Infinite Series Tests: Convergent or Divergent?
First, it is important to recall some key definitions and notations before discussing the individual series
aSequence: a function that takes on real number values and is defined on the set of positive integers n
= 1, 2, 3,… n
aaaaa！,,,,, Listed Notation: (；nn123
n1234a！！,,,(；n For example: for = 1, 2, 3, 4, ... nn？12345
Series: a summation of a sequence of numbers
aaaaa！？？？？？?nn123 Notation: n1！
Note that a sequence is a list of values, each value occurring for different value of n
However a series is the sum of these values.
Convergence - whenever a sequence or a series has a limit
Divergence - whenever a sequence or a series does not have a limit
As a matter of notation, in the following pages, let for some integer aa！c~0??nnnc！
The first type of series test is used only for a series with a specific form:
TEST FOR CONVERGENCE OF A GEOMETRIC SERIES:
A geometric series is a series which follows the pattern:
：nn2araararar！？？？？？ ?n0 ！
135where is the initial term and is a constant ratio (i.e., , , , etc. ). ra243
？1，r，1If (that is, ), then the series converges. r，1
If , the series diverges. r~1
aSum of the series: Further, when convergent, this series converges to: 1？r
11 Determine whether the series converges. 2？1？？？24
To find r divide any term by the term preceeding it.
1a！2 The series follows the pattern of a geometric series with and . r！2
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n：111?，？？？？！212 So, ?(?242)：0n！
11！，1 By the test for convergence, this series does converge, since . 22
It follows that the series converges to 4 since:
An infinite series with no negative terms is often referred to as a positive series.
There are several tests used to determine whether or not a positive series converges or diverges.
THE NTH-TERM TEST:
a If , then the series diverges. lima?0?nnn；：
Otherwise, the test is inconclusive.
This test is useful in quickly determining whether a given series diverges.
：23n？ Does the following series converge or diverge: ?51n？n1！
Notice that the terms (the sequence values) get smaller as gets larger. However, the terms do n
not approach zero. In fact:
2Therefore, for very large values of , the series is (more or less) adding on with each n5
successive term. And so, the sum (series) will grow infinitely large.
：223n？a！?lim0Therefore, since: , we have that the series diverges. n?；：n551n？n1！
THE INTEGRAL TEST:
f(x)，0f(x)x~1Let for , and is a continuous decreasing function.
a！f(n) Given , n
f(x)dx if is convergent, then so is a; ?n?1！n1
f(x)dxa if is divergent, then so is . ?n?1！n1
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This test is useful for the determining the convergence or divergence of a series that has a sequence
that looks like it might be easy enough to integrate. a(；n
Determine whether or not the series converges
(This special series is called the Harmonic Series)
The nth-term test doesn’t help here since (the terms approach a value of zero). lima！0nn；：
11 Since , Let f(x)！a！(；nxn
Using the integral test ：b11limdxdx！??b；：xx11
Since the integral is divergent, then the series is divergent.
1 Given a series of the form: ?pn
0，p，1p，1If , then the series diverges. If, then the seriesconverges.
This test is derived from the integral test and is useful only for the series which contain this specific form.
However, it will be quite helpful during the second portion of the comparison tests.
DIRECT COMPARISON TEST:
bcd Given three series, ,, : nnn
0，，cd if for each and converges, then converges; dcnnn??nn
0，，bc if for each and diverges, then diverges. bcnnn??nn
This test is used quite often and is one of the more important ones to know. It requires familiarity with
the convergence or divergence of other series, in addition to being able to compare them with a given series. This test proves very useful when a series appears “untestable.”
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Example #1: ：1n？2 Does the series converge or diverge? ??3nn？3n！1
1This series closely resembles the p-series, , which converges. In order to compare the two 3n
n？2，1series using the direct comparison test, determine which series is larger. Since for n？3
1n？2111any n, . (Multiplying each side by preserves the inequality since is a ，?3333nnnn？3n：1n？2positive term for all n greater than or equal to 1). So, by the comparison test, ??3nn？3n！1
Example #2: ：n？5 Does the series converge or diverge? ?2？3nnn！1
Solution: ：n？5 Factor the denominator, so the series can be represented as ?n(n？3)n！1：n？5n？5111n？5，1?，Then, split up the fraction into . Note that , so . Thus, by ??n？3nnn？3nn？3n！1：n？5comparing this series with the divergent harmonic series, diverges. ?2？3nnn！1
LIMIT COMPARISON TEST:
Suppose that and are series with positive terms. ab??nn
and Kis a non-zero, finite number 0，，：K;；
then either both series converge or both series diverge.
Similar to the Direct Comparison Test, this test is also helpful when confronted with a series that appears
untestable. Most often, this test can be applied to rational functions or with a combination of polynomials
Example: ：2n？3n？7 Determine the convergence or divergence of the series ?42？3n？104nn！1
As n increases, the highest powers of the numerator and the denominator become the significant
22：n11nn？？3724！n4nterms, in this case and . So is similar to , and is a ?4224244nn4n4310nn？？！1n
1convergent p-series. Dividing the series by (i.e., multiply by the reciprocal ) results in 24n
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432432412284nnn？？4n？12n？28nlim1！！. Now . So, since the limit is a positive, 4242n；：4n？3n？1043104nn？？：2：n？3n？71behaves the same way as . That is to say, finite value, the series ??422？3n？104n4nn！1！1n：2n？3n？7converges. ?42？3n？104nn！1
For a given series : a?n
an？1lim if < 1, then the series converges; n；：an
an？1lim if > 1 or goes to ：, then the series diverges; and n；：an
an？1lim if = 1, then the test is inconclusive. n；：an
This test proves particularly helpful for series in which exponentials and/or factorials appear. The
absolute-ratio test below will prove to be a more encompassing test, but the ratio test can certainly be
Example: ：n3 Determine the convergence for the series, . ?n!n！1
This series has both a factorial and an exponential. First, it is important to determine the n？1
n？13a！term, which is, . Now, place it in the ratio test format, such that n？1n？(1)!
a3n？1！！，limlim01 So, thus the series converges. nn；：；：？an1n
aFor a given series, : ?n
nnlimalima： if < 1, then the series converges; if > 1 or , then the series diverges; nn；：；：nn
nlima and if = 1, then the test is inconclusive. n；：n
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This test will assist in determining the convergence or divergence of monomials of degree , especially nnn appears in the series. The only terms that appear in such a series should be various when the term
degrees of alone and no other terms- this test’s effectiveness is limited to a small array of series. n
Example: 2n：2 Determine the convergence of the series, . ?nnn！1
Since the series has both exponentials in the top and bottom, the root test should be applied. So,
2nnn222n！！：limand . Thus, by the root test, this series diverges. nn；：nnn
SERIES WITH BOTH POSITIVE AND NEGATIVE TERMS
The next set of series tests are those that apply to series with both negative and positive terms (known as alternating series). All the previous tests apply only to those tests whose terms are all positive. An alternating series is a series of the form
：n(1)？！？？？？？？aaaaaa , ?12345n1n！
ain other words, a series whose terms alternate between positive and negative values. Note that is a nn(1)？sequence of positive numbers and the determines the sign.
ALTERNATING SERIES TEST:
n(1)？a For a given alternating series, consider (the sequence of positive numbers without the ) n
a ， aIf (i) for all , and nn？1n
(ii) lima！0 nn；：
nThen the series converges. (1)？a?n
This test will immediately give us the convergence of an alternating series, just by looking at its basic behavior. If each term gets progressively smaller, and the terms approach 0, then the series converges. If an alternating series fails one of the two criteria, the test is inconclusive, however a version of the N-th Term Test can often be used to determine the behavior of the series.
n：(1)？ Determine whether the series converges or diverges. ?n1n！
For this series, the terms continually decrease:
111 (for all ) and the terms decrease toward zero: lim n，！0n；：nn？1n
Therefore, because these 2 criteria are met, the series converges.
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ABSOLUTE CONVERGENCE TEST:
If converges, then so does aa??nn
Like the alternating series test, this test checks only for the convergence of an alternating series and says nothing about the divergence.
Example: ：nn(？1)2 Does the series converge? ?2(n？2)n！1
Solution: ：n2Take the absolute value of the series, and get . By applying the nth term test, this ?2(n？2)n！1
series diverges (the series continues to grow without bound), so the absolute convergence test
does not conclude whether the original series converges or diverges.
Before explaining the last two tests involved with alternating series, a few definitions will help
clarify certain types of convergence.
aa A convergent series is said to converge absolutely if converges as well. ??nn
aa A convergent series is said to converge conditionally if diverges. ??nn
For example, the alternating harmonic series converges conditionally because:
nn：：：(1)？(1)1？ converges, while diverges. ！???nnn1n！11nn！！
While any of the above tests may be used when determining whether a given alternating series is absolutely convergent or conditionally convergent, the following two tests are very helpful, especially when determining the Interval of Convergence and Radius of Convergence of a power series.
aGiven the series , (whose terms may be positive or mixed positive and negative) ?n
an？1lim1，aIf , then the series is absolutely convergent. ?nn；：an
an？1lim1，a：If or goes to , then the series is divergent. ?nn；：an
an？1lim1！If , the test is inconclusive. n；：an
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The absolute-ratio test can be used in similar situations to the ratio test. (i.e., factorials and exponentials)
3：nn Test the series for absolute convergence. (1)？?n3n！1
3nnUse the Absolute-Ratio Test with (1)a！？nn3
3a111n？?，n？1 So, limlim1！！，(?nn；：；：an33)：n
3：nn Therefore, the series is absolutely convergent. (1)？?n3n！1
a Given the series , (whose terms may be positive or mixed positive and negative) ?n
nlimaa If < 1, then the series is absolutely convergent. ?nn；：n
nlimaa If > 1 or ：, then the series is divergent. ?nn；：n
nlima If = 1, the test is inconclusive. n；：n
n：23n？?，n Test the convergence of the series: (1)？?(?32n？)：1！n
n23n？?，nUse the Absolute-Root Test with a！？(1)n(?32n？)：
232n？?，n So, limlim1a！！，n(?；：；：nn323n？)：
n：23n？?，n Therefore, the series is absolutely convergent. (1)？?(?32n？)：1！n
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