8.4: The Divergence and Integral Tests

Section Preview

The following is a list of learning objectives for this section.

Learning Objectives (click to expand)

• Use the Divergence Test to determine whether a series converges or diverges.
• Use the Integral Test to determine the convergence of a series.
• Estimate the value of a series by finding bounds on its remainder term.

In the previous section, we determined the convergence or divergence of several series by explicitly calculating the limit of the sequence of partial sums \(\{S_k\}\). In practice, explicitly calculating this limit can be difficult or impossible. Luckily, several tests exist that allow us to determine convergence or divergence for many types of series. This section discusses two tests: the Divergence Test and the Integral Test. We will examine several other tests in the rest of this chapter and then summarize how and when to use them.

Divergence Test

We begin our exploration of tests for the convergence or divergence of series with the test that is the quickest to use.

Theorem: Divergence Test

If \(\displaystyle \lim_{n \to \infty}a_n \neq 0\), then the series \(\displaystyle \sum_{n=1}^ \infty a_n\) diverges.

Proof by Contraposition

Suppose \(\displaystyle \sum_{n=1}^ \infty a_n\) converges. Let \(S_n = a_1 + a_2 + \cdots + a_n\). Then \(a_n = S_n - S_{n - 1}\) (a fact that can be very useful in proofs). Since we have assumed \(\displaystyle \sum_{n=1}^ \infty a_n\) converges, the sequence of partial sums, \(\{S_n \}\) must be convergent. Let \(\displaystyle \lim_{n \to \infty} S_n = S\). Since \(n - 1 \to \infty\) as \(n \to \infty\), we also have \(\displaystyle \lim_{n \to \infty} S_{n - 1} = S\). Therefore\[\lim_{n \to \infty} a_n = \lim_{n \to \infty} \left( S_n - S_{n - 1} \right) = \lim_{n \to \infty} S_n - \lim_{n \to \infty} S_{n - 1} = S - S = 0.\nonumber\]Hence, the limit \(\displaystyle \lim_{n \to \infty} a_n\) exists and is \(0\).

Proofs by Contraposition

Most theorems can be stated symbolically as the conditional proposition \(P \implies Q\), where \(P\) is a statement called the antecedent, and \(Q\) is a statement called the consequent. Instead of directly proving \(P \implies Q\), it is sometimes easier to prove it indirectly. There are two kinds of indirect proofs: proof by contradiction, and proof by contraposition.

You have already seen proofs by contradiction (see the uses of Rolle's Theorem in Calculus I for proving that a given equation has exactly one root). In a proof by contraposition, we use the fact that an implication, "if \(P\), then \(Q\)," is equivalent to its contrapositive, "if not \(Q\), then not \(P\)" (the notation, \(\sim P\), is called the negation of \(P\)). Therefore, instead of proving \(P \implies Q\), we may prove its contrapositive \(\sim Q \implies \sim P\). Since it is an implication, we could use a direct proof at this point:

Assume \(\sim Q\) is true (hence, assume \(Q\) is false). Show that \(\sim P\) is true (that is, show that \(P\) is false).

As with all of the proofs in Calculus, it is worthwhile to read through the proof of the Divergence Test and try to understand it. The method (proof by contraposition) is a wonderful indirect proof style.

Caution: The Converse is NOT True

It is important to note that the converse of the Divergence Test is not true. That is, if \(\displaystyle \lim_{n \to \infty}a_n=0\), we cannot make any conclusion about the convergence of \(\displaystyle \sum_{n=1}^ \infty a_n\).

To illustrate that cautionary statement, consider the harmonic series \(\displaystyle \sum^ \infty _{n=1}\frac{1}{n}\).

\(\displaystyle \lim_{n \to 0}\frac{1}{n}=0\), but the harmonic series \(\displaystyle \sum^ \infty _{n=1}\frac{1}{n}\) diverges. From this point forward, we show many more examples of such series. Consequently, although we can use the Divergence Test to show that a series diverges, we cannot use it to prove that a series converges. Specifically, if \(a_n \to 0\), the Divergence Test is inconclusive.

Example \(\PageIndex{1}\): Using the Divergence Test

For each of the following series, apply the Divergence Test. If the Divergence Test proves that the series diverges, state so. Otherwise, indicate that the Divergence Test is inconclusive.

1. \(\displaystyle \sum^ \infty _{n=1}\frac{n}{3n-1}\)
2. \(\displaystyle \sum^ \infty _{n=1}\frac{1}{n^3}\)
3. \(\displaystyle \sum^ \infty _{n=1}e^{1/n^2}\)

Solutions

1. Since \(\displaystyle \lim_{n \to \infty} \frac{n}{3n-1}=\frac{1}{3} \neq 0\), by the Divergence Test, we can conclude that \(\displaystyle \sum_{n=1}^ \infty \frac{n}{3n-1}\) diverges.
2. Since \(\displaystyle \lim_{n \to \infty} \frac{1}{n^3}=0\), the Divergence Test is inconclusive.
3. Since \(\displaystyle \lim_{n \to \infty} e^{1/n^2}=1 \neq 0\), by the Divergence Test, the series \(\displaystyle \sum_{n=1}^ \infty e^{1/n^2}\) diverges.

Checkpoint \(\PageIndex{1}\)

What does the Divergence Test tell us about the series \(\displaystyle \sum_{n=1}^ \infty \cos(1/n^2)\)?

Answer

The series diverges.

Integral Test

In the previous section, we proved that the harmonic series diverges by looking at the sequence of partial sums \(\{S_k\}\) and showing that \(S_{2^k} \gt 1+k/2\) for all \(k \in \mathbb{N}\). In this section, we use a different technique to prove the divergence of the harmonic series. This technique is important because it is used to prove the divergence or convergence of many other series. This test, called the Integral Test, compares an infinite sum to an improper integral. It is important to note that this test can only be applied when considering a series with positive terms.

To illustrate how the Integral Test works, use the harmonic series as an example. In Figure \(\PageIndex{1}\), we depict the harmonic series by sketching a sequence of rectangles with areas \(1, 1/2, 1/3, 1/4, \ldots\) along with the function \(f(x)=1/x\).

Figure \(\PageIndex{1}\): The sum of the areas of the rectangles is greater than the area between the curve \(f(x)=1/x\) and the \(x\)-axis for \(x \geq 1\). Since the area bounded by the curve is infinite (as calculated by an improper integral), the sum of the areas of the rectangles is also infinite.

From the graph, we see that\[\sum_{n=1}^k\dfrac{1}{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+ \cdots +\dfrac{1}{k} \gt \int ^{k+1}_1\dfrac{1}{x}\,dx.\nonumber\]Therefore, for each \(k\), the \(k^{\text{th}}\) partial sum \(S_k\) satisfies\[\begin{array}{rcl} S_k & = & \displaystyle \sum_{n=1}^k\dfrac{1}{n} \\[6pt] & \gt & \displaystyle \int^{k+1}_1\dfrac{1}{x}\,dx \\[6pt] & = & \ln x \big| ^{k+1}_1 \\[6pt] & = & \ln (k+1)-\ln (1) \\[6pt] & = & \ln (k+1). \\[6pt] \end{array}\nonumber\]Since \(\displaystyle \lim_{k \to \infty}\ln(k+1)= \infty\), we see that the sequence of partial sums \(\{S_k\}\) is unbounded. Therefore, \(\{S_k\}\) diverges, and, consequently, the series \(\displaystyle \sum_{n=1}^ \infty \frac{1}{n}\) also diverges.

Interactive Element: Integral Test: Comparing a Series to an Integral

The following Interactive Element might help you visualize how the sum of the rectangle areas compares to the area under the curve \( f(x)=1/x \).

Interact: Drag the green point along the \( x \)-axis to change \( N \), the number of rectangles.

Observation: Because \( f(x)=1/x \) is decreasing, every left-endpoint rectangle rises above the curve, so \( S_N \) always exceeds the area \( \ln(N+1) \) under the curve from \( 1 \) to \( N+1 \). Since \( \ln(N+1)\to\infty \), the partial sums are unbounded and the harmonic series diverges.

Now consider the series \(\displaystyle \sum_{n=1}^ \infty \frac{1}{n^2}\). We show how an integral can be used to prove that this series converges. In Figure \(\PageIndex{2}\), we sketch a sequence of rectangles with areas \(1, 1/2^2, 1/3^2, \ldots\) along with the function \(f(x)=\frac{1}{x^2}\).

Figure \(\PageIndex{2}\): The sum of the areas of the rectangles is less than the sum of the area of the first rectangle and the area between the curve \(f(x)=1/x^2\) and the \(x\)-axis for \(x \geq 1\). Since the area bounded by the curve is finite, the sum of the areas of the rectangles is also finite.

From the graph we see that\[\sum_{n=1}^k\dfrac{1}{n^2}=1+\dfrac{1}{2^2}+\dfrac{1}{3^2}+ \cdots +\dfrac{1}{k^2} \lt 1+ \int ^k_1\dfrac{1}{x^2}\,dx.\nonumber\]Therefore, for each \(k\), the \(k^{\text{th}}\) partial sum \(S_k\) satisfies\[\begin{array}{rcl} S_k & = & \sum_{n=1}^k\dfrac{1}{n^2} \\[6pt] & \lt & 1+ \int ^k_1\dfrac{1}{x^2}\,dx \\[6pt] & = & 1 - \dfrac{1}{x} \bigg|^k_1 \\[6pt] & = & 1 - \dfrac{1}{k}+1 \\[6pt] & = & 2-\dfrac{1}{k} \\[6pt] & \lt & 2. \\[6pt] \end{array}\nonumber\]We conclude that the sequence of partial sums \(\{S_k\}\) is bounded. Moreover, since\[S_k = S_{k-1}+\dfrac{1}{k^2}\nonumber\]for \(k \geq 2\), the sequence of partial sums is also increasing. Therefore, by the Monotone Convergence Theorem, it converges. Thus, the series \(\displaystyle \sum_{n=1}^ \infty \frac{1}{n^2}\) converges.

We can extend this idea to prove convergence or divergence for many different series. Suppose \(\displaystyle \sum^ \infty _{n=1}a_n\) is a series with positive terms \(a_n\) such that there exists a continuous, positive, decreasing function \(f\) where \(f(n)=a_n\) for all positive integers. Then, as in Figure \(\PageIndex{3a}\), for any integer \(k\), the \(k^{\text{th}}\) partial sum \(S_k\) satisfies\[S_k=a_1+a_2+a_3+ \cdots +a_k \lt a_1+ \int ^k_1f(x)\,dx \lt 1+ \int ^ \infty _1f(x)\,dx.\nonumber\]

Figure \(\PageIndex{3a}\): If we can inscribe rectangles inside a region bounded by a curve \(y=f(x)\) and the \(x\)-axis, and the area bounded by those curves for \(x \geq 1\) is finite, then the sum of the areas of the rectangles is also finite.
Figure \(\PageIndex{3b}\): If a set of rectangles circumscribes the region bounded by \(y=f(x)\) and the \(x\)-axis for \(x \geq 1\) and the region has infinite area, then the sum of the areas of the rectangles is also infinite.

Therefore, if \(\displaystyle \int ^ \infty _1f(x)\,dx\) converges, then the sequence of partial sums \(\{S_k\}\) is bounded. Since \(\{S_k\}\) is an increasing sequence, it converges by the Monotone Convergence Theorem if it is also a bounded sequence. We conclude that if \(\displaystyle \int ^ \infty _1f(x)\,dx\) converges, then the series \(\displaystyle \sum^ \infty _{n=1}a_n\) also converges.

On the other hand, from Figure \(\PageIndex{3b}\), for any integer \(k\), the \(k^{\text{th}}\) partial sum \(S_k\) satisfies\[S_k=a_1+a_2+a_3+ \cdots +a_k> \int ^{k+1}_1f(x)\,dx.\nonumber\]If\[\lim_{k \to \infty} \int ^{k+1}_1f(x)\,dx= \infty ,\nonumber\]then \(\{S_k\}\) is an unbounded sequence and therefore diverges. As a result, the series \(\displaystyle \sum_{n=1}^ \infty a_n\) also diverges. Since \(f\) is a positive function, if \(\displaystyle \int ^ \infty _1f(x)\,dx\) diverges, then\[\lim_{k \to \infty} \int ^{k+1}_1f(x)\,dx= \infty .\nonumber\]We conclude that if \(\displaystyle \int ^ \infty _1f(x)\,dx\) diverges, then \(\displaystyle \sum_{n=1}^ \infty a_n\) diverges.

Theorem: Integral Test

Suppose \(\displaystyle \sum_{n=1}^ \infty a_n\) is a series with positive terms \(a_n\). Suppose there exists a function \(f\) and a positive integer \(N\) such that the following three conditions are satisfied:

1. \(f\) is continuous,
2. \(f\) is decreasing, and
3. \(f(n)=a_n\) for all integers \(n \geq N\).

Then\[\sum_{n=1}^ \infty a_n\nonumber\]and\[\int ^ \infty _Nf(x)\,dx\nonumber\]both converge or both diverge (Figure \(\PageIndex{3}\)).

Caution: The Sequence Must EVENTUALLY be Non-Negative

The base sequence, \(\{a_n \}\), in the Integral Test must eventually consist of non-negative terms only. That is, to be able to use the Integral Test, there must exist \(N\) such that \(f(n) = a_n \geq 0\) for all \(n \geq N\). This (along with the need to evaluate an improper integral) is a weakness of the Integral Test and a reason we will eventually need to consider other tests for convergence.

Although the convergence of \(\displaystyle \int ^ \infty _Nf(x)\, dx\) implies convergence of the related series \(\displaystyle \sum_{n=1}^ \infty a_n\), it does not imply that the value of the integral and the series are the same. They may be different and often are. For example,\[\sum_{n=1}^ \infty \left(\dfrac{1}{e}\right)^n=\dfrac{1}{e}+\left(\dfrac{1}{e}\right)^2+\left(\dfrac{1}{e}\right)^3+ \cdots\nonumber\]is a geometric series with initial term \(a=1/e\) and ratio \(r=1/e\), which converges to\[\dfrac{1/e}{1-(1/e)}=\dfrac{1/e}{(e-1)/e}=\dfrac{1}{e-1}.\nonumber\]However, the related integral \(\displaystyle \int ^ \infty _1(1/e)^x\,dx\) satisfies\[\int ^ \infty _1\left(\dfrac{1}{e}\right)^x\,dx= \int ^ \infty _1e^{-x}\,dx=\lim_{b \to \infty} \int ^b_1e^{-x}\,dx=\lim_{b \to \infty}-e^{-x}\big|^b_1=\lim_{b \to \infty}[-e^{-b}+e^{-1}]=\dfrac{1}{e}.\nonumber\]

Example \(\PageIndex{2}\): Using the Integral Test

For each of the following series, use the Integral Test to determine whether the series converges or diverges.

1. \(\displaystyle \sum_{n=1}^ \infty \frac{1}{n^3}\)
2. \(\displaystyle \sum^ \infty _{n=1}\frac{1}{\sqrt{2n-1}}\)

Solutions

1. Compare\[\sum_{n=1}^ \infty \dfrac{1}{n^3} \text{ and } \int ^ \infty _1\dfrac{1}{x^3}\,dx.\nonumber\]We have\[\begin{array}{rcl} \displaystyle \int^\infty_1 \dfrac{1}{x^3}\,dx & = & \displaystyle \lim_{b \to \infty} \int ^b_1\dfrac{1}{x^3}\,dx \\[6pt] & = & \displaystyle \lim_{b \to \infty}\left[-\dfrac{1}{2x^2}\bigg|^b_1\right] \\[6pt] & = & \displaystyle \lim_{b \to \infty}\left[-\dfrac{1}{2b^2}+\dfrac{1}{2}\right] \\[6pt] & = & \dfrac{1}{2}. \\[6pt] \end{array}\nonumber\]Thus the integral \(\displaystyle \int ^ \infty _1\frac{1}{x^3}\,dx\) converges, and therefore so does the series\[\sum_{n=1}^ \infty \dfrac{1}{n^3}.\nonumber\]
2. Compare\[\sum_{n=1}^ \infty \dfrac{1}{\sqrt{2n-1}} \text{ and } \int ^ \infty _1\dfrac{1}{\sqrt{2x-1}}\,dx.\nonumber\]Since\[\begin{array}{rcl} \displaystyle \int ^ \infty _1\dfrac{1}{\sqrt{2x-1}}\,dx & = & \displaystyle \lim_{b \to \infty} \int ^b_1\dfrac{1}{\sqrt{2x-1}}\,dx \\[6pt] & = & \displaystyle \lim_{b \to \infty}\sqrt{2x-1}\bigg|^b_1 \\[6pt] & = & \displaystyle \lim_{b \to \infty}\left[\sqrt{2b-1}-1\right] \\[6pt] & = & \infty, \\[6pt] \end{array}\nonumber\]the integral \(\displaystyle \int ^ \infty _1\frac{1}{\sqrt{2x-1}}\,dx\) diverges, and therefore\[\sum_{n=1}^ \infty \dfrac{1}{\sqrt{2n-1}}\nonumber\]diverges.

Checkpoint \(\PageIndex{2}\)

Use the Integral Test to determine whether the series \(\displaystyle \sum^ \infty _{n=1}\frac{n}{3n^2+1}\) converges or diverges.

Answer

The series diverges.

The \(p\)-Series

The harmonic series \(\displaystyle \sum^ \infty _{n=1}1/n\) and the series \(\displaystyle \sum^ \infty _{n=1}1/n^2\) are both examples of a type of series called a \(p\)-series.

Definition: \(p\)-series

For any real number \(p\), the series\[\sum_{n=1}^ \infty \dfrac{1}{n^p}\nonumber\]is called a \(p\)-series.

We know the \(p\)-series converges if \(p=2\) and diverges if \(p=1\). What about other values of \(p\)? In general, it is difficult, if not impossible, to compute the exact value of most \(p\)-series. However, we can use the tests presented thus far to prove whether a \(p\)-series converges or diverges.

If \(p \lt 0\), then \(1/n^p \to \infty\), and if \(p=0\), then \(1/n^p \to 1\). Therefore, by the Divergence Test,\[\sum_{n=1}^ \infty \dfrac{1}{n^p}\nonumber\]diverges if \(p \leq 0\).

If \(p \gt 0\), then \(f(x)=1/x^p\) is a positive, continuous, decreasing function. Therefore, for \(p \gt 0\), we use the Integral Test, comparing\[\sum_{n=1}^ \infty \dfrac{1}{n^p} \text{ and } \int ^ \infty _1\dfrac{1}{x^p}\,dx.\nonumber\]We have already considered the case when \(p=1\). We consider the case when \(p \gt 0, p \neq 1\). For this case,\[\begin{array}{rcl} \displaystyle \int ^ \infty _1\dfrac{1}{x^p}\,dx & = & \displaystyle \lim_{b \to \infty} \int ^b_1\dfrac{1}{x^p}\,dx \\[6pt] & = & \displaystyle \lim_{b \to \infty}\dfrac{1}{1-p}x^{1-p}\bigg|^b_1 \\[6pt] & = & \displaystyle \lim_{b \to \infty}\dfrac{1}{1-p}[b^{1-p}-1]. \\[6pt] \end{array}\nonumber\]Because \(b^{1-p} \to 0\) if \(p \gt 1\) and \(b^{1-p} \to \infty\) if \(p \lt 1\), we conclude that\[\int ^ \infty _1\dfrac{1}{x^p}\,dx=\begin{cases}\dfrac{1}{p-1}, \text{if}\;p>1\\[6pt] \infty , \text{if}\;p<1.\end{cases}\nonumber\]Therefore, \(\displaystyle \sum^ \infty _{n=1}1/n^p\) converges if \(p>1\) and diverges if \(0<p<1\). We summarize this as a theorem.

Theorem: \(p\)-series Test

The \(p\)-series\[\sum_{n = 1}^{\infty} \dfrac{1}{n^p}\nonumber\]converges if and only if \(p \gt 1\). It diverges otherwise.

Example \(\PageIndex{3}\): Testing for Convergence of p-series

For each of the following series, determine whether it converges or diverges.

1. \(\displaystyle \sum^ \infty _{n=1}\frac{1}{n^4}\)
2. \(\displaystyle \sum^ \infty _{n=1}\frac{1}{n^{2/3}}\)

Solutions

1. This is a \(p\)-series with \(p=4>1\), so the series converges.
2. Since \(p=2/3<1\), the series diverges.

Checkpoint \(\PageIndex{3}\)

Does the series \(\displaystyle \sum^ \infty _{n=1}\frac{1}{n^{5/4}}\) converge or diverge?

Answer

The series converges.

Estimating the Value of a Series

Suppose we know that a series \(\displaystyle \sum_{n=1}^ \infty a_n\) converges, and we want to estimate the sum of that series. Certainly we can approximate that sum using any finite sum \(\displaystyle \sum_{n=1}^N a_n\) where \(N \in \mathbb{N}\). The question we address here is, for a convergent series \(\displaystyle \sum^ \infty _{n=1}a_n\), how good is the approximation \(\displaystyle \sum^N_{n=1}a_n\)?

More specifically, if we let\[R_N = \sum_{n=1}^ \infty a_n-\sum_{n=1}^N a_n\nonumber\]be the remainder when the sum of an infinite series is approximated by the \(N^{\text{th}}\) partial sum, how large is \(R_N\)? For some types of series, we can use the ideas from the Integral Test to estimate \(R_N\).

Theorem: Remainder Estimate from the Integral Test

Suppose \(\displaystyle \sum^ \infty _{n=1}a_n\) is a convergent series with positive terms (remember, when dealing with the Integral Test, the terms of your base sequence, \(\{a_n \}\) must eventually be non-negative). Also suppose there exists a function \(f\) satisfying the following three conditions:

1. \(f\) is continuous,
2. \(f\) is decreasing, and
3. \(f(n)=a_n\) for all integers \(n \geq 1\).

Let \(S_N\) be the \(N^{\text{th}}\) partial sum of \(\displaystyle \sum^ \infty _{n=1}a_n\). For all \(N \in \mathbb{N}\),\[S_N + \int^\infty_{N+1} f(x) \, dx \lt \sum_{n=1}^\infty a_n \lt S_N + \int ^ \infty _Nf(x)\,dx.\nonumber\]In other words, the remainder \(\displaystyle R_N=\sum^ \infty _{n=1}a_n-S_N=\sum^ \infty _{n=N+1}a_n\) satisfies the following estimate:\[\int ^ \infty _{N+1}f(x)\,dx<R_N< \int ^ \infty _Nf(x)\,dx.\nonumber\]This is known as the remainder estimate.

The remainder estimate can be considered the error in approximating the series value.

We illustrate this theorem in Figure \(\PageIndex{4}\).

Figure \(\PageIndex{4}\): Given a continuous, positive, decreasing function \(f\) and a sequence of positive terms \(a_n\) such that \(a_n=f(n)\) for all \(n \in \mathbb{N}\), (a) the areas \(\displaystyle a_{N+1}+a_{N+2}+a_{N+3}+ \cdots \lt \int ^ \infty _Nf(x)\,dx\), or (b) the areas \(\displaystyle a_{N+1}+a_{N+2}+a_{N+3}+ \cdots \gt \int ^ \infty _{N+1}f(x)\,dx\). Therefore, the integral is either an overestimate or an underestimate of the error.

In particular, by representing the remainder \(R_N=a_{N+1}+a_{N+2}+a_{N+3}+ \cdots\) as the sum of areas of rectangles, we see that the area of those rectangles is bounded above by \(\displaystyle \int ^ \infty _Nf(x)\, dx\) and bounded below by \(\displaystyle \int ^ \infty _{N+1}f(x)\, dx\). In other words,\[R_N=a_{N+1}+a_{N+2}+a_{N+3}+ \cdots \gt \int ^ \infty _{N+1}f(x)\,dx\nonumber\]and\[R_N=a_{N+1}+a_{N+2}+a_{N+3}+ \cdots \lt \int ^ \infty _Nf(x)\,dx.\nonumber\]We conclude that\[\int ^ \infty _{N+1}f(x)\,dx \lt R_N \lt \int ^ \infty _Nf(x)\,dx.\nonumber\]Since\[\sum_{n=1}^ \infty a_n=S_N+R_N,\nonumber\]where \(S_N\) is the \(N^{\text{th}}\) partial sum, we conclude that\[S_N+ \int ^ \infty _{N+1}f(x)\,dx \lt \sum_{n=1}^ \infty a_n \lt S_N+ \int ^ \infty _Nf(x)\,dx.\nonumber\] The ability to approximate a series's value is important, and we will use different approximation methods for different convergence tests as we move forward.

Example \(\PageIndex{4}\): Estimating the Value of a Series

Consider the series \(\displaystyle \sum^ \infty _{n=1}\frac{1}{n^3}\).

1. Calculate \(\displaystyle S_{10}=\sum^{10}_{n=1}\frac{1}{n^3}\) and estimate the error.
2. Determine the least value of \(N\) necessary such that \(S_N\) will estimate \(\displaystyle \sum^ \infty _{n=1}\frac{1}{n^3}\) to within \(0.001\).

Solutions

1. Using technology, we have\[S_{10}=1+\dfrac{1}{2^3}+\dfrac{1}{3^3}+\dfrac{1}{4^3}+ \cdots +\dfrac{1}{10^3} \approx 1.19753.\nonumber\]By the remainder estimate, we know\[R_N \lt \int ^ \infty _N\dfrac{1}{x^3}\,dx.\nonumber\]We have\[\begin{array}{rcl} \displaystyle \int ^ \infty _{10}\dfrac{1}{x^3}\,dx & = & \displaystyle \lim_{b \to \infty} \int ^b_{10}\dfrac{1}{x^3}\,dx \\[6pt] & = & \displaystyle \lim_{b \to \infty}\left[-\dfrac{1}{2x^2}\right]^b_{10} \\[6pt] & = & \displaystyle \lim_{b \to \infty}\left[-\dfrac{1}{2b^2}+\dfrac{1}{2(10)^2}\right] \\[6pt] & = & \dfrac{1}{200}. \\[6pt] \end{array}\nonumber\]Therefore, the error is \(R_{10} \lt \frac{1}{200} = 0.005\).
2. We want to find \(N\) such that \(R_N \lt 0.001\). In part a. we showed that \(R_N \lt 1/2N^2\). Therefore, the remainder \(R_N \lt 0.001\) as long as \(\frac{1}{2N^2} \lt 0.001\). That is, we need \(2N^2 \gt 1000\). Solving this inequality for \(N\), we see that we need \(N \gt 22.36\). We need to round up to the nearest integer to ensure the remainder is within the desired amount. Therefore, the minimum necessary value is \(N=23\).

Checkpoint \(\PageIndex{4}\)

For \(\displaystyle \sum^ \infty _{n=1}\frac{1}{n^4}\), calculate \(S_5\) and estimate the error \(R_5\).

Answer

\(S_5 \approx 1.09035, R_5 \lt 0.00267\)

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Success in Mathematics: Mathematics in Other Courses

In other courses, we use mathematics to understand and explore the big ideas of the subject. Professors may use different vocabulary, variables, and notations than you learned in your math classes. The mathematics is the same. The way of talking about it may be different.

Also, some courses will use formulas based on mathematics that you have not yet learned. You may not be ready to understand why these formulas work. Sometimes you have to accept that the formula is correct.

Making connections between different courses takes work. It would be best if you were willing to ask questions. You may need to ask your math instructor about topics in your other classes. You may need to ask your other professors why they use different words or notations. It is your responsibility to make these connections. Your ability to do critical thinking in other classes depends on it.

1. There are many formulas that we hear about in daily news or history. Of all these famous formulas, name one.
2. Can you describe the use of the formula you mentioned and what each variable stands for?

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Miscellaneous Topics: Teachers and Role Models

Parents, teachers, and other role models influence our attitudes about math. What they say and do can change our view of the importance of math, how hard math is, and whether we are good at math. Here are some questions about this topic that may help you succeed in this class (and possibly many other classes as well):

1. Describe the attitudes towards math of your parent, teacher (not me, of course), or another personal role model.
2. Describe how the opinion of this role model has influenced your attitudes about math.