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In mathematics, the nth-term test for divergence^[1] is a simple test for the divergence of an infinite series:

If $\lim _{n\to \infty }a_{n}\neq 0$ or if the limit does not exist, then $\sum _{n=1}^{\infty }a_{n}$ diverges.

Many authors do not name this test or give it a shorter name.^[2]

When testing if a series converges or diverges, this test is often checked first due to its ease of use.

In the case of p-adic analysis the term test is a necessary and sufficient condition for convergence due to the non-Archimedean ultrametric triangle inequality.

Usage

[edit]

Unlike stronger convergence tests, the term test cannot prove by itself that a series converges. In particular, the converse to the test is not true; instead all one can say is:

If $\lim _{n\to \infty }a_{n}=0,$ then $\sum _{n=1}^{\infty }a_{n}$ may or may not converge. In other words, if $\lim _{n\to \infty }a_{n}=0,$ the test is inconclusive.

The harmonic series is a classic example of a divergent series whose terms approach zero in the limit as $n\rightarrow \infty$ .^[3] The more general class of p-series,

\sum _{n=1}^{\infty }{\frac {1}{n^{p}}},

exemplifies the possible results of the test:

If p ≤ 0, then the nth-term test identifies the series as divergent.
If 0 < p ≤ 1, then the nth-term test is inconclusive, but the series is divergent by the integral test for convergence.
If 1 < p, then the nth-term test is inconclusive, but the series is convergent by the integral test for convergence.

Proofs

[edit]

The test is typically proven in contrapositive form:

If $\sum _{n=1}^{\infty }a_{n}$ converges, then $\lim _{n\to \infty }a_{n}=0.$

Limit manipulation

[edit]

If s_n are the partial sums of the series, then the assumption that the series converges means that

\lim _{n\to \infty }s_{n}=L

for some number L. Then^[4]

\lim _{n\to \infty }a_{n}=\lim _{n\to \infty }(s_{n}-s_{n-1})=\lim _{n\to \infty }s_{n}-\lim _{n\to \infty }s_{n-1}=L-L=0.

Cauchy's criterion

[edit]

Assuming that the series converges implies that it passes Cauchy's convergence test: for every $\varepsilon >0$ there is a number N such that

\left|a_{n+1}+a_{n+2}+\cdots +a_{n+p}\right|<\varepsilon

holds for all n > N and p ≥ 1. Setting p = 1 recovers the claim^[5]

\lim _{n\to \infty }a_{n}=0.

Scope

[edit]

The simplest version of the term test applies to infinite series of real numbers. The above two proofs, by invoking the Cauchy criterion or the linearity of the limit, also work in any other normed vector space^[6] or any additively written abelian group.

Notes

[edit]

^ Kaczor p.336
^ For example, Rudin (p.60) states only the contrapositive form and does not name it. Brabenec (p.156) calls it just the nth term test. Stewart (p.709) calls it the Test for Divergence. Spivak (p.473) calls it the Vanishing Condition.
^ Rudin p.60
^ Brabenec p.156; Stewart p.709
^ Rudin (pp.59-60) uses this proof idea, starting with a different statement of Cauchy criterion.
^ Hansen p.55; Șuhubi p.375

References

[edit]

Brabenec, Robert (2005). Resources for the study of real analysis. MAA. ISBN 0883857375.
Hansen, Vagn Lundsgaard (2006). Functional Analysis: Entering Hilbert Space. World Scientific. ISBN 9812565639.
Kaczor, Wiesława and Maria Nowak (2003). Problems in Mathematical Analysis. American Mathematical Society. ISBN 0821820508.
Rudin, Walter (1976) [1953]. Principles of mathematical analysis (3e ed.). McGraw-Hill. ISBN 0-07-054235-X.
Spivak, Michael (2008). Calculus (4th ed.). Houston, TX: Publish or Perish. ISBN 978-0-914098-91-1.
Stewart, James (1999). Calculus: Early transcendentals (4e ed.). Brooks/Cole. ISBN 0-534-36298-2.
Șuhubi, Erdoğan S. (2003). Functional Analysis. Springer. ISBN 1402016166.

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The nth-term test, also known as the divergence test or term test for divergence, is a fundamental criterion in calculus for assessing the convergence or divergence of an infinite series

\sum_{n=1}^\infty a_n

. It states that if

\lim_{n \to \infty} a_n

does not exist or is not equal to zero, then the series diverges.^[1]^[2] This test is often the first applied when evaluating series, as it provides a quick way to rule out convergence without further analysis.^[3] Conversely, if

\lim_{n \to \infty} a_n = [0](/page/zero)

, the test is inconclusive, meaning the series may converge or diverge; for example, the harmonic series

\sum_{n=1}^\infty \frac{1}{n}

has terms approaching zero but diverges.^[1]^[2] The necessity of this condition arises from the definition of convergence: if the partial sums

s_n = \sum_{k=1}^n a_k

converge to a finite limit

s

, then

a_n = s_n - s_{n-1}

must approach zero as

n \to \infty

, since both

s_n

and

s_{n-1}

approach

s

.^[1] Thus, failure of the limit to be zero implies the partial sums cannot converge.^[2] The test applies to any series of real or complex numbers without restrictions on the signs of the terms, making it versatile for initial screening before more advanced tests like the ratio, root, integral, or comparison tests.^[3] For instance, consider

\sum_{n=1}^\infty \frac{n}{n+1}

; here,

\lim_{n \to \infty} \frac{n}{n+1} = 1 \neq 0

, so the series diverges by the nth-term test.^[1] Despite its simplicity, the test highlights a key insight into series behavior: terms must vanish in the limit for any chance of convergence.^[2]

Statement

Formal Statement

The nth-term test, also known as the divergence test or term test for divergence, is a criterion in mathematical analysis for assessing the convergence or divergence of an infinite series

\sum_{n=1}^{\infty} a_n

, where

\{a_n\}

is a sequence of real or complex numbers.^[1] For the series to converge, the sequence of partial sums

s_k = \sum_{n=1}^{k} a_n

must converge to a finite limit as

k \to \infty

, assuming the reader is familiar with the basic concepts of sequences and their limits.^[4] A necessary condition for this convergence is that the terms of the series approach zero:

\lim_{n \to \infty} a_n = 0

.^[5] The formal statement of the nth-term test is the contrapositive of this condition: if

\lim_{n \to \infty} a_n \neq 0

or if the limit does not exist, then the series

\sum_{n=1}^{\infty} a_n

diverges.^[6] Equivalently, if

\lim_{n \to \infty} a_n = L

where

L \neq 0

, or if the limit fails to exist, the series cannot converge.^[7]

Interpretation

The nth-term test, also known as the divergence test, provides an intuitive foundation for understanding series convergence by highlighting that if the terms

a_n

of a series

\sum a_n

do not approach zero as

n

approaches infinity, the partial sums cannot settle to a finite limit. This occurs because non-negligible additions continue indefinitely, causing the sum to grow without bound or oscillate indefinitely, preventing stabilization.^[8]^[1] In mathematical analysis, the test functions as a necessary condition for convergence, serving as a rapid preliminary screening tool before applying more sophisticated methods like the ratio or integral tests. A failure of the condition—where

\lim_{n \to \infty} a_n \neq 0

—immediately establishes divergence, streamlining the investigative process, whereas success (the limit equaling zero) merely permits further examination without confirming convergence.^[8]^[1] Originating in 19th-century calculus, the test emerged from efforts to establish rigorous criteria for infinite series, with Augustin-Louis Cauchy articulating the necessary condition in his 1821 Cours d'Analyse, stating that for convergence, "the general term

u_n

decreases indefinitely" as

n

increases.^[9] Central to the test is the distinction between necessary and sufficient conditions: while the limit of

a_n

to zero is required for convergence, it alone does not ensure it, as demonstrated by divergent series like the harmonic series where terms vanish but the sum diverges. This nuance underscores the test's utility as a gatekeeper rather than a definitive arbiter in series evaluation.^[8]

Application

Procedure

To apply the nth-term test, also known as the divergence test, begin by recalling its formal statement: for an infinite series

\sum a_n

, if

\lim_{n \to \infty} a_n \neq 0

or the limit does not exist, then the series diverges.^[1] The procedure involves the following steps:

Identify the general term $a_n$ of the series, which is the expression for the $n$ th term in the summation.^[10]
Compute the limit $\lim_{n \to \infty} a_n$ using appropriate techniques.^[11]
Interpret the result: if the limit equals 0, the test is inconclusive regarding convergence or divergence; if the limit is not 0 or fails to exist (including oscillating limits), conclude that the series diverges.^[12]

For computing the limit in step 2, employ standard techniques such as algebraic simplification for rational functions, L'Hôpital's rule for indeterminate forms like

\frac{\infty}{\infty}

\frac{0}{0}

, or direct substitution when applicable.^[13] A common pitfall is mistakenly interpreting a limit of 0 as proof of convergence; the test only detects divergence and provides no information when the limit is 0, requiring other tests for further analysis.^[1] The test applies to series with positive, negative, or alternating terms, provided the limit of the general term is evaluated correctly.^[14]

Examples

To illustrate the application of the nth-term test, consider the series

\sum_{n=1}^\infty 1

. Here, the general term is

a_n = 1

, and

\lim_{n \to \infty} a_n = 1 \neq 0

. Thus, the series diverges by the nth-term test.^[15] Another example of divergence is the series

\sum_{n=1}^\infty \frac{n}{n+1}

. The general term is

a_n = \frac{n}{n+1}

, and

\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{n}{n(1 + 1/n)} = \lim_{n \to \infty} \frac{1}{1 + 1/n} = 1 \neq 0.

Therefore, the series diverges by the nth-term test.^[1] An example where the limit does not exist is the series

\sum_{n=1}^\infty (-1)^n

. The general term is

a_n = (-1)^n

, and

\lim_{n \to \infty} (-1)^n

does not exist, as it oscillates between -1 and 1. Thus, the series diverges by the nth-term test.^[11] The nth-term test can be inconclusive when the limit is zero. For the harmonic series

\sum_{n=1}^\infty \frac{1}{n}

, the general term is

a_n = \frac{1}{n}

, and

\lim_{n \to \infty} a_n = 0

. The test provides no conclusion about convergence or divergence (though the series diverges by the integral test).^[1] Similarly, for the series

\sum_{n=1}^\infty \frac{1}{n^2}

a_n = \frac{1}{n^2}

and

\lim_{n \to \infty} a_n = 0

. The nth-term test is inconclusive (though the series converges by the p-series test with p=2>1).^[1] An example involving an alternating series is

\sum_{n=1}^\infty \frac{(-1)^{n+1}}{\sqrt{n}}

∑n=1∞n

(−1)n+1. The general term is $a_n = \frac{(-1)^{n+1}}{\sqrt{n}}$

(−1)n+1. Since $|a_n| = \frac{1}{\sqrt{n}} \to 0$

1→0 as $n \to \infty$ , it follows that $\lim_{n \to \infty} a_n = 0$ by the squeeze theorem. The nth-term test is inconclusive (though the series converges by the alternating series test).^[16] For a series with oscillatory terms, consider $\sum_{n=1}^\infty \frac{\sin n}{n}$ . The general term is $a_n = \frac{\sin n}{n}$ . Since $|\sin n| \leq 1$ , $|a_n| \leq \frac{1}{n} \to 0$ , so $\lim_{n \to \infty} a_n = 0$ by the squeeze theorem. The nth-term test is inconclusive (though the series converges by the Dirichlet test).^[17]

Proofs

Limit-Based Proof

The limit-based proof of the nth-term test demonstrates that if the infinite series

\sum_{n=1}^\infty a_n

of real numbers converges, then

\lim_{n \to \infty} a_n = 0

. This establishes a necessary condition for convergence, with the contrapositive forming the test itself: if

\lim_{n \to \infty} a_n \neq 0

or the limit fails to exist, the series diverges.^[18] Assume the series converges to a finite sum

S \in \mathbb{R}

. The partial sums are defined as

s_n = \sum_{k=1}^n a_k

for

n \geq 1

, with

s_0 = 0

. Convergence of the series implies that the sequence

\{s_n\}_{n=1}^\infty

converges to

S

, so

\lim_{n \to \infty} s_n = S

. For

n \geq 2

a_n = s_n - s_{n-1}.

Taking the limit as

n \to \infty

\lim_{n \to \infty} a_n = \lim_{n \to \infty} (s_n - s_{n-1}) = S - S = 0,

since the limits of

s_n

and

s_{n-1}

both equal

S

. This algebraic manipulation of limits confirms the necessity of

\lim_{n \to \infty} a_n = 0

for convergence.^[18] To see why the contrapositive holds, suppose

\lim_{n \to \infty} a_n \neq 0

or the limit does not exist. In either case,

\lim_{n \to \infty} a_n \neq 0

, meaning the sequence

\{a_n\}

does not converge to 0. Equivalently, the successive differences

s_{n+1} - s_n = a_{n+1}

do not converge to 0. However, if

\{s_n\}

converged to some limit, then

\lim_{n \to \infty} (s_{n+1} - s_n) = 0

would hold, as the difference of two sequences converging to the same limit converges to 0. This contradiction implies that

\{s_n\}

cannot converge, so the series diverges.^[18] If the limit exists and equals

L \neq 0

, then for any

\varepsilon = |L|/2 > 0

, there exists

N

such that for all

n > N

|a_n - L| < \varepsilon

, so

|a_n| > |L| - \varepsilon = |L|/2 > 0

. Thus, the partial sums

s_n

change by amounts bounded away from zero for all sufficiently large

n

, preventing

\{s_n\}

from converging, as the terms would not settle into a Cauchy-like stability required for limit existence.

Cauchy Criterion Proof

The Cauchy criterion for the convergence of an infinite series

\sum_{n=1}^\infty a_n

states that the series converges if and only if for every

\epsilon > 0

, there exists a positive integer

N

such that for all integers

m > n \geq N

|s_m - s_n| < \epsilon,

where

s_k = \sum_{i=1}^k a_i

denotes the

k

th partial sum of the series.^[19] To establish the nth-term test via this criterion, consider its contrapositive form: if

\sum a_n

converges, then

\lim_{n \to \infty} a_n = 0

. Assume the series converges. By the Cauchy criterion, for every

\epsilon > 0

, there exists

N

such that

|s_m - s_n| < \epsilon

whenever

m > n \geq N

. Specializing to

m = n+1

yields

|s_{n+1} - s_n| = |a_{n+1}| < \epsilon

for all

n \geq N

, which implies

\lim_{n \to \infty} a_n = 0

.^[19] For the direct implication when the limit exists but is nonzero, suppose

\lim_{n \to \infty} a_n = L \neq 0

. Let

\epsilon = |L|/2 > 0

. By the definition of the limit, there exists

N

such that for all

n \geq N

|a_n - L| < \epsilon

, so

|a_n| \geq |L| - |a_n - L| > |L| - \epsilon = |L|/2 = \epsilon

. Thus, for all

n \geq N

|a_{n+1}| > \epsilon

. This implies

|s_{n+1} - s_n| = |a_{n+1}| > \epsilon

for all

n \geq N

, violating the Cauchy criterion since no such

N

can satisfy the condition for this fixed

\epsilon > 0

. Therefore, the series diverges.^[19] If

\lim_{n \to \infty} a_n

does not exist (equivalently,

a_n

fails to approach 0), then

a_n \not\to 0

. By the negation of the limit definition, there exists

\epsilon > 0

such that for every positive integer

N

, there is some

n > N

with

|a_n| \geq \epsilon

; in particular, there are infinitely many such

n

, yielding a subsequence

\{a_{n_k}\}

where

|a_{n_k}| \geq \epsilon

for all

k

. For each such

n_k \geq N

(choosing

N

large),

|s_{n_k} - s_{n_k - 1}| = |a_{n_k}| \geq \epsilon

, which again prevents the partial sums from satisfying the Cauchy condition, as the differences remain bounded away from zero infinitely often. Hence, the series diverges.^[19] This proof underscores the nth-term test's reliance on the

\epsilon

N

definition of limits, linking the behavior of individual terms directly to the completeness properties of the real numbers embedded in the Cauchy criterion.^[19] The limit-based proof serves as a simpler precursor by focusing on partial sum increments without invoking the full criterion.^[19]

Scope and Limitations

Inconclusive Cases

The nth-term test, also known as the divergence test, fails to provide a definitive conclusion regarding the convergence of an infinite series

\sum a_n

when

\lim_{n \to \infty} a_n = 0

. In such cases, the series may either converge or diverge, necessitating the application of additional convergence tests. For instance, the p-series

\sum_{n=1}^\infty \frac{1}{n^2}

converges despite the terms approaching zero, while the harmonic series

\sum_{n=1}^\infty \frac{1}{n}

diverges even though its terms also tend to zero.^[20]^[1] This inconclusiveness arises because the condition

\lim_{n \to \infty} a_n = 0

is a necessary requirement for convergence but not a sufficient one. While the partial sums of a convergent series must approach a finite limit, and thus the individual terms must diminish to zero, the converse does not hold: terms approaching zero merely permit the possibility of convergence without ensuring that the partial sums remain bounded.^[20]^[1] A common misconception is that

\lim_{n \to \infty} a_n = 0

implies the series converges, but counterexamples like the harmonic series demonstrate this error. More generally, for p-series

\sum_{n=1}^\infty \frac{1}{n^p}

, the limit of the terms is zero for all

p > 0

, yet the series converges only when

p > 1

and diverges otherwise.^[20]^[1] When the nth-term test yields an inconclusive result, mathematicians typically proceed to more advanced tests, such as the integral test, comparison test, or ratio test, to ascertain convergence or divergence.^[20]^[1]

Relation to Other Tests

The nth-term test, also known as the divergence test, serves as a foundational preliminary check in the evaluation of infinite series convergence, often applied as the initial step in systematic testing algorithms outlined in standard calculus resources. If the limit of the general term

a_n

n

approaches infinity is not zero, the series diverges immediately, allowing practitioners to rule out convergence without further analysis. This positions the test as a gatekeeper, filtering out obviously divergent series before proceeding to more involved methods, a strategy emphasized in convergence testing protocols where it precedes examinations of term ratios, roots, or integrals.^[21]^[22] In comparison to the ratio test, the nth-term test is simpler and quicker for initial divergence detection, particularly when terms do not exhibit clear exponential growth or decay patterns suited to ratio analysis. The ratio test, which examines

\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|

limn→∞

anan+1

, provides decisive results for series involving factorials or exponentials—converging if the limit is less than 1 and diverging if greater than 1—but becomes inconclusive when the limit equals 1, at which point the nth-term test can still identify divergence if $\lim_{n \to \infty} a_n \neq 0$ . Thus, the nth-term test complements the ratio test by handling cases where the latter fails to resolve, though it cannot confirm convergence in any scenario.^[23] The root test bears a structural similarity to the nth-term test, as both involve limits of transformed terms, but the root test computes $\lim_{n \to \infty} \sqrt{|a_n|}$

, making it more effective for series with polynomial-like powers where the nth-term limit might approach zero slowly. For such terms, the nth-term test remains easier to apply directly without root extraction, serving as a preliminary filter before the root test's absolute convergence assessment (convergent if the limit is less than 1, divergent if greater than 1, inconclusive if equal to 1). Like the ratio test, the root test's inconclusiveness often reverts to the nth-term test for divergence confirmation.^[23] The integral test links closely to the nth-term test, as both are essential for positive-term series and the integral test presupposes that $\lim_{n \to \infty} a_n = 0$ (a condition verified by the nth-term test passing without divergence). When the nth-term test is inconclusive, the integral test can resolve convergence by comparing the series to the improper integral of a corresponding decreasing function, proving absolute convergence or divergence for cases like the harmonic series where terms vanish but partial sums grow. This pairing highlights the nth-term test's role in enabling subsequent checks for conditional convergence scenarios unresolved by simpler limits.^[18]

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