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nth-term test

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Test for the divergence of an infinite series
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In mathematics, the nth-term test for divergence is a simple test for the divergence of an infinite series:

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

Many authors do not name this test or give it a shorter name.

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

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 a n = 0 , {\displaystyle \lim _{n\to \infty }a_{n}=0,} then n = 1 a n {\displaystyle \sum _{n=1}^{\infty }a_{n}} may or may not converge. In other words, if lim n a n = 0 , {\displaystyle \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 {\displaystyle n\rightarrow \infty } . The more general class of p-series,

n = 1 1 n p , {\displaystyle \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

The test is typically proven in contrapositive form:

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

Limit manipulation

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

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

for some number L. Then

lim n a n = lim n ( s n s n 1 ) = lim n s n lim n s n 1 = L L = 0. {\displaystyle \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

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

| a n + 1 + a n + 2 + + a n + p | < ε {\displaystyle \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

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

Scope

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 or any additively written abelian group.

Notes

  1. Kaczor p.336
  2. 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.
  3. Rudin p.60
  4. Brabenec p.156; Stewart p.709
  5. Rudin (pp.59-60) uses this proof idea, starting with a different statement of Cauchy criterion.
  6. Hansen p.55; Șuhubi p.375

References

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