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# On Certain Conditionally Convergent Series

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ON CERTAIN CONDITIONALLY CONVERGENT SERIES

LUCA GOLDONI

Abstract. In this paper we investigate the problem of the con-vergence of a very special kind of non absolutely convergent series which can not be solved by means of traditional tests as Dirichlet test.

1. Introduction We investigate the behavior of the series

+∞

X

n=0

(−1)n( mod p)an.

where p is an odd prime number and an is not negative for each n. We

could call ‘almost alternating series’ because the sequence of the signs is of the kind

+ − · · · − + | {z } p−terms + − · · · − + | {z } p−terms . . .

We observe that the Dirichlet’s test is not applicable even in the case of further assumptions on an because the partial sums of the sequence

bn= (−1)

n( mod p)

are not bounded. Indeed, if we indicate with σn the

sequence of this partial sums we have that σpk = k + 1.

2. The theorem Lemma 1. Let be +∞ X n=0 (−1)n( mod p)an. where

(a): an≥ 0 for each n ∈ N.

(b): +∞ P n=0 an = +∞. Date: April 7, 2013.

2000 Mathematics Subject Classification. 40A05, 30B50.

Key words and phrases. Infinite series, Conditionally convergence tests, Elemen-tary proof.

Dipartimento di Matematica. Universit`a di Trento. 1

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(c): lim

n→+∞an = 0.

and let be (sn)n the sequence of the partial sums. If there exists the

(1) lim

k→+∞spk = s ∈ R.

then

lim

k→+∞spk+1= limk→+∞spk+2· · · limk→+∞sp(k+1)−1 = s

so that the given series converges. Proof. Since (1) holds, it follows that

∀ε > 0 ∃k1(ε) : ∀k > k1(ε) ⇒ s − ε 2 < spk < s+ ε 2. Let be 1 ≤ h ≤ p − 1 then |spk+h− spk| = |apk+1+ · · · apk+h| 6 |apk+1| + · · · |apk+h| .

Since hypothesis (c) holds, it follows that

∀ε > 0 ∃n (ε) : ∀n > n (ε) ⇒ |an| 6

ε 2h. Let be k such that pk + 1 > n (ε) i.e.

k > n(ε) − 1 p = k2(ε) . then |apk+1| + · · · |apk+h| 6 ε(h − 1) 2h < ε 2. thus |spk+h− spk| < ε 2. If k > maxk1(ε) , k2(ε) then    s− ε 2 < spk < s+ ε 2 spk− ε 2 < spk+h < spk+ ε 2 so that s − ε < spk+h < s+ ε. Hence lim k→∞spk+h = s.

Since it holds for each 1 ≤ h ≤ p the thesis follows.  Lemma 2. If

+∞

X

n=0

(−1)n( mod p)an

satisfies the hypothesis of Lemma 1 and if (d): dk = apk+p+ p−1 P h=1 (−1)hapk+h >0 for each k ∈ N. (e): +∞ P k=0 dk <+∞.

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then ∃ lim k→∞spk = s < +∞. Proof. Since spk+p = spk+(−apk+1+ apk+2− apk+3+ · · · − apk+p−2+ apk+p−1+ apk+p) we have that spk = s0+ k−1 X j=0 dj.

from hypothesis (d) it follows that the sequence spk in not decreasing

so it has limit. Moreover, since

k−1 X h=0 dh 6 +∞ X h=0 dh <+∞

the limit belongs to R. 

So we have that Theorem 1. If +∞ X n=0 (−1)n( mod p)an. where

(a): an≥ 0 for each n ∈ N.

(b): +∞ P n=0 an = +∞. (c): lim n→+∞an = 0. (d): dk = apk+p+ p−1 P h=1 (−1)hapk+h >0 for each k ∈ N. (e): +∞ P k=0 dk <+∞.

(f): p is an odd prime number. then the given series is simply convergent.

In particular we have the following

Corollary 1. If there exist A > 0 and δ > 0 so that 0 6 dk 6

A kδ.

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References

[1] T.J. Bromwich “An introduction to the theory of infinite series” Macmillan; 2d ed. rev. 1947.

[2] G.H. Hardy “A course in Pure Mathematics” Cambridge University Press, 2004. [3] K. Knopp “Theory and Application of Infinite Series” Dover 1990.

[4] C.J. Tranter “Techniques of Mathematical analysis” Hodder and Stoughton,1976.

Universit`a di Trento, Dipartimento di Matematica, v. Sommarive 14, 56100 Trento, Italy

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