• Theorem 1 remains true if \(a_n\ge 0\). It is not necessary to assume that \(a_n>0\).
• It also remains true if \(a_n >0\) for all \(n\ge M\) for some \(M\), because the convergence of a series is not affected by the first \(M\) terms.

THEOREM 1 Dichotomy for Positive Series

If \(\displaystyle S=\sum_{n=1}^\infty a_n\) is a positive series, then either:

1. The partial sums \(S_N\) are bounded above. In this case, \(S\) converges. Or,
2. The partial sums \(S_N\) are not bounded above. In this case, \(S\) diverges.

The Integral Test is valid for any series \(\displaystyle\sum^\infty_{n=K} f(n)\), provided that for some \(M>0\), \(f(x)\) is positive, decreasing, and continuous for \(x\ge M\). The convergence of the series is determined by the convergence of \[ \int_M^{\infty}~f(x)\, dx \]

THEOREM 2 Integral Test

Let \(a_n = f(n)\), where \(f(x)\) is positive, decreasing, and continuous for \(x\ge 1\).

1. If \(\displaystyle\int_1^{\infty} f(x)\, dx\) converges, then \(\displaystyle\sum^\infty_{n=1} a_n\) converges.
2. If \(\displaystyle\int_1^{\infty} f(x)\, dx\) diverges, then \(\displaystyle\sum^\infty_{n=1} a_n\) diverges.

Proof

Because \(f(x)\) is decreasing, the shaded rectangles in Figure 10.17 lie below the graph of \(f(x)\), and therefore for all \(N\) \[ \underbrace{a_2 + \cdots +a_N}_{\textrm{Area of shaded rectangles in Figure 10.17}} \le \int_1^{N}f(x)\,dx\le \int_1^{\infty} f(x)\, dx \]

Figure 10.17: The shaded rectangles lie below the graph.

If the improper integral on the right converges, then the sums \(a_2 + \cdots + a_N\) remain bounded. In this case, \(S_N\) also remains bounded, and the infinite series converges by the Dichotomy Theorem (Theorem 1). This proves (i).

On the other hand, the rectangles in Figure 10.18 lie above the graph of \(f(x)\), so \[ \begin{equation*} \int_1^N f(x)\,dx\le \underbrace{a_1 + a_2 + \cdots + a_{N-1}}_{\textrm{Area of shaded rectangles in Figure 10.18}}\tag {1} \end{equation*} \]

Figure 10.18: The shaded rectangles lie above the graph.

If \(\displaystyle\sum^\infty_{n=1} f(x)\, dx\) diverges, then \({\int_1^N f(x)\,dx}\) tends to \(\infty\), and Eq. (1) shows that \( S_N\) also tends to \(\infty\). This proves (ii).

EXAMPLE 1 The Harmonic Series Diverges

Show that \(\displaystyle\sum^\infty_{n=1} \frac1n\) diverges.

Solution Let \(f(x)=\tfrac1x\). Then \(f(n) = \frac1n\), and the Integral Test applies because \(f\) is positive, decreasing, and continuous for \(x \geq 1\). The integral diverges:

\[ \int_1^{\infty} \frac{dx}x = \lim_{R\to\infty} \int_1^R \frac{dx}{x} = \lim_{R\to\infty}\ln R = \infty \] Therefore, the series \(\sum\limits_{n=1}^\infty \frac1n\) diverges.

EXAMPLE 2

Does \(\displaystyle\sum^\infty_{n=1} \dfrac{n}{(n^2 + 1)^2} = \frac{1}{2^2}+\frac{2}{5^2}+\frac{3}{10^2}+\cdots\) converge?

Solution The function \(f(x) = \frac{x}{(x^2+1)^2}\) is positive and continuous for \(x \geq 1\). It is decreasing because \(f'(x)\) is negative: \[ f'(x) = \frac{1-3x^2}{(x^2+1)^3} <0\qquad\textrm{for \(x\ge 1\)} \] Therefore, the Integral Test applies. Using the substitution \(u = x^2 + 1\), \(du = 2x\, dx\), we have \[ \begin{align*} \int_1^{\infty} \frac{x}{(x^2+1)^2}\, dx &= \lim_{R\to\infty} \int_1^{R} \frac{x}{(x^2+1)^2}\, dx =\lim_{R\to\infty} \frac12 \int_2^{R} \frac{du}{u^2} \\ &=\lim_{R\to\infty} \frac{-1}{2u} \bigg|_2^{R} = \lim_{R\to\infty} \left(\frac1{4} - \frac1{2R} \right) = \frac14 \end{align*} \] The integral converges. Therefore, \(\displaystyle\sum_{n=1}^{\infty} \dfrac{n}{(n^2 + 1)^2}\) also converges.

THEOREM 3 Convergence of \(p\)-Series

The infinite series \(\displaystyle\sum^\infty_{n=1} \frac1{n^p}\) converges if \(p>1\) and diverges otherwise.

Proof

If \(p\le 0\), then the general term \(n^{-p}\) does not tend to zero, so the series diverges. If \(p>0\), then \(f(x) = x^{-p}\) is positive and decreasing, so the Integral Test applies. According to Theorem 1 in Section 7.6, \[ \int_1^{\infty} \frac{1}{x^p}\, dx = \left\{\begin{array}{@{}l@{\qquad}l} \frac1{p-1} & \text{if } p>1\\ \infty & \text{if } p\le 1 \end{array}\right. \] Therefore, \(\displaystyle\sum^\infty_{n=1} \frac1{n^p}\) converges for \(p>1\) and diverges for \(p\le 1\).

THEOREM 4 Comparison Test

Assume that there exists \(M>0\) such that \(0 \leq a_n \leq b_n\) for \(n \geq M\).

1. If \(\displaystyle\sum^\infty_{n=1} b_n\) converges, then \(\displaystyle\sum^\infty_{n=1} a_n\) also converges.
2. If \(\displaystyle\sum^\infty_{n=1} a_n\) diverges, then \(\displaystyle\sum^\infty_{n=1} b_n\) also diverges.

Proof

We can assume, without loss of generality, that \(M=1\). If \(\displaystyle\sum^\infty_{n=1} b_n\) converges, then the partial sums of \(\displaystyle\sum^\infty_{n=1} a_n\) are bounded above by \(S\) because \[ \begin{equation*} a_1 + a_2 + \cdots + a_N \le b_1+b_2+\cdots +b_N \le \sum_{n = 1}^{\infty} b_n = S\tag{2} \end{equation*} \] Therefore, \(\sum\limits_{n=1}^\infty a_n\) converges by the Dichotomy Theorem (Theorem 1). This proves (i). On the other hand, if \(\displaystyle\sum^\infty_{n=1} a_n\) diverges, then \(\displaystyle\sum^\infty_{n=1} b_n\) must also diverge. Otherwise we would have a contradiction to (i).

EXAMPLE 3

Does \(\displaystyle\sum^\infty_{n=1} \frac{1}{\sqrt{n}\,3^n}\) converge?

Solution For \(n\ge 1\), we have \[ \frac{1}{\sqrt{n}\,3^n} \le \frac{1}{3^n} \] The larger series \(\displaystyle\sum^\infty_{n=1} \frac{1}{3^{n}}\) converges because it is a geometric series with \({r=\frac13<1}\). By the Comparison Test, the smaller series \(\displaystyle\sum^\infty_{n=1} \frac{1}{\sqrt{n}\,3^n}\) also converges.

In words, the Comparison Test states that for positive series:

• Convergence of the larger series forces convergence of the smaller series.
• Divergence of the smaller series forces divergence of the larger series.

EXAMPLE 4

Does \(\displaystyle\sum^\infty_{n=1} \frac1{(n^2+3)^{1/3}}\) converge?

Solution Let us show that \[ \frac{1}{n}\le \frac1{(n^2+3)^{1/3}} \qquad\textrm{for \(n\ge 2\)} \] This inequality is equivalent to \((n^2+3)\le n^3\), so we must show that \[f(x) = x^3-(x^2+3)\ge 0 \qquad\textrm{for \(x\ge 2\)}\] The function \(f(x)\) is increasing because its derivative \(f'(x) = 3x\big(x-\frac23\big)\) is positive for \(x\ge 2\). Since \(f(2)=1\), it follows that \(f(x)\ge 1\) for \(x\ge 2\), and our original inequality follows. We know that the smaller harmonic series \(\displaystyle\sum^\infty_{n=2} \frac{1}{n}\) diverges. Therefore, the larger series \(\displaystyle\sum^\infty_{n=2} \frac{1}{(n^2+1)^{1/3}}\) also diverges.

EXAMPLE 5 Using the Comparison Correctly

Study the convergence of \[ \sum_{n = 2}^{\infty} \frac{1}{n(\ln n)^2} \]

Solution We might be tempted to compare \(\displaystyle\sum^\infty_{n=2} \frac{1}{n(\ln n)^2}\) to the harmonic series \(\displaystyle\sum^\infty_{n=2} \frac{1}{n}\) using the inequality (valid for \(n \geq 3\)) \[ \frac1{n(\ln n)^2}\le \frac1n \] However, \(\displaystyle\sum^\infty_{n=2} \frac{1}{n}\) diverges, and this says nothing about the smaller series \(\sum \frac{1}{n(\ln n)^2}\). Fortunately, the Integral Test can be used. The substitution \(u = \ln x\) yields \[ \int_2^\infty \frac{dx}{x(\ln x)^2} = \int_{\ln 2}^\infty\frac{du}{u^2} = \lim_{R\to\infty} \left(\frac1{\ln 2} -\frac1{R} \right) = \frac1{\ln 2} <\infty \] The Integral Test shows that \(\displaystyle\sum^\infty_{n=2} \frac{1}{n(\ln n)^2}\) converges.

THEOREM 5 Limit Comparison Test

Let \(\{a_n\}\) and \(\{b_n\}\) be positive sequences. Assume that the following limit exists: \[ L = \lim\limits_{n\to\infty}\frac{a_n}{b_n} \]

• If \(L > 0\), then \({\displaystyle\sum a_n}\) converges if and only if \({\displaystyle\sum b_n}\) converges.
• If \(L=\infty\) and \({\displaystyle\sum a_n}\) converges, then \({\displaystyle\sum b_n}\) converges.
• If \(L = 0\) and \({\displaystyle\sum b_n}\) converges, then \({\displaystyle\sum a_n}\) converges.

CAUTION

The Limit Comparison Test is not valid if the series are not positive. See Exercise 44 in Section 10.4.

Proof

Assume first that \(L\) is finite (possibly zero) and that \(\displaystyle\sum b_n\) converges. Choose a positive number \(R > L\). Then \(0 \leq a_n/b_n \leq R\) for all \(n\) sufficiently large because \(a_n/b_n\) approaches \(L\). Therefore \(a_n \leq Rb_n\). The series \(\displaystyle\sum Rb_n\) converges because it is a multiple of the convergent series \(\displaystyle\sum b_n\). Therefore \(\sum a_n\) converges by the Comparison Test.

Next, suppose that \(L\) is nonzero (positive or infinite) and that \(\sum a_n\) converges. Let \(L^{-1} = \lim_{n\to\infty} {b_n}/{a_n}\). Then \(L^{-1}\) is finite and we can apply the result of the previous paragraph with the roles of \(\{a_n\}\) and \(\{b_n\}\) reversed to conclude that \({\displaystyle\sum b_n}\) converges.

CONCEPTUAL INSIGHT

To remember the different cases of the Limit Comparison Test, you can think of it this way. If \(L>0\), then \(a_n\approx Lb_n\) for large \(n\). In other words, the series \({\displaystyle\sum a_n}\) and \({\displaystyle\sum b_n}\) are roughly multiples of each other, so one converges if and only if the other converges. If \(L=\infty\), then \(a_n\) is much larger than \(b_n\) (for large \(n\)), so if \({\displaystyle\sum a_n}\) converges, \({\displaystyle\sum b_n}\) certainly converges. Finally, if \(L=0\), then \(b_n\) is much larger than \(a_n\) and the convergence of \({\displaystyle\sum b_n}\) yields the convergence of \(\sum a_n\).

EXAMPLE 6

Show that \(\displaystyle\sum^\infty_{n=2} \frac{n^2}{n^4 - n-1}\) converges.

Solution Let \[{a_n ={\frac{n^2}{n^4 - n-1}}}\quad \text{and}\quad {b_n = \frac{1}{{n^{2}}}}\]

We observed above that \(a_n\approx b_n\) for large \(n\). To apply the Limit Comparison Test, we observe that the limit \(L\) exists and \(L>0\): \[ \begin{eqnarray*} L &=& \lim_{n\to \infty}\frac{a_n}{b_n} = \lim_{n\to \infty}\frac{n^2}{n^4 -n -1}\cdot\frac{n^2}{1} = \lim_{n\to \infty} \frac{1}{1 - n^{-3}-n^{-4}} =1 \end{eqnarray*} \]

Since \(\displaystyle\sum^\infty_{n=2} \frac1{n^2}\) converges, our series \(\displaystyle\sum^\infty_{n=2} \frac{n^2}{n^4 - n-1}\) also converges by Theorem 5.

EXAMPLE 7

Determine whether \(\displaystyle\sum^\infty_{n=3} \frac{1}{\sqrt{n^2+4}}\) converges.

Solution Apply the Limit Comparison Test with \(a_n = \frac{1}{\sqrt{n^2+4}}\) and \(b_n = \frac1n\). Then \[ L = \lim_{n\to\infty}\frac{a_n}{b_n} = \lim_{n\to\infty}\frac{n}{\sqrt{n^2+4}} = \lim_{n\to\infty}\frac1{\sqrt{1+4/n^2}} = 1 \] Since \(\displaystyle\sum^\infty_{n=3} \frac1n\) diverges and \(L > 0\), the series \(\displaystyle\sum^\infty_{n=3} \frac{1}{\sqrt{n^2+4}}\) also diverges.