Comparison Tests: Learn It 1

  • Compare series to determine convergence using the comparison test
  • Use the limit comparison test to figure out if a series converges

We’ve seen that the integral test determines convergence by comparing a series to a related improper integral. Now we’ll learn how to determine convergence by comparing a series to another series whose behavior we already know. Comparison tests are particularly useful for series that resemble geometric series or p-series—two types whose convergence patterns we understand completely.

The Comparison Test

In previous sections, we studied geometric series and p-series and learned exactly when they converge or diverge. The comparison test lets us use the known behavior of these series to determine convergence for other, similar series.

Before diving into the test, recall this important inequality behavior:

If [latex]0 < a_n \le b_n[/latex], then [latex]\frac{1}{a_n} \ge \frac{1}{b_n}[/latex]

In other words, a larger denominator corresponds to a smaller fraction.

Comparing Similar Series

Consider the series:

[latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{2}+1}[/latex].

This series looks similar to the convergent series:

[latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}[/latex].

Since the terms in each of the series are positive, the sequence of partial sums for each series is monotone increasing.

Notice that for all positive integers [latex]n[/latex]:

[latex]0<\frac{1}{{n}^{2}+1}<\frac{1}{{n}^{2}}[/latex].
This inequality holds because [latex]n^2 + 1 > n^2[/latex], making [latex]\frac{1}{n^2 + 1}[/latex] smaller than [latex]\frac{1}{n^2}[/latex].

Therefore, the [latex]k[/latex]th partial sum [latex]S_k[/latex] of [latex]\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}[/latex] satisfies:

[latex]{S}_{k}=\displaystyle\sum _{n=1}^{k}\frac{1}{{n}^{2}+1}<\displaystyle\sum _{n=1}^{k}\frac{1}{{n}^{2}}<\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{2}}[/latex].

(See Figure 1 (a) and the following table.)

Comparing a series with a p-series (p = 2)
[latex]k[/latex] [latex]1[/latex] [latex]2[/latex] [latex]3[/latex] [latex]4[/latex] [latex]5[/latex] [latex]6[/latex] [latex]7[/latex] [latex]8[/latex]
[latex]\displaystyle\sum _{n=1}^{k}\frac{1}{{n}^{2}+1}[/latex] [latex]0.5[/latex] [latex]0.7[/latex] [latex]0.8[/latex] [latex]0.8588[/latex] [latex]0.8973[/latex] [latex]0.9243[/latex] [latex]0.9443[/latex] [latex]0.9597[/latex]
[latex]\displaystyle\sum _{n=1}^{k}\frac{1}{{n}^{2}}[/latex] [latex]1[/latex] [latex]\begin{array}{l}1.25\hfill \end{array}[/latex] [latex]1.3611[/latex] [latex]1.4236[/latex] [latex]1.4636[/latex] [latex]1.4914[/latex] [latex]1.5118[/latex] [latex]1.5274[/latex]

Since the series [latex]\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2}[/latex] converges, the sequence [latex]{S_k}[/latex] is bounded above. We now have a sequence that is both monotone increasing and bounded above.

By the Monotone Convergence Theorem, [latex]{S_k}[/latex] must converge. Therefore, [latex]\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2 + 1}[/latex] converges.

Now consider the series:

[latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{n - \frac{1}{2}}[/latex].

This series looks similar to the divergent series:

[latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{n}[/latex].

The sequence of partial sums for each series is monotone increasing, and we can compare the terms:

[latex]\frac{1}{n - \frac{1}{2}}>\frac{1}{n}>0[/latex]

This inequality holds because [latex]n - \frac{1}{2} < n[/latex], making [latex]\frac{1}{n - \frac{1}{2}}[/latex] larger than [latex]\frac{1}{n}[/latex].

Therefore, the [latex]k[/latex]th partial sum [latex]S_k[/latex] of [latex]\displaystyle\sum_{n=1}^{\infty} \frac{1}{n - \frac{1}{2}}[/latex] satisfies:

[latex]{S}_{k}=\displaystyle\sum _{n=1}^{k}\frac{1}{n - \frac{1}{2}}>\displaystyle\sum _{n=1}^{k}\frac{1}{n}[/latex].

(See Figure 1 (a) and the following table.)

Comparing a series with the harmonic series
[latex]k[/latex] [latex]1[/latex] [latex]2[/latex] [latex]3[/latex] [latex]4[/latex] [latex]5[/latex] [latex]6[/latex] [latex]7[/latex] [latex]8[/latex]
[latex]\displaystyle\sum _{n=1}^{k}\frac{1}{n - \frac{1}{2}}[/latex] [latex]2[/latex] [latex]2.6667[/latex] [latex]3.0667[/latex] [latex]3.3524[/latex] [latex]3.5746[/latex] [latex]3.7564[/latex] [latex]3.9103[/latex] [latex]4.0436[/latex]
[latex]\displaystyle\sum _{n=1}^{k}\frac{1}{n}[/latex] [latex]1[/latex] [latex]1.5[/latex] [latex]\begin{array}{l}1.8333\hfill \end{array}[/latex] [latex]2.0933[/latex] [latex]2.2833[/latex] [latex]2.45[/latex] [latex]2.5929[/latex] [latex]2.7179[/latex]

Since the harmonic series [latex]\displaystyle\sum_{n=1}^{\infty} \frac{1}{n}[/latex] diverges to infinity, the sequence of partial sums [latex]\displaystyle\sum_{n=1}^{k} \frac{1}{n}[/latex] is unbounded.

Consequently, [latex]{S_k}[/latex] is also unbounded and therefore diverges. We conclude that [latex]\displaystyle\sum_{n=1}^{\infty} \frac{1}{n - \frac{1}{2}}[/latex] diverges.

This shows two graphs side by side. The first shows plotted points for the partial sums for the sum of 1/n^2 and the sum 1/(n^2 + 1). Each of the partial sums for the latter is less than the corresponding partial sum for the former. The second shows plotted points for the partial sums for the sum of 1/(n - 0.5) and the sum 1/n. Each of the partial sums for the latter is less than the corresponding partial sum for the former.
Figure 1. (a) Each of the partial sums for the given series is less than the corresponding partial sum for the converging [latex]p-\text{series}[/latex]. (b) Each of the partial sums for the given series is greater than the corresponding partial sum for the diverging harmonic series.

comparison test

  1. Suppose there exists an integer [latex]N[/latex] such that [latex]0\le {a}_{n}\le {b}_{n}[/latex] for all [latex]n\ge N[/latex].
    If [latex]\displaystyle\sum _{n=1}^{\infty }{b}_{n}[/latex] converges, then [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] converges.
  2. Suppose there exists an integer [latex]N[/latex] such that [latex]{a}_{n}\ge {b}_{n}\ge 0[/latex] for all [latex]n\ge N[/latex].
    If [latex]\displaystyle\sum _{n=1}^{\infty }{b}_{n}[/latex] diverges, then [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] diverges.

Proof

We prove part i. The proof of part ii. is the contrapositive of part i. Let [latex]\left\{{S}_{k}\right\}[/latex] be the sequence of partial sums associated with [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex], and let [latex]L=\displaystyle\sum _{n=1}^{\infty }{b}_{n}[/latex]. Since the terms [latex]{a}_{n}\ge 0[/latex],

[latex]{S}_{k}={a}_{1}+{a}_{2}+\cdots +{a}_{k}\le {a}_{1}+{a}_{2}+\cdots +{a}_{k}+{a}_{k+1}={S}_{k+1}[/latex].

 

Therefore, the sequence of partial sums is increasing. Further, since [latex]{a}_{n}\le {b}_{n}[/latex] for all [latex]n\ge N[/latex], then

[latex]\displaystyle\sum _{n=N}^{k}{a}_{n}\le \displaystyle\sum _{n=N}^{k}{b}_{n}\le \displaystyle\sum _{n=1}^{\infty }{b}_{n}=L[/latex].

 

Therefore, for all [latex]k\ge 1[/latex],

[latex]{S}_{k}=\left({a}_{1}+{a}_{2}+\cdots +{a}_{N - 1}\right)+\displaystyle\sum _{n=N}^{k}{a}_{n}\le \left({a}_{1}+{a}_{2}+\cdots +{a}_{N - 1}\right)+L[/latex].

 

Since [latex]{a}_{1}+{a}_{2}+\cdots +{a}_{N - 1}[/latex] is a finite number, we conclude that the sequence [latex]\left\{{S}_{k}\right\}[/latex] is bounded above. Therefore, [latex]\left\{{S}_{k}\right\}[/latex] is an increasing sequence that is bounded above. By the Monotone Convergence Theorem, we conclude that [latex]\left\{{S}_{k}\right\}[/latex] converges, and therefore the series [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] converges.

[latex]_\blacksquare[/latex]

How to: Apply the Comparison Test

To use the comparison test for a series [latex]\displaystyle\sum_{n=1}^{\infty} a_n[/latex], you need to find a suitable comparison series. Since we know the convergence properties of geometric series and [latex]p[/latex]-series, these are the most commonly used comparison series.

  • Strategy for convergence: If you can show that [latex]a_n \leq b_n[/latex] for all [latex]n \geq N[/latex] (for some integer [latex]N[/latex]), where [latex]\displaystyle\sum_{n=1}^{\infty} b_n[/latex] is a known convergent series, then [latex]\displaystyle\sum_{n=1}^{\infty} a_n[/latex] converges.
  • Strategy for divergence: If you can show that [latex]a_n \geq b_n[/latex] for all [latex]n \geq N[/latex], where [latex]\displaystyle\sum_{n=1}^{\infty} b_n[/latex] is a known divergent series, then [latex]\displaystyle\sum_{n=1}^{\infty} a_n[/latex] diverges.

For each of the following series, use the comparison test to determine whether the series converges or diverges.

  1. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{3}+3n+1}[/latex]
  2. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{2}^{n}+1}[/latex]
  3. [latex]\displaystyle\sum _{n=2}^{\infty }\frac{1}{\text{ln}\left(n\right)}[/latex]