ABSOLUTE CONVERGENCE AND THE RATIO TEST

You should note that, outside of the Alternating Series Test presented in section 7.4, our other tests for convergence of series (i.e., the Integral Test and the two comparison tests) apply only to series all of whose terms are positive. So, what do we do if we’re faced with a series that has both positive and negative terms, but that is not an alternating series? For instance, look at the series

∞ k=1

sin k

k³ = sin 1 +1

8sin 2+ 1

27sin 3+ 1

64sin 4+ · · · .

This has both positive and negative terms, but the terms do not alternate signs. (Calculate the first five or six terms of the series to see this for yourself.) For any such series^∞

k=1ak, we can get around this problem by checking if the series of absolute values^∞

k=1|ak| is convergent.

When this happens, we say that the original series ^∞

k=1ak is absolutely convergent (or converges absolutely). You should note that to test the convergence of the series of absolute values^∞

k=1|ak| (all of whose terms are positive), we have all of our earlier tests for positive term series available to us.

EXAMPLE 5.1 Testing for Absolute Convergence Determine if^∞

k=1

(−1)^k+1

2^k is absolutely convergent.

Solution It is easy to show that this alternating series is convergent. (Try it!) From the graph of the first 20 partial sums in Figure 7.34, it appears that the series converges to approximately 0.35. To determine absolute convergence, we need to determine whether or not the series of absolute values, ^∞

k=1

(−1)^k+1 2^k

 is convergent. We have

∞ k=1

(−1)^k+1 2^k

 =^∞

k=1

1 2^k =^∞

k=1

1 2

k

,

5 10 15 20

0.1 0.2 0.3 0.4 0.5 S_n

n

FIGURE 7.34 Sn=ⁿ

k=1

(−1)^k+1 2^k .

which you should recognize as a convergent geometric series (|r| = ¹₂ < 1). This says that the original series^∞

k=1

(−1)^k+1

2^k converges absolutely.



You might be wondering about the relationship between convergence and absolute con-vergence. We’ll prove shortly that every absolutely convergent series is also convergent (as in example 5.1). However, the reverse is not true; there are many series that are convergent, but not absolutely convergent. These are called conditionally convergent series. Can you think of an example of such a series? If so, it’s probably the series in example 5.2.

EXAMPLE 5.2 A Conditionally Convergent Series Determine if the alternating harmonic series^∞

k=1

(−1)^k+1

k is absolutely convergent.

Solution In example 4.2, we showed that this series is convergent. To test this for absolute convergence, we consider the series of absolute values,

∞ k=1

(−1)^k+1 k

 =^∞

k=1

1 k,

which is the harmonic series. We showed in section 7.2 (example 2.7) that the harmonic series diverges. This says that ^∞

k=1

(−1)^k+1

k converges, but does not converge absolutely (i.e., it converges conditionally). 

THEOREM 5.1 If^∞

k=1|ak| converges, then^∞

k=1akconverges.

This result says that if a series converges absolutely, then it must also converge. Because of this, when we test series, we first test for absolute convergence. If the series converges absolutely, then we need not test any further to establish convergence.

PROOF

Notice that for any real number, x, we can say that −|x| ≤ x ≤ |x|. So, for any k, we have

−|ak| ≤ ak≤ |ak|.

Adding|ak| to all the terms, we get

0≤ ak+ |ak| ≤ 2|ak|. (5.1)

Since ^∞

k=1ak is absolutely convergent, we have that ^∞

k=1|ak| and hence, also ^∞

k=12|ak| = 2^∞

k=1|ak| is convergent. Define bk= ak+ |ak|. From (5.1), 0≤ bk≤ 2|ak|

and so, by the Comparison Test,^∞

k=1bkis convergent. Observe that we may write

∞ k=1

ak =^∞

k=1

(ak+ |ak| − |ak|) =^∞

k=1

(ak+ |ak|)  

bk

−^∞

k=1

|ak|

=^∞

k=1

bk−^∞

k=1

|ak|.

Since the two series on the right-hand side are convergent, it follows that^∞

k=1akmust also be convergent.

EXAMPLE 5.3 Testing for Absolute Convergence Determine whether^∞

k=1

sin k

k³ is convergent or divergent.

5 10 15 20

0.86 0.90 0.94 0.98

S_n

n

FIGURE 7.35 Sn=ⁿ

k=1

sin k k³ .

Solution Notice that while this is not a positive-term series, it is also not an alternating series. Because of this, our only choice (given what we know) is to test the series for absolute convergence. From the graph of the first 20 partial sums seen in Figure 7.35, it appears that the series is converging to some value around 0.94. To test for absolute convergence, we consider the series of absolute values,^∞

k=1

sin k k³

. Notice that

sin k k³

 = |sin k|

k³ ≤ 1

k³, (5.2)

since|sin k| ≤ 1, for all k. Of course,^∞

k=1

1

k³is a convergent p-series ( p= 3 > 1). By the Comparison Test and (5.2),^∞

k=1

sin k k³

 converges, too. Consequently, the original series

∞ k=1

sin k

k³ converges absolutely (and hence, converges).



The Ratio Test

We now introduce a very powerful tool for testing a series for absolute convergence. This test can be applied to a wide range of series, including the extremely important case of power series that we discuss in section 7.6. As you’ll see, this test is also remarkably easy to use.

THEOREM 5.2 (Ratio Test) Given^∞

k=1ak, with ak = 0 for all k, suppose that

klim→∞

ak+1

ak

 = L.

Then,

(i) if L< 1, the series converges absolutely, (ii) if L> 1 (or L = ∞), the series diverges and (iii) if L= 1, there is no conclusion.

Because the proof of Theorem 5.2 is somewhat involved, we omit it.

EXAMPLE 5.4 Using the Ratio Test Test^∞

k=1

(−1)^kk

2^k for convergence.

5 10 15 20

1.0 1.5 2.0

0.5 S_n

n

FIGURE 7.36 S_n=ⁿ

k=1

k 2^k.

Solution From the graph of the first 20 partial sums of the series of absolute values,

∞ k=1

k

2^k, seen in Figure 7.36, it appears that the series of absolute values converges to about 2. From the Ratio Test, we have

klim→∞

ak+1

ak

 = limk→∞

k+ 1 2^k+1

k 2^k

= lim

k→∞

k+ 1 2^k+1

2^k k =1

2 lim

k→∞

k+ 1

k = 1

2 < 1 ^Since

2^k+1= 2^k· 2¹.

and so, the series converges absolutely, as expected from Figure 7.36. 

The Ratio Test is particularly useful when the general term of a series contains an exponential term, as in example 5.4 or a factorial, as in example 5.5.

EXAMPLE 5.5 Using the Ratio Test Test^∞

k=0

(−1)^kk!

e^k for convergence.

5 10 15 20

4  10⁸

6  10⁸

2  10⁸ 2  10⁸ S_n

n

FIGURE 7.37 S_n=ⁿ⁻¹

k=0

(−1)^kk!

e^k .

Solution From the graph of the first 20 partial sums of the series seen in Figure 7.37, it appears that the series is diverging. (Look closely at the scale on the y-axis and compute a table of values for yourself.) We can confirm this suspicion with the Ratio Test. We have

klim→∞

ak+1

ak

 = lim_k→∞

(k+ 1)!

e^k+1 k!

e^k

= lim

k→∞

(k+ 1)!

e^k+1 e^k k!

= lim

k→∞

(k+ 1) k!

ek! = 1

e lim

k→∞

k+ 1

1 = ∞. ^{Since (k}+ 1)! = (k + 1) · k!

and e^k+1= e^k· e¹.

By the Ratio Test, the series diverges, as we suspected. 

Recall that in the statement of the Ratio Test (Theorem 5.2), we said that if

klim→∞

ak+1

ak

 = 1,

then the Ratio Test yields no conclusion. By this, we mean that in such cases, the series may or may not converge and further testing is required.

HISTORICAL NOTES

Srinivasa Ramanujan (1887–1920) Indian

mathematician whose incredible discoveries about infinite series still mystify mathematicians.

Largely self-taught, Ramanujan filled notebooks with conjectures about series, continued fractions and the Riemann-zeta function.

Ramanujan rarely gave a proof or even justification of his results.

Nevertheless, the famous English mathematician G. H. Hardy said,

“They must be true because, if they weren’t true, no one would have had the imagination to invent them.” (See Exercise 39.)

EXAMPLE 5.6 A Divergent Series for Which the Ratio Test Fails Use the Ratio Test for the harmonic series^∞

k=1

1 k. Solution We have

klim→∞

ak+1

ak

 = limk→∞

1 k+ 1

1 k

= lim

k→∞

k k+ 1 = 1.

In this case, the Ratio Test yields no conclusion, although we already know that the harmonic series diverges. 

EXAMPLE 5.7 A Convergent Series for Which the Ratio Test Fails Use the Ratio Test to test the series^∞

k=0

1 k². Solution Here, we have

klim→∞

ak+1

ak

 = lim_k→∞ 1 (k+ 1)²

k² 1

= lim

k→∞

k²

k²+ 2k + 1= 1.

So again, the Ratio Test yields no conclusion, although we already know that this is a convergent p-series (with p= 2 > 1). 

Look carefully at examples 5.6 and 5.7. You should recognize that the Ratio Test will be inconclusive for any p-series. Fortunately, we don’t need the Ratio Test for these series.

We now present one final test for convergence of series.

THEOREM 5.3 (Root Test) Given^∞

k=1ak, suppose that lim

k→∞

√k

|ak| = L. Then, (i) if L< 1, the series converges absolutely, (ii) if L> 1 (or L = ∞), the series diverges and (iii) if L= 1, there is no conclusion.

Notice how similar the conclusion is to the conclusion of the Ratio Test.

EXAMPLE 5.8 Using the Root Test

Use the Root Test to determine the convergence or divergence of the series

∞ k=1

2k+ 4 5k− 1

k

.

Solution In this case, we consider

klim→∞

k

|ak| = lim

k→∞

k

2k+ 4 5k− 1

^k= lim

k→∞

2k+ 4 5k− 1 = 2

5 < 1.

By the Root Test, the series is absolutely convergent. 

By this point in your study of series, it may seem as if we have thrown at you a dizzying array of different series and tests for convergence or divergence. Just how are you to keep all of these straight? The only suggestion we have is that you work through many problems.

We provide a good assortment in the exercise set that follows this section. Some of these require the methods of this section, while others are drawn from the preceding sections ( just to keep you thinking about the big picture). For the sake of convenience, we summarize our convergence tests in the table that follows.

Test When to use Conclusions Section

Geometric Series ^∞

k=0

ar^k Converges to a

1− r if|r| < 1; 7.2 diverges if|r| ≥ 1.

kth-Term Test All series If lim

k→∞a_k = 0, the series diverges. 7.2

Integral Test ^∞

k=1akwhere f (k)= akand ^∞

k=1akand_∞

1 f (x) d x 7.3

f is continuous, decreasing and f (x)≥ 0 both converge or both diverge.

p-series ^∞

k=1

1

k^p Converges for p> 1; diverges for p ≤ 1. 7.3

Comparison Test ^∞

k=1akand^∞

k=1bk, where 0 ≤ ak≤ bk If^∞

k=1bkconverges, then^∞

k=1akconverges. 7.3 If^∞

k=1

a_kdiverges, then^∞

k=1

b_kdiverges.

Limit Comparison Test ^∞

k=1

a_kand^∞

k=1

b_k, where ^∞

k=1

a_kand^∞

k=1

b_k 7.3

ak, bk> 0 and lim

k→∞

ak

bk = L > 0 both converge or both diverge.

Alternating Series Test ^∞

k=1(−1)^k+1akwhere ak> 0 for all k If lim

k→∞ak= 0 and a^k+1≤ a^kfor all k, 7.4 then the series converges.

Absolute Convergence Series with some positive and some negative terms (including alternating series)

If^∞

k=1|a^k| converges, then 7.5

∞ k=1

a_kconverges (absolutely).

Ratio Test Any series (especially those involving exponentials and/or factorials)

For lim

k→∞

a_k+1 a_k

 = L, 7.5

if L< 1,^∞

k=1

a_kconverges absolutely if L> 1,^∞

k=1

a_kdiverges, if L= 1, no conclusion.

Root Test Any series (especially those involving exponentials)

For lim

k→∞

√k|ak| = L, 7.5

if L< 1,^∞

k=1

a_kconverges absolutely if L> 1,^∞

k=1

akdiverges, if L= 1, no conclusion.

EXERCISES 7.5

WRITING EXERCISES

1. Suppose that two series have identical terms except that in

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