Calculus II 09.04 Series Comparisons

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9.4 Comparisons of Series

• Use the Direct Comparison Test to determine whether a series converges or diverges.
• Use the Limit Comparison Test to determine whether a series converges or diverges.

Direct Comparison Test

In the pairs listed below, the second series cannot be tested by the same convergence test as the first series, even though it is similar to the first. $$1.\:\: \sum_{n=0}^{\infty} \frac{1}{2^{n}} \text{ is geometric, but } \sum_{n=0}^{\infty} \frac{n}{2^{n}} \text{ is not. }$$ $$2.\:\: \sum_{n=1}^{\infty} \frac{1}{n^{3}} \text{ is a }p\text{-series, but } \sum_{n=1}^{\infty} \frac{1}{n^{3}+1} \text{ is not. }$$ $$3.\:\:a_{n}=\frac{n}{(n^{2}+3)^{2}} \text{ is easily integrated, but } b_{n}= \frac{n^{2}}{(n^{2}+3)^{2}} \text{ is not. }$$ The Direct Comparison Test and the Limit Comparison Test can compare a complex series with a simpler series whose convergence or divergence is known.

Theorem 9.4.1 Direct Comparison Test

Let \(0< a_{n} \leqslant b_{n}\) for all \(n\). $$1. \text{ If } \sum_{n=1}^{\infty} b_{n} \text{ converges, then } \sum_{n=1}^{\infty} a_{n} \text{ converges. }$$ $$2. \text{ If } \sum_{n=1}^{\infty} a_{n} \text{ diverges, then } \sum_{n=1}^{\infty} b_{n} \text{ diverges. }$$ Proof To prove the first property, let

$$ L = \sum_{n=1}^{\infty} b_{n}$$

and let

$$ S_{n}=a_{1}+a_{2}+a_{3}+\cdots+a_{n}.$$

Because \(0<a_{n} \leqslant b_{n}\), the sequence \(S_{1},\:S_{2},\:(S_{3},\cdots\) is nondecreasing and bounded above by \(L\); therefore it must converge. Because

$$ \lim_{n \to \infty} S_{n} = \sum_{n=1}^{\infty} a_{n} $$

it follows that \( \sum_{n=1}^{\infty} a_{n} \) converges. The second property is logically equivalent to the first.

Example 9.4.1 Using the Direct Comparison Test with Powers

Determine the convergence or divergence for

$$ \sum_{n=1}^{\infty} \frac{1}{2+3^{n}}.$$

Solution This series resembles

$$ \sum_{n=1}^{\infty} \frac{1}{3^{n}}.\:\:\:\:\: \color{red}{ \text{ Convergent geometric series} }$$

Term-by-term comparison yields

$$ a_{n} = \frac{1}{2+3^{n}} < \frac{1}{3^{n}} = b_{n},\:\:\:\: n \leqslant 1.$$

By the Direct Comparison Test, the series converges.

Example 9.4.2 Using the Direct Comparison Test with Roots

Determine the convergence or divergence for

$$ \sum_{n=1}^{\infty} \frac{1}{2+\sqrt{n}}.$$

Solution This series resembles

$$ \sum_{n=1}^{\infty} \frac{1}{n^{1/2}}.\:\:\:\:\: \color{red}{ \text{ Divergent } p \text{-series} }$$

Term-by-term comparison yields

$$\frac{1}{2+\sqrt{n}} \leqslant \frac{1}{\sqrt{n}},\:\:\: n \geqslant 1.$$

which does not meet the requirements for divergence. When term-by-term comparison reveals a series that is less than a divergent series, the Direct Comparison Test tells yields nothing. For another divergent test, compare the series with

$$ \sum_{n=1}^{\infty} \frac{1}{n}.\:\:\:\:\: \color{red}{ \text{ Divergent harmonic series}}$$

The new term-by-term comparison yields

$$ a_{n} = \frac{1}{n} \leqslant \frac{1}{2+\sqrt{n}} = b_{n},\:\:\: n \geqslant 4.$$

this shows the given series diverges by the Direct Comparison Test.

Limit Comparison Test

The Limit Comparison Test is used for \(p\)-series and geometric series where term-by-term comparisons can not be established and thus the Direct Comparison Test can not be applied.

Theorem 9.4.2 Limit Comparison Test

If \(a_{n} > 0, \: b_{n} > 0\), and

$$ \lim_{n \to \infty} \frac{a_{n}}{b_{n}} =L $$

where \(L\) is finite and positive, then

$$ \sum_{n=0}^{\infty} a_{n} \:\:\text{ and } \:\: \sum_{n=0}^{\infty} b_{n} $$

either both converge or both diverge.
Proof Because \(a_{n} > 0, \: b_{n} > 0\), and

$$ \lim_{n \to \infty} \frac{a_{n}}{b_{n}} =L $$

there exists \(N>0\) such that

$$ 0 < \frac{a_{n}}{b_{n}} < L + 1, \:\:\: n \geqslant N.$$

This implies that

$$ 0 < a_{n} < (L + 1)b_{n}.$$

By the Direct Comparison Test, the convergence for \( \sum b_{n}\) implies the convergence for \( \sum a_{n}\). Similarly, the fact that

$$ \lim_{n \to \infty} \frac{a_{n}}{b_{n}} = \frac{1}{L} $$

can be used to show that the convergence for \( \sum a_{n}\) implies the convergence for for \( \sum b_{n}\).

Example 9.4.3 Using the Limit Comparison Test

Show that the general harmonic series

$$ \sum_{n=1}^{\infty} \frac{1}{an+b}, \: a>0, \: b > 0$$

diverges.
Solution Comparison with

$$ \sum_{n=1}^{\infty} \frac{1}{n}\:\:\:\: \color{red}{ \text{ Divergent harmonic series} }$$

yields

$$ \lim_{n \to \infty} \frac{ 1/(an+b)}{1/n} = \lim_{n \to \infty} \frac{n}{an+b} = \frac{1}{a}. $$

Because this limit is greater than 0, Limit Comparison Test says the series diverges.

The Limit Comparison Test works well for comparing a complex algebraic series with a \(p\)-series. In choosing an appropriate \(p\)-series, choose one with an \(n\)th term with the same magnitude as the \(n\)th term for the given series.

Given Series	Comparison Series	Conclusion
$$ \sum_{n=1}^{\infty} \frac{1}{3n^{2}-4n+5} $$	$$ \sum_{n=1}^{\infty} \frac{1}{n^{2}} $$	Both series converge.
$$ \sum_{n=1}^{\infty} \frac{1}{\sqrt{3n-2}} $$	$$ \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} $$	Both series diverge.
$$ \sum_{n=1}^{\infty} \frac{n^{2}-10}{4n^{5}+n^{3}} $$	$$ \sum_{n=1}^{\infty} \frac{n^{2}}{n^{5}} = \sum_{n=1}^{\infty} \frac{1}{n^{3}} $$	Both series converge.

When choosing a series for comparison disregard all but the highest powers for \(n\) in both the numerator and the denominator.

Example 9.4.4 Using the Limit Comparison Test with Roots

Determine the convergence or divergence for

$$ \sum_{n=1}^{\infty} \frac{\sqrt{n}}{n^{2}+1}. $$

Solution Disregarding all but the highest powers for \(n\) in the numerator and the denominator allows a comparison with

$$ \sum_{n=1}^{\infty} \frac{\sqrt{n}}{n^{2}} = \sum_{n=1}^{\infty} \frac{1}{n^{3/2}} .\:\:\:\:\: \color{red}{ \text{ Convergent }p \text{-series} }$$

Because

$$ \lim_{n \to \infty} \frac{a_{n}}{b_{n}} $$	$$= \lim_{n \to \infty} \left ( \frac{\sqrt{n}}{n^{2}+1} \right) \left ( \frac{n^{3/2}}{1} \right)$$
	$$ \lim_{n \to \infty} \frac{n^{2}}{n^{2}+1} $$
	$$ = 1$$

the series converges by the Limit Comparison Test.

Example 9.4.5 Using the Limit Comparison Test with \(2^{n}\)

Determine the convergence or divergence for

$$ \sum_{n=1}^{\infty} \frac{n2^{n}}{4n^{3}+1}. $$

Solution A reasonable comparison can be made with the series

$$ \sum_{n=1}^{\infty} \frac{2^{n}}{n^{2}}.\:\:\:\:\: \color{red}{ \text{ Divergent series} }$$

Note that this series diverges by the \(n\)th-Term Test. From the limit

$$ \lim_{n \to \infty} \frac{a_{n}}{b_{n}} $$	$$= \lim_{n \to \infty} \left ( \frac{n2^{n}}{4n^{3}+1} \right) \left ( \frac{n^{2}}{2^{n}} \right)$$
	$$ \lim_{n \to \infty} \frac{1}{4+(1/n^{3})} $$
	$$ = \frac{1}{4}$$

the series diverges.

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Parent Article: Calculus II 09 Infinite Series