Chapter 8: Infinite Sequences and Series
Section 8.3: Convergence Tests
Essentials
Table 8.3.1 details several tests for the convergence (or divergence) of infinite series.
Test Name
Test Details
nth-term test
If limn→∞an≠0, then Σ an diverges.
Integral test
f is a continuous on n0,∞, positive, decreasing function on k,∞, for some k≥n0
an=fn
⇒
∑n=n0an converges or diverges accordingly as does ∫k∞fx dx
Alternating-Series test (Leibniz test)
The alternating series Σ −1nan , with an>0, converges (at least conditionally) if an is a monotonically decreasing sequence whose limit is zero. (Leibniz)
Ratio test: L=limn→∞an+1an
L<1 ⇒ Σ an converges absolutely
L>1 ⇒ Σ an diverges
L=1 ⇒ Test is inconclusive: Σ an could converge or diverge
Root test: ρ=limn→∞⁡ann
ρ<1 ⇒ Σ an converges absolutely
ρ>1 ⇒ Σ an diverges
ρ=1 ⇒ Test is inconclusive: Σ an could converge or diverge
Comparison test
The terms in Σ an and Σ bn are positive and an≤bn for all n . Then:
Σ bn converges ⇒ Σ an converges
Σ an diverges ⇒ Σ bn diverges
Limit-Comparison test:
c=limn→∞anbn
The terms in Σ an and Σ bn are positive. Then:
c>0 ⇒ both series converge or diverge
c=0 and Σ bn converges ⇒ Σ an converges
c=∞ and Σ bn diverges ⇒ Σ an diverges
Table 8.3.1 Some tests for the convergence (or divergence) of series
The nth-term test is sometimes called the "Divergence test." Actually, it is nothing more than the contrapositive of Theorem 8.2.2 in Table 8.2.1. Theorem 8.2.2 states that the nth-term in a convergent series goes to zero; it's contrapositive therefore states that if the nth-term does not go to zero, then the series does not converge.
If the Integral test declares a series to be convergent, the value of the integral and the sum of the series are not necessarily the same!
The conclusion of the Alternating-Series test is that the tested series is conditionally convergent. But the series might actually be absolutely convergent by some other test. To see that this is so, take an absolutely convergent series whose terms satisfy the hypotheses of the Alternating-Series test, and alternate the signs. The Alternating-Series test will declare as conditionally convergent a series that actually converges absolutely.
It can be shown that whenever L exists in the Ratio test, so does ρ in the Root test, and the two numbers will be equal! Hence, if L=1 in the Ratio test (so the test fails), then ρ will be 1 in the Root test (and it will likewise fail). However, there are series that are undecided by the Ratio test (necessarily, it must be that L does not exist), but decided by the Root test. Hence, the Root test is deemed to be "stronger" than the Ratio test, even though it is often harder to find the value of ρ than it is to find L.
Examples
Determine if each of the series in Table 8.3.2 diverges, converges absolutely, or converges conditionally. For series that converge conditionally, determine whether they also converge absolutely. Each series in the table is summed to infinity, but that notation is not repeated to save vertical space.
Example 8.3.1
∑n=1arctannn2+4
Example 8.3.9
∑n=0e−nn!
Example 8.3.17
∑n=2−1nn lnn
Example 8.3.2
∑n=21n lnn
Example 8.3.10
∑n=2lnnn2
Example 8.3.18
∑n=2−1n lnnn
Example 8.3.3
∑n=01n3+5
Example 8.3.11
∑n=1cosnn2
Example 8.3.19
∑n=12⋅4⋅⋯⋅2 nn!
Example 8.3.4
∑n=1nn2+2
Example 8.3.12
∑n=11/nn
Example 8.3.20
∑n=2−1n nlnn
Example 8.3.5
∑n=2n3−1n!
Example 8.3.13
∑n=112+n
Example 8.3.21
∑n=010n/n!
Example 8.3.6
Example 8.3.14
∑n=2−1nlnn
Example 8.3.22
∑n=1−1n+1n n
Example 8.3.7
∑n=1lnn5 n+2
Example 8.3.15
∑n=1−1n+1n+1
Example 8.3.23
∑n=03n3 n+2!
Example 8.3.8
∑n=1sin3 nn2
Example 8.3.16
∑n=2lnnn
Example 8.3.24
∑n=1−1n+12 n+5
Table 8.3.2 Infinite series to be tested for convergence or divergence
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