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.}
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