convergent series where not onlyan but also nan tends to 0


Proposition.  If the terms (http://planetmath.org/Series) an of the convergent seriesMathworldPlanetmathPlanetmath

a1+a2+

are positive and form a monotonically decreasing sequenceMathworldPlanetmath, then

limnnan= 0. (1)

Proof.  Let ε be any positive number.  By the Cauchy criterion for convergence and the positivity of the terms, there is a positive integer m such that

0<am+1++am+p<ε2  (p= 1, 2,).

Since the sequence  a1,a2,  is decreasing, this implies

0<pam+p<ε2  (p= 1, 2,). (2)

Choosing here especially  p:=m,  we get

0<mam+m<ε2,

whence again due to the decrease,

0<mam+p<ε2  (p=m,m+1,). (3)

Adding the inequalitiesMathworldPlanetmath (2) and (3) with the common values  p=m,m+1,  then yields

0<(m+p)am+p<ε  forpm.

This may be written also in the form

0<nan<ε  forn 2m

which means that  limnnan= 0.

Remark.  The assumption of monotonicity in the Proposition is essential.  I.e., without it, one cannot gererally get the limit result (1).  A counterexample would be the series a1+a2+ where  an:=1n  for any perfect squareMathworldPlanetmath n but 0 for other values of n.  Then this series is convergentMathworldPlanetmath (cf. the over-harmonic series), but  nan=1  for each perfect square n; so  nan↛0  as  n.

Title convergent series where not onlyan but also nan tends to 0
Canonical name ConvergentSeriesWhereNotOnlyanButAlsoNanTendsTo0
Date of creation 2013-03-22 19:03:29
Last modified on 2013-03-22 19:03:29
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 14
Author pahio (2872)
Entry type Theorem
Classification msc 40A05
Synonym Olivier’s theorem
Related topic NecessaryConditionOfConvergence
Related topic AGeneralisationOfOlivierCriterion