}{k} \nonumber\]

\[= 1 + \dfrac{1}{2} + \dfrac{1}{3}+...\dfrac{1}{n} \nonumber\]

\[1(1) + (1)(\dfrac{1}{2})+(1)(\dfrac{1}{3})+...+(1)\dfrac{1}{n}. \nonumber\]

Through this last expression for \(S_n\), we can visualize this partial sum as a sum of areas of rectangles with heights \(\dfrac{1}{m}\) and bases of length 1, as shown in Figure 8.3, which uses the 9th partial sum. The graph of the continuous function \(f\) defined by \(f (x) = \frac{1}{x}\) is overlaid on this plot.

- Explain how this picture represents a particular Riemann sum.
- What is the definite integral that corresponds to the Riemann sum you considered in (a)?

*Figure 8.3: A picture of the 9th partial sum of the harmonic series as a sum of areas of rectangles.*

- Which is larger, the definite integral in (b), or the corresponding partial sum \(S_9\) of the series? Why?
- If instead of considering the 9th partial sum, we consider the nth partial sum, and we let \(n\) go to infinity, we can then compare the series \(\sum_{k=1}^\infty \dfrac{1}{k}\) to the improper integral \(\int_{1}^{\infty} \frac{1}{x} dx\). Which of these quantities is larger? Why?
- Does the improper integral \(\int_{1}^{\infty} \frac{1}{x} dx\) converge or diverge? What does that result, together with your work in (d), tell us about the series \(\sum_{k=1}^\infty \dfrac{1}{k}\) ?

The ideas from Activity \(\PageIndex{5}\) hold more generally. Suppose that \(f\) is a continuous decreasing function and that \(a_k = f (k)\) for each value of \(k\). Consider the corresponding series \(\sum_{k=1}^\infty a_k\). The partial sum

\[S_n = \sum_{k=1}^n a_k \nonumber\]

can always be viewed as a left hand Riemann sum of \(f (x)\) using rectangles with heights given by the values \(a_k\) and bases of length 1. A representative picture is shown at left in Figure 8.4. Since \(f\) is a decreasing function, we have that

\[S_n > \int_{1}^{n} (x) dx. \nonumber\]

Taking limits as \(n\) goes to infinity shows that

\[ \sum_\infty^{k=1} a_k >\int_{1}^{\infty} f(x) dx. \nonumber\]

Therefore, if the improper integral \( \int_{1}^{\infty} f(x) dx. \) diverges, so does the series \(\sum_{k=1}^\infty a_k\).

*Figure 8.4: Comparing an improper integral to a series*

What’s more, if we look at the right hand Riemann sums of f on [1, \(n\)] as shown at right in Figure 8.4, we see that

\[ \int_{1}^{\infty} f(x) dx > \sum_{k=2}^\infty a_k. \nonumber\]

So if \(\int_{1}^{\infty} f(x) dx\) converges, then so does \(\sum_{k=2}^\infty a_k\), which also means that the series \(\sum_{k=1}^\infty a_k\) converges. Our preceding discussion has demonstrated the truth of the Integral Test.

### The Integral Test

Let \(f\) be a real valued function and assume \(f\) is decreasing and positive for all \(x\) larger than some number \( c\). Let \(a_k = f (k)\) for each positive integer \(k\).

1. If the improper integral \(\int_{c}^{\infty} f(x) dx\) converges, then the series \(\sum_{k=1}^\infty a_k \) converges.

2. If the improper integral \(\int_{c}^{\infty} f(x) dx\) diverges, then the series \(\sum_{k=1}^\infty a_k \) diverges.

Note that the Integral Test compares a given infinite series to a natural, corresponding improper integral and basically says that the infinite series and corresponding improper integral both have the same convergence status. In the next activity, we apply the Integral Test to determine the convergence or divergence of a class of important series.

Activity \(\PageIndex{6}\):

The series \(\sum \frac{1}{k^p}\) are special series called \(p\)-series. We have already seen that the \(p\)-series with \(p = 1\) (the harmonic series) diverges. We investigate the behavior of other p-series in this activity.

- Evaluate the improper integral \(\int_{1}^{\infty} \frac{1}{x^2} dx\). Does the series \(\sum_{k=1}^\infty \frac{1}{k^P}\) converge or diverge? Explain.
- Evaluate the improper integral \(\int_{1}^{\infty} \frac{1}{x^P} dx\) where \(p > 1\). For which values of \(p\) can we conclude that the series \(\sum_{k=1}^\infty \frac{1}{k^P}\) converges?
- Evaluate the improper integral \int_{1}^{\infty} \frac{1}{x^P} dx where \(p < 1\). What does this tell us about the corresponding p-series \(\sum_{k=1}^\infty \frac{1}{k^P}\) ?
- Summarize your work in this activity by completing the following statement. The \(p\)-series \(\sum_{k=1}^\infty \frac{1}{k^P}\) converges if and only if ___________________________.

## The Limit Comparison Test

The Integral Test allows us to determine the convergence of an entire family of series: the p-series. However, we have seen that it is, in general, difficult to integrate functions, so the Integral Test is not one that we can use all of the time. In fact, even for a relatively simple series like \(\sum \frac{k^2+1}{k^4_2k+2'}\), the Integral Test is not an option. In this section we will develop a test that we can use to apply to series of rational functions like this by comparing their behavior to the behavior of \(p\)-series.

Activity \(\PageIndex{7}\):

Consider the series \(\sum \frac{k+1}{k^3+ 2}\). Since the convergence or divergence of a series only depends on the behavior of the series for large values of \(k\), we might examine the terms of this series more closely a \(k\) gets large.

a. By computing the value of \( \frac{k+1}{k^3+ 2}\)for \(k = 100\) and \(k = 1000\), explain why the terms \( \frac{k+1}{k^3+ 2}\) are essentially \( \dfrac{k}{k^3}\) when \(k\) is large.

b. Let’s formalize our observations in (a) a bit more. Let \(a_k = \frac{k+1}{k^3+2}\) and \(b_k = \frac{k}{ k^3}\).

Calculate

\[\lim_{k \rightarrow \infty} \frac{a_k}{b_k}. \nonumber\]

What does the value of the limit tell you about \(a_k\) and \(b_k\) for large values of \(k\)? Compare your response from part (a).

c. Does the series \( \sum \dfrac {k}{k^3}\) converge or diverge? Why? What do you think that tells us about the convergence or divergence of the series \(\sum \dfrac{k+1}{k^3+2} \)? Explain.

Activity \(\PageIndex{7}\) illustrates how we can compare one series with positive terms to another whose behavior (that is, whether the series converges or diverges) we know. More generally, suppose we have two series \( \sum a_k\) and \(\sum b_k\) with positive terms and we know the behavior of the series \( \sum a_k\). Recall that the convergence or divergence of a series depends only on what happens to the terms of the series for large values of k, so if we know that ak and bk are essentially proportional to each other for large \(k\), then the two series \( \sum a_k\) and \(\sum b_k\) should behave the same way. In other words, if there is a positive finite constant c such that

\[\lim_{k \rightarrow \infty} \frac{b_k}{a_k} = c , \nonumber\]

then \(b_k \approx ca_k\) for large values of \(k\). So

\[ \sum b_k \approx \sum ca_k = c \sum a_k. \nonumber\]

Since multiplying by a nonzero constant does not affect the convergence or divergence of a series, it follows that the series P ak and P bk either both converge or both diverge. The formal statement of this fact is called the Limit Comparison Test.

### The Limit Comparison Test.

Let \(\sum a_k\) and \(\sum b_k\) be series with positive terms. If

\[\lim_{k \rightarrow \infty} \frac{b_k}{a_k} = c \nonumber\]

for some positive (finite) constant \(c\), then \( \sum a_k\) and \(\sum b_k\)either both converge or both diverge.

In essence, the Limit Comparison Test shows that if we have a series \( \sum \dfrac{p(k)}{q(k)}\) of rational functions where \(p(k)\) is a polynomial of degree \(m\) and \(q(k)\) a polynomial of degree \(l\), then the series \( \sum \dfrac{p(k)}{q(k)}\) will behave like the series \( \sum \frac{k^m}{k^l}\). So this test allows us to quickly and easily determine the convergence or divergence of series whose summands are rational functions.

Activity \(\PageIndex{8}\):

Use the Limit Comparison Test to determine the convergence or divergence of the series

\[ \sum \dfrac{3k^2 + 1}{5k^4 + 2k + 2} . \nonumber\]

by comparing it to the series \(\sum \dfrac{1}{k^2}\).

## The Ratio Test

The Limit Comparison Test works well if we can find a series with known behavior to compare. But such series are not always easy to find. In this section we will examine a test that allows us to examine the behavior of a series by comparing it to a geometric series, without knowing in advance which geometric series we need.

Activity \(\PageIndex{9}\):

Consider the series defined by

\[ \sum_{k=1}^{\infty} \dfrac{2^k}{3^k - k}.\tag{4.18} \label{8.14}\]

This series is **not **a geometric series, but this activity will illustrate how we might compare this series to a geometric one. Recall that a series \( \sum a_k\) is geometric if the ratio \( \dfrac{a_{k+1}}{ak}\) is always the same. For the series in Equation \(\ref{8.14}\), note that \(a_k = \dfrac{2^k}{3^k−k}\).

a. To see if \( \sum \dfrac{2^k}{3^k - k }\) is comparable to a geometric series, we analyze the ratios of successive terms in the series. Complete Table 8.6, listing your calculations to at least 8 decimal places.

k | \( \dfrac{a_{k+1}}{a_k}\) |

5 | |

10 | |

20 | |

21 | |

22 | |

23 | |

24 | |

25 |

Table 8.6: Ratios of successive terms in the series \( \sum \dfrac{2^k}{3^k - k }\)

b. Based on your calculations in Table 8.6, what can we say about the ratio \( \dfrac{a_{k+1}}{a_k}\) if \(k\) is large?

c. Do you agree or disagree with the statement: “the series \(\sum \dfrac{2^k}{3^k - k}\) is approximately geometric when \(k\) is large”? If not, why not? If so, do you think the series \(\sum \dfrac{2^k}{3^k - k}\) converges or diverges? Explain.

We can generalize the argument in Activity 8.14 in the following way. Consider the series \( \sum a_k\). If

\(\dfrac{a_{k+1}}{a_k} \approx r \)

for large values of \(k\), then \(a_{k+1} \approx ra_k\) for large \(k\) and the series \( \sum ak\) is approximately the geometric series \( \sum ar^k \)for large \(k\). Since the geometric series with ratio r converges only for \(−1 < r < 1\), we see that the series \( \sum a_k\) will converge if

\[l\lim_{k \rightarrow \infty}\dfrac{a_{k+1}}{a_k} = r \nonumber\]

for a value of \(r\) such that \(|r| < 1\). This result is known as the Ratio Test.

### The Ratio Test

Let \( \sum a_k\) be an infinite series. Suppose

\[l\lim_{k \rightarrow \infty}\dfrac{a_{k+1}}{a_k} = r \nonumber\]

1. If \(0 ≤ r < 1\), then the series \( \sum a_k\) converges.

2. If \(1 < r\), then the series \( \sum a_k\) diverges.

3. If \(r = 1\), then the test is inconclusive.

Note well: The Ratio Test takes a given series and looks at the limit of the ratio of consecutive terms; in so doing, the test is essentially asking, “is this series approximately geometric?” If the series can be thought of as essentially geometric, the test use the limiting common ratio to determine if the given series converges.

We have now encountered several tests for determining convergence or divergence of series. The Divergence Test can be used to show that a series diverges, but never to prove that a series converges. We used the Integral Test to determine the convergence status of an entire class of series, the \(p\)-series. The Limit Comparison Test works well for series that involve rational functions and which can therefore by compared to \(p\)-series. Finally, the Ratio Test allows us to compare our series to a geometric series; it is particularly useful for series that involve nth powers and factorials. Two other tests, the Direct Comparison Test and the Root Test, are discussed in the exercises. Now it is time for some practice.

Activity \(\PageIndex{10}\):

Determine whether each of the following series converges or diverges. Explicitly state which test you use.

a. \( \sum \dfrac{k}{2^k}\)

b. \(\sum \dfrac{k^3+2}{k^2+1}\)

c. \(\sum \dfrac{10^k}{k!}\)

d. \(\sum \dfrac{k^3-2k^2+1}{k^6+4}\)

## Summary

In this section, we encountered the following important ideas:

- An infinite series is a sum of the elements in an infinite sequence. In other words, an infinite series is a sum of the form

\[a_1 + a_2 + · · · + a_n + · · · = \sum_{k=1}^\infty a_k \nonumber\]

where \(a_k\) is a real number for each positive integer \(k\).

- The \(n\)t partial sum \(S_n\) of the series \(\sum_{k=1}^\infty a_k\) is the sum of the first \(n\) terms of the series. That is,

\[S_n = a_1 + a_2 + · · · + a_n = \sum_{k=1}^n a_k \nonumber\]

- The sequence {\(S_n\)} of partial sums of a series \(\sum_{k=1}^\infty a_k\) tells us about the convergence or divergence of the series. In particular
- The series \(\sum_{k=1}^\infty a_k\) converges if the sequence {\(S_n\)} of partial sums converges. In this case we say that the series is the limit of the sequence of partial sums and write

\[\sum_{k=1}^\infty a_k = \lim_{n \rightarrow \infty} Sn. \nonumber\]

- The series \(\sum_{k=1}^\infty a_k\)diverges if the sequence {\(S_n\)} of partial sums diverges.