Knowing whether or not a series converges is very important, especially when we discuss Power Series in Section 8.6. Theorems 8.2.5 and 8.2.10 give criteria for when Geometric and $p$-series converge, and Theorem 8.2.20 gives a quick test to determine if a series diverges. There are many important series whose convergence cannot be determined by these theorems, though, so we introduce a set of tests that allow us to handle a broad range of series. We start with the Integral Test.

Subsection8.3.1Integral Test

We stated in Section 8.1 that a sequence $\{a_n\}$ is a function $a(n)$ whose domain is $\mathN\text{,}$ the set of natural numbers. If we can extend $a(n)$ to $\mathbb{R}\text{,}$ the real numbers, and it is both positive and decreasing on $[1,\infty)\text{,}$ then the convergence of $\ds \infser a_n$ is the same as $\ds\int_1^\infty a(x)\ dx\text{.}$

Theorem 8.3.1 does not state that the integral and the summation have the same value.

We can demonstrate the truth of the Integral Test with two simple graphs. In Figure 8.3.2.(a), the height of each rectangle is $a(n)=a_n$ for $n=1,2,\ldots\text{,}$ and clearly the rectangles enclose more area than the area under $y=a(x)\text{.}$ Therefore we can conclude that $$\ds \int_1^\infty a(x)\ dx \lt \infser a_n. \label{eq_integral_testa}\tag{8.3.1}$$

In Figure 8.3.2.(b), we draw rectangles under $y=a(x)$ with the Right-Hand rule, starting with $n=2\text{.}$ This time, the area of the rectangles is less than the area under $y=a(x)\text{,}$ so $\ds\sum_{n=2}^\infty a_n \lt \int_1^\infty a(x)\ dx\text{.}$ Note how this summation starts with $n=2\text{;}$ adding $a_1$ to both sides lets us rewrite the summation starting with $n=1\text{:}$ $$\infser a_n \lt a_1 +\int_1^\infty a(x)\ dx. \label{eq_integral_testb}\tag{8.3.2}$$

Combining Equations (8.3.1) and (8.3.2), we have $$\infser a_n\lt a_1 +\int_1^\infty a(x)\ dx \lt a_1 + \infser a_n. \label{eq_integral_testc}\tag{8.3.3}$$

From Equation (8.3.3) we can make the following two statements:

1. If $\ds \infser a_n$ diverges, so does $\ds\int_1^\infty a(x)\ dx$ (because $\ds \infser a_n \lt a_1 +\int_1^\infty a(x)\ dx$)

2. If $\ds \infser a_n$ converges, so does $\ds\int_1^\infty a(x)\ dx$ (because $\ds \ds \int_1^\infty a(x)\ dx \lt \infser a_n\text{.}$)

Therefore the series and integral either both converge or both diverge. Theorem 8.2.21 allows us to extend this theorem to series where $a(n)$ is positive and decreasing on $[b,\infty)$ for some $b>1\text{.}$ A formal proof of the Integral Test is shown below.

Example8.3.3Using the Integral Test

Determine the convergence of $\ds\infser \frac{\ln(n) }{n^2}\text{.}$ (The terms of the sequence $\{a_n\} = \{\ln(n) /n^2\}$ and the n$^{\text{ th } }$ partial sums are given in Figure 8.3.4.)

Solution

Theorem 8.2.10 was given without justification, stating that the general $p$-series $\ds \infser \frac 1{(an+b)^p}$ converges if, and only if, $p>1\text{.}$ In the following example, we prove this to be true by applying the Integral Test.

Example8.3.5Using the Integral Test to establish Theorem 8.2.10

Use the Integral Test to prove that $\ds \infser \frac1{(an+b)^p}$ converges if, and only if, $p>1\text{.}$

Solution

We consider two more convergence tests in this section, both comparison tests. That is, we determine the convergence of one series by comparing it to another series with known convergence.

Subsection8.3.2Direct Comparison Test

A sequence $\{a_n\}$ is a positive sequence if $a_n>0$ for all $n\text{.}$

Because of Theorem 8.2.21, any theorem that relies on a positive sequence still holds true when $a_n>0$ for all but a finite number of values of $n\text{.}$

Example8.3.7Applying the Direct Comparison Test

Determine the convergence of $\ds\infser \frac1{3^n+n^2}\text{.}$

Solution
Example8.3.8Applying the Direct Comparison Test

Determine the convergence of $\ds\infser \frac{1}{n-\ln(n) }\text{.}$

Solution

The concept of direct comparison is powerful and often relatively easy to apply. Practice helps one develop the necessary intuition to quickly pick a proper series with which to compare. However, it is easy to construct a series for which it is difficult to apply the Direct Comparison Test.

Consider $\ds\infser \frac1{n+\ln(n) }\text{.}$ It is very similar to the divergent series given in Example 8.3.8. We suspect that it also diverges, as $\ds \frac 1n \approx \frac1{n+\ln(n) }$ for large $n\text{.}$ However, the inequality that we naturally want to use “goes the wrong way”: since $n\leq n+\ln(n)$ for all $n\geq 1\text{,}$ $\ds\frac1n \geq \frac{1}{n+\ln(n) }$ for all $n\geq 1\text{.}$ The given series has terms less than the terms of a divergent series, and we cannot conclude anything from this.

Fortunately, we can apply another test to the given series to determine its convergence.

Subsection8.3.3Limit Comparison Test

Theorem 8.3.9 is most useful when the convergence of the series from $\{b_n\}$ is known and we are trying to determine the convergence of the series from $\{a_n\}\text{.}$

We use the Limit Comparison Test in the next example to examine the series $\ds\infser \frac1{n+\ln(n) }$ which motivated this new test.

Example8.3.10Applying the Limit Comparison Test

Determine the convergence of $\ds\infser \frac1{n+\ln(n) }$ using the Limit Comparison Test.

Solution
Example8.3.11Applying the Limit Comparison Test

Determine the convergence of $\ds\infser \frac1{3^n-n^2}$

Solution

As mentioned before, practice helps one develop the intuition to quickly choose a series with which to compare. A general rule of thumb is to pick a series based on the dominant term in the expression of $\{a_n\}\text{.}$ It is also helpful to note that factorials dominate exponentials, which dominate algebraic functions (e.g., polynomials), which dominate logarithms. In the previous example, the dominant term of $\ds\frac{1}{3^n-n^2}$ was $3^n\text{,}$ so we compared the series to $\ds \infser \frac1{3^n}\text{.}$ It is hard to apply the Limit Comparison Test to series containing factorials, though, as we have not learned how to apply L'Hôpital's Rule to $n!\text{.}$

Example8.3.12Applying the Limit Comparison Test

Determine the convergence of $\ds\infser \frac{\sqrt{n}+3}{n^2-n+1}\text{.}$

Solution

We mentioned earlier that the Integral Test did not work well with series containing factorial terms. The next section introduces the Ratio Test, which does handle such series well. We also introduce the Root Test, which is good for series where each term is raised to a power.

Subsection8.3.4Exercises

Terms and Concepts

In the following exercises, use the Integral Test to determine the convergence of the given series.