A NOTE ON SEMI- CONVERGENT SERIES

Transcription

1 International J. of Math. Sci. & Engg. Appls. (IJMSEA) ISSN , Vol. 10 No. II (August, 016), pp A NOTE ON SEMI- CONVERGENT SERIES SREEJA S. NAIR 1 AND DR. V. MADHUKAR MALLAYYA 1 Assistant Professor of Mathematics, Govt. Arts College, Thiruvananthapuram, Kerala, India Former Professor and Head, Department of Mathematics, Mar Ivanios College, Thiruvananthapuram, India Abstract In this paper we shall give some properties of a semi-convergent series. We also distinguish a semi- convergent and an absolutely convergent series. 1. Introduction The study of infinite series includes two important class of series namely, absolute convergent series and conditionally convergent series. A conditionally convergent series is also named as a semi- convergent series. The study of a semi-convergent series plays a vital role in mathematical analysis. Here we shall prove the re-arrangement theorems for a semi-convergent series. Definition 1 : An infinite series is a sum of infinite number of terms a 1 +a +a 3 +. An infinite series is usually denoted by a n. If the sum of infinite number of terms is a finite number, say S, then we say that the series a n is convergent and S is called the sum of the series. c http: // 49

2 50 SREEJA S. NAIR 1 & DR. V. MADHUKAR MALLAYYA We denote it as a n = S. If a n = ±, we say that the series is divergent. Definition : An alternating series is of the form ( 1) n 1 a n where a n > 0. Definition 3 : The sequence of partial sums of the series a n is the sum to n terms of the series. It is denoted by S n. That is S n = a k. Theorem : A series a n is convergent if and only if its sequence of partial sums S n is convergent. Definition 4 : A series of positive terms a n is convergent. a n is called an absolutely convergent series, if the series If a series a n is absolutely convergent, then the series itself is also convergent. Example : 1. The series. The series ( 1) n 1 1 ( 1) n 1 1 n(n+1). n p where p > 1 Definition 5 : A series a n is called a semi-convergent (conditionally conver- gent) series, if the series a n is convergent, but the series of positive terms a n is not convergent. Example : The series ( 1) n 1 1 n p where 0 p 1. Definition 6 : Let a n be a series of positive and negative terms. Define a n, if a n > 0 a + n = 0, if a n 0 a n, if a n < 0 = 0, if a n 0

3 A NOTE ON SEMI- CONVERGENT SERIES 51 Then the series a + n is called the series of positive terms and series is called the series of negative terms. It follows that a n = a + n a n = a + n + That is a + n = an +an, = an an. Note : For the series a n, the series a + n and the series is called the series of negative terms. Theorem : For an absolutely convergent series, the series convergent. Proof : Let convergent. Let a n = S and a n = T. a n be an absolutely convergent series. Then Let S n and T n be the sequence of partial sums of is called the series of positive terms a + n + and a n and a n and a n respectively. Let P n and Q n denote the sequence of partial sums of Then we have S n = P n Q n and T n = P n + Q n. are a n are a + n and respectively. That is P n = Tn+Sn and Q n = Tn Sn. Now lim P T n = lim n+s n 1 n n = lim n (T n + S n ) = T +S, a finite number. Also lim Q T n = lim n S n 1 n n = lim n (T n S n ) = T S, a finite number. Hence the sequence of partial sums of a + n and are convergent. Thus the two series a + n and are convergent. Theorem : For a semi-convergent series a n, the two series divergent. Proof : Let is not convergent. Let a n = S, a n =. a + n and a n be a semi-convergent series. Then a n is convergent, but a n Let S n and T n be the sequence of partial sums of Let lim n S n = S. We have lim n T n =. a n and a n respectively. are

4 5 SREEJA S. NAIR 1 & DR. V. MADHUKAR MALLAYYA Let P n and Q n denote the sequence of partial sums of Then we have S n = P n Q n and T n = P n + Q n. That is P n = Tn+Sn and Q n = Tn Sn. Now Also lim P T n + S n n = lim n n lim Q T n S n n = lim n n Hence the sequence of partial sums of a + n and respectively. 1 = lim n (T n + S n ) = + S =. 1 = lim n (T n S n ) = S =. a + n and are divergent. Thus the two series a + n and a n are divergent. Lemma : For a semi-convergent series a n, the two series a + n and infinitely many positive terms. Proof : Assume that has only a finite number of positive terms. Then a n = = (a + n + ) (a + n ) + a n + We have a n is convergent and by our assumption, have is convergent. It follows that a n is convergent which is a contradiction. Hence our assumption is wrong. Thus must have infinite number of positive terms. Similarly assume that a + n has only finite number of terms. Since a n = (a + n + ) ( a + n ) + a + n = ( 1) a n + a + n.

5 A NOTE ON SEMI- CONVERGENT SERIES 53 Since a n is convergent, we have ( 1) a n is also convergent. By our assumption a + n is convergent. It follows that a n is convergent, which is a contradiction. Hence a + n must have infinite number of positive terms. Lemma : The series a + n and are unbounded below. Proof : We have is a series of non-positive terms and so it is monotone decreasing. If the series is bounded below, then by monotone convergence theorem, the series will converge which is a contradiction. Thus the series is unbounded below. Similarly we can prove that a + n + is unbounded below. Theorem : A semi-convergent series a n, the terms can be rearranged so that the re-arranged series diverges to. Proof : In this proof we shall denote u n = a + n and v n =. Then clearly u n and v n are non-negative real numbers. Now we make a re-arrangement of a n as follows. (u 1 + u + + u m1 )v 1 + (u m1+1 + u m u m ) v + (u m+1 + u m+ + + u m3 )v 3 + where a group of positive terms is followed by a single negative term. We denote the re-arranged series as sums of b n. b n and let S n denote the sequence of partial Since u n is divergent, its sequence of partial sums is unbounded. Let us choose m 1 so large that u 1 + u + + u m1 > 1 + v 1. Now choose m > m 1 such that u 1 + u + + u m1 + u m1+1 + u m u m > + v 1 + v.

6 54 SREEJA S. NAIR 1 & DR. V. MADHUKAR MALLAYYA In general, choose m n > m n 1 such that u 1 +u + +u m1 +u m1+1 +u m1+ + +u m + +u mn > n+v 1 +v + +v n for n N. Since each of the partial sums S m1+1, S m+, of the rearranged series, whose last term is a negative term v n, is greater than n, these partial sums are unbounded above and hence the series b n diverges to. Theorem : The terms of a semi-convergent series a n can be rearranged so that the rearranged series diverges to. Proof : The non negative real numbers u n and v n are defined as in the proof of previous theorem. Here we make another re-arrangement of as follows. ( v 1 v v m1 ) + u 1 + (v m1+1 v m1+ v m ) +u + ( vm +1 v m+ v m3 ) + u 3 + where a group of negative terms is followed by a single positive term. We denote the re-arranged series as sum of b n. b n and let S n denote the sequence of partial Since v n is divergent, its sequence of partial sums is unbounded. Let us choose m 1 so large that v 1 v v m1 < 1 u 1. Now choose m > m 1 such that v 1 v v m1 v m1+1 v m1+ v m < u 1 u. In general, choose m n > m n 1 such that v 1 v v m1 v m1+1 v m1+ v m v mn < n u 1 u u n for n N. Since each of the partial sums S m1+1, S m+, of the rearranged series b n, whose last term is a positive term u n is less than n, these partial sums are unbounded below. Hence the re-arranged series b n diverges to.

7 A NOTE ON SEMI- CONVERGENT SERIES 55 Theorem (Riemanns re-arrangement theorem) : Let a n be a semi-convergent series and let r R. Then there exists a rearrangement of the terms of to r. a n that converges Proof : The non negative real numbers u n and v n are defined as in the proof of previous theorem. Here we make another re-arrangement of a n as follows. Choose first m 1 positive terms such that their sum exceeds r. i.e. (u 1 +u + +u m1 ) > r. Then add first m negative terms such that the sum is less than r. i.e. (u 1 + u + + u m1 ) + ( v m1+1 v m1+ v m ) < r.. Again add next m 3 positive terms so that the sum exceeds r. i.e. (u 1 +u + +u m1 )+( v m1+1 v m1+ v m )+(u m+1 +u m+ + +u m3 ) > r. Next we add m 4 negative terms so that the sum is less than r. i.e. (u 1 + u + + u m1 ) + ( v m1+1 v m1+ v m ) +(u m+1 + u m+ + + u m3 ) + ( v m1+1 v m1+ v m4 ) < r. We continue the above process, where a group of positive terms is followed by another group of negative terms. We denote the re-arranged series as sum of b n. b n and let S n denote the sequence of partial We can see that the partial sums S m1+m+m3 + of the rearranged seris b n brackets the real number r. We have u n 0 and v n 0 as n. Hence for given ɛ > 0, there exists a natural number N such that S n r < ɛ for n N. ThusS n r as n. Hence the rearranged series b n converges to the real number r. References [1] Knopp Konra, Theoryand Application of Infinite Series, Blackie and Son Limited (London and Glasgow).

8 56 SREEJA S. NAIR 1 & DR. V. MADHUKAR MALLAYYA [] Hardy G. H., A Course of Pure Mathematics, Cambridge at the University Press, (Tenth edition), (1963). [3] Knopp K., Infinite Sequences and Series, Dover, (1956). [4] Apostol Tom M., Mathematical Analysis, Narosa Publishing House, (Second Edition), ISBN [5] Galanor Stewart, Reimann s rearrangement theorem, mathematics teacher, 80(8), (Novermber 1987),