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n t h - t e r m t e s t
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T a l k
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L a n g u a g e
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( R e d i r e c t e d f r o m T e r m t e s t )
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.[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 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
→
∞
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 limit to zero.[3 ] 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 term test identifies the series as divergent.
If 0 < p ≤ 1, then the term test is inconclusive, but the series is divergent by the integral test for convergence .
If 1 < p , then the term test is inconclusive, but the series is convergent, again by the integral test for convergence.
Proofs
edit
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
edit
If s n 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[4 ]
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
edit
The assumption that the series converges means 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 definition of the statement[5 ]
lim
n
→
∞
a
n
=
0.
{\displaystyle \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
^ 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 .
^ 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
R e t r i e v e d f r o m " https://en.wikipedia.org/w/index.php?title=Nth-term_test&oldid=1111045388 "
L a s t e d i t e d o n 1 9 S e p t e m b e r 2 0 2 2 , a t 0 1 : 1 6
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