The Error in an Alternating Series

Transcription

1 The Error in an Alternating Series arxiv: v1 [math.ca] 27 Nov Introduction Mathematicians have studied the alternating series ( 1) n 1 a n = a 1 a 2 +a 3 a 4 + (1) n=1 since the dawn of analysis. In January of 1714, the great Gottfried Leibniz wrote a letter to Johann Bernoulli in which he explicitly stated his famous criterion for the convergence of (1) as well as the corresponding error estimate [3, p. 33]. Theorem 1. If a n 0 for n = 0,1,... and if the sequence (a n ) decreases monotonically to zero, then the series (1) converges. Let L be its sum. Moreover, let S n := a 1 a 2 +a 3 a 4 + +( 1) n 1 a n, (2) R n := L S n, denote its n th partial sum and remainder, respectively. Then and R n has the sign ( 1) n. R n a n+1, (3) It is amazing that the error estimate (8) remained virtually unimproved for almost 250 years! Then, in 1962, Philip Calabrese, a sophomore (!) at the University of Illinois, proved the following significant refinement [2]. Theorem 2. Let a n := a n a n+1. If, additionally, the sequence ( a n ) converges monotonically to zero, then a n+1 < R n < a n 2 2. (4) Calabrese s refined error estimate allowed him to prove the very pretty result that the first partial sum of Leibniz s series ln2 = which approximates ln2 with four decimal places of accuracy is S Seventeen years later, in 1979, Richard Johnsonbaugh published the following refinement [4] of Calabrese s result. 1

2 Theorem 3. Let r a n := r 1 a n r 1 a n+1 for r = 1,2,3,...,k. If all the sequences ( r a n ) for r = 1,2,3,...,k decrease monotonically to zero, then a n+1 + a n k a n+1 < R k+1 n < a { } n 2 an + + k a n. (5) k+1 For example, Johnsonbaugh uses the sharper upper bound estimate (5) with k = 2 to prove the remarkably precise result that the first partial sum, S n, of the series π 4 = that approximates π with four decimal places of accuracy is S Here the precision means that although 4999 terms of the series do not give four decimal places of accuracy, subtracting 1 just one more term,, does give it. This underlines the slowness of the convergence of 9999 Leibniz s formula for π. However, we will show below how to spectacularly accelerate the convergence of this same series. The lower bound for R n in Johnsonbaugh s theorem (and, in particular, in Calabrese s) had already been found by L.D. Ames in 1901 [1], but the upper bound is new. Finally, in1985,robertm.young[6]usedcantor stheoremonnestedintervalstogivean elegant new proof of Calabrese s refinement which makes the error estimates almost intuitive. It does not seem to have been noticed that Young s method can be adapted to give a new and transparent proof of Johnsonbaugh s refinement. We elaborate such a proof in this note. 2 Proof of Johnsonbaugh s Theorem Young points out that the crux of Leibniz s original proof of his theorem is this: if the numbers S 1,S 2,... satisfy the relation S n+1 S n = ( 1) n a n+1, (6) then, since the sequence a n+1 decreases to zero, the sequence of closed intervals [S 2,S 1 ], [S 4,S 3 ], [S 6,S 5 ], is nested. Therefore, Cantor s theorem on nested intervals shows that there is a number L common to all these intervals, and that L = lim n S n, whereby Leibniz s error estimate is immediate. To prove Calabrese s theorem, Young defines T n := S n +( 1) n an+1 2. (7) 2

3 Then T n+1 T n = ( 1) n a n+1, which has the same form as (6). We note that [S 2r,S 2r 1 ] [T 2r,T 2r 1 ] for r = 1,2,... Since a n+1 decreases to zero, Cantor s theorem on nested intervals shows that there is a number, L 1 = L, common to all the intervals [T 2,T 1 ], [T 4,T 3 ], [T 6,T 5 ], and that Now and L = lim n T n. T n S n = ( 1) n an+1 2 Therefore, if n is even, T n 1 S n = S n 1 +( 1) n 1 an 2 S n = ( 1) n 1 an 2 ( 1)n 1 a n = ( 1) n a n. L S n T n S n = a n+1 2, L S n T n 1 S n = a n 2, which are Calabrese s inequalities. A similar argument holds if n is odd. To prove Johnsonbaugh s theorem, we have to suitably generalize (7). Let T n := S n +( 1) n an+1 2, T n := T n +( 1) n a n+1 2 2, T n := T n +( 1) n 2 a n+1 2 3,.. T (k) n := T (k 1) n +( 1) n k a n+1 2 k+1. Now we apply the reasoning we had already applied to S n and T n. We note that [S 2r,S 2r 1 ] [T 2r,T 2r 1 ] [T 2r,T 2r 1 ] [T(k) 2r,T 2r 1] (k) 3

4 for r = 1,2,... Since k a n+1 decreases to zero, Cantor s theorem on nested intervals shows that there is a number, L k = L, common to all the intervals and that [T (k) 2,T (k) 1 ], [T (k) 4,T (k) 3 ], [T (k) 6,T (k) 5 ], L = lim T n (k). n Substituting recursively in the definitions of T n (r) for r = 1,2,..., some simple algebra leads us to the equations Therefore, if n is even, ( T n (k) S n = ( 1) n an+1 + a n k a n ) k+1. ( T (k) n 1 S n = ( 1) n an 2 a n k a n k+1 ), L S n T n (k) S n = a n+1 + a n k a n+1, k+1 L S n T (k) n 1 S n = a n 2 a n k a n 2 2 2, k+1 which are Johnsonbaugh s inequalities. A similar argument holds for n odd. This completes the proof. 3 Euler s transformation We saw that Leibniz s series for log2 and π, while esthetically pleasing, are useless for 4 practical computation because of the slowness of convergence. To rectify this situation, Euler stated the following important transformation formula: Theorem 4. If all the sequences ( r a n ) for r = 1,2,3,... decrease monotonically to zero, then the Euler transform series a a a 1 + (8) of the alternating series a 1 a 2 +a 3 + also converges and, indeed, to the same sum, L. If E n := a a a n 1 a 1 (9) n be the n-th partial sum, then the error, r n := L E n, in the approximation L E n satisfies 0 < r n n a 1 2 n. (10) which, shows that E n underestimates L. 4

5 Proof. A simple computation shows that that T (k) n+1 T(k) n = ( 1) n k+1 a n+1 2 k+1. By Cantor s theorem the intervals with end-points T n (k) and T (k) n+1, in that or opposite order dependingontheparityofn, closedownonl, andinparticular, Lisinallofthem. Therefore L T (k) n < k+1 a n+1 2 k+1. (11) Taking n = 0 and then k = n 1, we obtain the inequality a n 1 a 1 2 n < L < a n 1 a 1 2 n + n a 1 2 n, (12) But, the left-hand side is E n, and therefore the error estimate (10) is valid. Asanexample ofeuler s transformationlet scompute π withfourdecimal digit accuracy 4 from Leibniz s series. We already know that the first partial sum of the Leibniz series that achieves this accuracy is S If we compute the differences in the Euler transformation we find that n a 1 = 2 n n n+1. The error estimate shows that n must satisfy n n and the first value of n which works is n = 13 with an upper bound for the error given by Therefore, our theorem states that E 13 = 1 2 { = = approximates π 4 with an error no larger than In fact the true error is which is impresively close to the predicted upper bound for the error. Moreover one only needs 13 summands instead of 5000 to reach the desired accuracy, an extraordinary acceleration of the rate of convergence. In practice one normally computes a partial sum of the series exactly, and then uses the Euler transform to compute the remainder. For example if we compute S 10 exactly and apply Euler s transform to the next 11 summands we obtain which is correct to nine decimal places! We note that the inequality (12) appears as a problem (without a solution) on p.270 of [5]. 5 }

6 Acknowledgment We thank the referees for suggestions which substantially improved the style and content of this paper. In particular we thank a referee for suggesting the relation of our method of proof to Euler s transformation. Support from the Vicerrectoría de Investigación of the University of Costa Rica is gratefully acknowledged. References [1] L. D. Ames, Evaluation of slowly convergent series, Ann. Math. 3 (1901/02) [2] P. Calabrese, A note on alternating series, Amer. Math. Monthly 69 (1962) [3] G. Ferraro, The Rise and Development of the Theory of Series up to the Early 1820s, Springer, New York, [4] R. Johnsonbaugh, Summing an alternating series, Amer. Math. Monthly 86 (1979) [5] Konrad Knopp, Theory and Application of Infinite Series, Dover, New York, [6] R. M. Young, The error in alternating series, Mathl. Gazette 69 (1985)