INFINITE SERIES FOR π/3 AND OTHER IDENTITIES ROBERT SCHNEIDER Abstract. We combine classical identities for π, log 2, and harmonic numbers, to arrive at a nice infinite series formula for π/3 that does not appear to be well known. In an appendix we give twenty-seven related identities involving π and other irrational numbers. (Note: This is a LaTeX transcription of the author’s first mathematics paper, written in 2006.)
1. Main identity and proof Recall the lovely identity known in the literature as the Gregory–Leibniz formula for π: 1 1 1 1 π = 1 − + − + − ... (1) 4 3 5 7 9 This identity is immediate from the Maclaurin series expansion of arctan x at x = 1. Here, we prove another simple summation formula that can be used to compute the value of π. Theorem 1. We have the identity ∞
π X 1 = . 3 n(2n − 1)(4n − 3) n=1 Proof. We begin with the Maclaurin series for the natural logarithm of 2: ∞ X 1 1 1 1 1 (−1)n+1 = lim 1 − + − + −... + − = log 2 (2) n→∞ n 2 3 4 2n − 1 2n n=1 We may rewrite (2) as ∞ X n=1
1 1 − 2n − 1 2n
=
∞ X n=1
1 = log 2, 2n(2n − 1)
thus (3)
∞ X n=1
∞
1 1X 1 1 = = log 2. 4n(4n − 2) 4 n=1 2n(2n − 1) 4
Euler [1] found that the difference between the kth harmonic number and log k approaches a constant γ = 0.5772... (the so-called Euler-Mascheroni constant), so we can write 1 1 1 1 (4) lim 1 + + + + ... + − log(2n) = γ. n→∞ 2 3 4 2n Adding limits from equations (2) and (4), then dividing by 2 (and writing log(2n) = log 2 + log n), we find 1 1 1 1 1 lim 1 + + + ... + − (log 2 + log n) = (log 2 + γ), n→∞ 3 5 2n − 1 2 2 1
2
ROBERT SCHNEIDER
thus (5)
lim
n→∞
1 1 1 1 1 1 + + + ... + − log n = log 2 + γ. 3 5 2n − 1 2 2
Splitting the sum on the left in half and reorganizing, equation (5) may be rewritten 1 1 1 + + ... + (6) lim n→∞ n+1 n+3 2n − 1 1 1 1 1 1 = log 2 + γ − lim 1 + + + ... + − log n . n→∞ 2 3 5 n−1 2 Now, substitute n/2 for n in (5), then subtract 1/2 · log 2 from both sides, to arrive at 1 1 1 1 1 1 log 2 + γ = lim 1 + + + ... + − log n . (7) n→∞ 2 2 3 5 n−1 2 Adding the corresponding sides of (6) and (7), then subtracting 1/2 · log 2 + γ/2 from the resulting equation, gives 1 1 1 1 (8) lim + + ... + = log 2. n→∞ n+1 n+3 2n − 1 2 Next we will use Leibniz’s formula (1), which we can write in the form 1 1 1 1 1 π (9) lim 1 − + − + −... + − = . n→∞ 3 5 7 2n − 3 2n − 1 4 If we add the limits from (5) and (9), then divide both sides by 2, we find π 1 1 1 1 1 1 lim 1 + + + ... + − log n = + log 2 + γ . n→∞ 5 9 2n − 3 4 8 2 2 Subtracting 1/2 times the limit in (5) from this equation gives 1 1 1 1 1 1 1 1 1 lim 1 − + − + −... + − − + + ... + n→∞ 2 5 6 2n − 3 2n − 2 2 n+1 n+3 2n − 1 π = . 8 Adding 1/2 times equation (8) to both sides of this expression yields 1 1 π 1 1 1 1 lim 1 − + − + −... + − = + log 2, n→∞ 2 5 6 2n − 3 2n − 2 8 4 which can be rewritten X ∞ ∞ X 1 1 1 π 1 (10) = − = + log 2. 4n − 3 4n − 2 (4n − 2)(4n − 3) 8 4 n=1 n=1 Subtracting equation (3) from (10) yields X ∞ ∞ X 3 1 1 π − = = . (4n − 2)(4n − 3) 4n(4n − 2) 4n(4n − 2)(4n − 3) 8 n=1 n=1 Finally, multiplication by 8/3 gives the theorem.
INFINITE SERIES FOR π/3 AND OTHER IDENTITIES
3
2. Appendix: Further identities The preceding section is a LaTeX transcription of the author’s first mathematics paper, written in 2006 at a time when I was almost entirely self taught in mathematics. During that period I also found a large number of related summation identities by combining Theorem 1 with other equations in the preceding proof, along with well-known zeta values such as Euler’s famous solution of the Basel problem (see [1]) ∞ X 1 π2 , = n2 6 n=1
and other classical summation identities from [2]. I give a selection of these identities below without proof, loosely organized by number of factors in the denominators of the summands. The first identity follows from the telescoping series (1 − 1/2) + (1/2 − 1/3) + (1/3−1/4)+..., the second and third identities follow by rewriting (2) and (9), respectively, and the rest arise from liberal use of partial fractions. ∞ X
(11)
n=1 ∞ X
(12)
n=1
1 = 2 log 2. (n + 1)(2n + 1)
∞ X
(13)
n=1
(14)
∞ X n=1
n=1 ∞ X
(16)
n=1
(17)
∞ X n=1
(18)
n=1
(19)
n=1
1 π + 6 log 2 = n(4n − 3) 6 1 6 log 2 − π = n(4n − 1) 2
1 log 2 = (2n − 1)(4n − 1)(4n − 3) 2
∞ X
∞ X
π 1 = (4n − 1)(4n − 3) 8
1 π + 2 log 2 = (2n − 1)(4n − 3) 4 ∞ X
(15)
1 1 = (2n − 1)(2n + 1) 2
1 2 log 2 − 1 = n(n + 1)(2n + 1) 2
1 π−2 = (4n + 1)(4n − 1)(4n − 3) 16
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ROBERT SCHNEIDER
∞ X
(20)
n=1 ∞ X
(21)
1 24 log 2 − π 2 = n2 (2n − 1) 6
72 log 2 − 12π − π 2 1 = n2 (4n − 1) 6
n=1
∞ X
(22)
n=1 ∞ X
(23)
n=1
1 π 2 + 24 log 2 − 24 = n2 (2n + 1) 6 ∞ X
(24)
n=1 ∞ X
(25)
n=1 ∞ X
(26)
n=1
(27)
∞ X n=1
1 π2 − 6 = n2 (n + 1) 6 1 π2 − 9 = n2 (n + 1)2 3
π−3 1 = (2n + 1)(2n − 1)(4n + 1)(4n − 1) 6
√ π(1 + 2 2) − 12 1 = (4n + 1)(4n − 1)(8n + 1)(8n − 1) 24 ∞ X
(28)
n=1
1 6 log 2 − π = n(2n − 1)(4n − 1)(4n − 3) 3
∞ X
(29)
n=1 ∞ X
(30)
n=1 ∞ X
(31)
n=1
(32)
1 12 − π 2 = n(n + 1)2 6
∞ X n=1
1 π 2 − 12 log 2 = n2 (n + 1)(2n + 1) 6
1 π 2 + 8π − 24 log 2 = n2 (2n − 1)(4n − 3) 18
1 π 2 + 24π − 120 log 2 = n2 (2n − 1)(4n − 1) 6
1 168 log 2 − 32π − π 2 = n2 (2n − 1)(4n − 1)(4n − 3) 18
INFINITE SERIES FOR π/3 AND OTHER IDENTITIES
∞ X
(33)
n=1 ∞ X
(34)
n=1
52 − 32 log 2 − 27 log 3 1 = n(2n + 1)(2n − 1)(6n + 1)(6n − 1) 16 1 19 + 16 log 2 − 27 log 3 = n(2n + 1)(2n − 1)(3n + 1)(3n − 1) 10 ∞ X
(35)
n=1 ∞ X
(36)
n=1
(37)
∞ X n=1
5
n3 (n
1 = 10 − π 2 3 + 1)
√ 45 − π(3 + 8 2) 1 = (2n + 1)(2n − 1)(4n + 1)(4n − 1)(8n + 1)(8n − 1) 6
1 64 log 2 + 27 log 3 − 74 = n(2n + 1)(2n − 1)(3n + 1)(3n − 1)(6n + 1)(6n − 1) 20 Acknowledgments
I am thankful to Paul Blankenship at Bluegrass Community and Technical College for encouraging my early efforts in analysis; and to David Leep at the University of Kentucky for mentoring me when I had little formal training in mathematics, and prompting me to write up the main result of this study. References [1] W. Dunham, Euler: the master of us all, The Dolciani Mathematical Expositions, 22. Mathematical Association of America, Washington, DC, 1999. [2] L. B. W. Jolley, ed. Summation of series. Courier Corporation, 2012.