Rapidly Convergent Summation Formulas Involving Stirling Series

Home , Convergent series, Series acceleration, Summation

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2 where the Gregory numbers Ck for k N [3, 5, 6, 7] are deﬁned by ∈ 0

1 1 Ck := (x)kdx k! Z0 1 1 = x(x 1) (x k + 1)dx k! Z0 − ··· − k 1 1 = S(1)(l). k! l +1 k Xl=0 The last one is a convergent version of Stirling’s formula for the logarithm of the factorial function [8], namely

n ∞ (−1)k k (−1)ll (1) 1 1 k l=1 l l Sk (l) log(k)= n log(n) n + log (2π)+ log(n)+ 2 ( +1)( +2) − 2 2 (n +P 1)(n + 2) (n + k) Xk=1 Xk=1 ··· ∞ 1 1 ak = n log(n) n + log (2π)+ log(n)+ − 2 2 (n + 1)(n + 2) (n + k) Xk=1 ··· 1 1 1 1 = n log(n) n + log (2π)+ log(n)+ + − 2 2 12(n + 1) 12(n + 1)(n + 2) 59 29 + + + ..., 360(n + 1)(n + 2)(n + 3) 60(n + 1)(n + 2)(n + 3)(n + 4) where the numbers ak for k N [8, 9, 10] are deﬁned by ∈ 1 1 1 ak := (x)k x dx k Z0 − 2 k 1 l (1) = Sk (l) 2k l (l + 1)(l + 2) X=1 k k l ( 1) ( 1) l = − − S(1)(l). 2k (l + 1)(l + 2) k Xl=1 and was also found around 1800.

2 Deﬁnitions

Deﬁnition 1. (Fractional part x of x) Let x denote the ﬂoor function{ evaluated} at the point x R, which is the largest integer less⌊ than⌋ or equal to x. Written out explicitly, this means that∈ x := max m x . ⌊ ⌋ m∈Z { ≤ }

3 We define by x the fractional part of x, that is { } x := x x . { } − ⌊ ⌋ Note that we have 0 x < 1. ≤{ } Definition 2. (Pochhammer symbol)[1] We define the Pochhammer symbol (or rising factorial function) (x)k by Γ(x + k) (x)k := x(x + 1)(x + 2)(x + 3) (x + k 1) = , ··· − Γ(x) where Γ(x) is the gamma function defined by ∞ Γ(x) := e−ttx−1dt. Z0 Definition 3. (Stirling Series)[1, 2] A Stirling series (or inverse factorial series) for a function f : C C, which vanishes at z + , is an expansion of the type → → ∞ ∞ k!ak 0!a 1!a 2!a 3!a f(x)= = 0 + 1 + 2 + 3 + ..., (x)k+1 x x(x + 1) x(x + 1)(x + 2) x(x + 1)(x + 2)(x + 3) Xk=0 where (x)k denotes the Pochhammer symbol. Definition 4. (Stirling numbers of the first kind)[1] Let k,l N0 be two non-negative integers such that k l 0. We define the Stirling ∈ (1) ≥ ≥ numbers of the first kind Sk (l) as the connecting coefficients in the identity

k k l (1) l (x)k =( 1) ( 1) Sk (l)x , − − Xl=0 where (x)k is the rising factorial. Definition 5. (Bernoulli numbers)[12] We define the k-th Bernoulli number Bk as the k-th coefficient in the generating function relation ∞ x Bk = xk x C with x < 2π. ex 1 k! ∀ ∈ | | − Xk=0 Definition 6. (Euler numbers)[13] ∞ We define the sequence of Euler numbers Ek by the generating function identity { }k=0 x ∞ 2e Ek = xk. e2x +1 k! Xk=0

4 Definition 7. (Tangent numbers)[14, 15] We define the k-th tangent number Tk by the equation 2k 2k 2 (2 1) B2k Tk := − | |, 2k where the Bk’s are the Bernoulli numbers. Definition 8. (Bernoulli polynomials)[16, Proposition 23.2, p. 86] We define the n-th Bernoulli polynomial Bn(x) by n n n−k Bn(x) := Bkx , k Xk=0 where the Bk’s are the Bernoulli numbers.

3 The Summation Formulas

In this section we will list 32 summation formulas for various ﬁnite sums of analytic functions. For this, we need the following Lemma 9. (Euler-Maclaurin summation formula)[13, 17] Let f be an analytic function. Then we have that n n m 1 Bk f(k)= f(x)dx + (f(n)+ f(1)) + f (k−1)(n) f (k−1)(1) Z 2 k! − Xk=1 1 Xk=2 m+1 n ( 1) (m) + − Bm( x )f (x)dx, m! Z1 { } where Bm(x) is the m-th Bernoulli polynomial and x denotes the fractional part of x. Therefore, for many functions f, we have the asymptotic{ } expansion n n ∞ B Bk f(k) f(x)dx + C 1 f(n)+ f (k−1)(n) as n , ∼ Z − 1! k! →∞ Xk=1 1 Xk=2 for some constant C C. ∈ As well as the next key result found by J. Weniger: Lemma 10. (Weniger transformation)[1] ∞ ak For every inverse power series k=1 xk+1 , the following transformation formula holds

∞ P ∞ k k ak ( 1) l (1) l k+1 = − ( 1) a Sk (l) x (x)k+1 − Xk=1 Xk=1 Xl=1 ∞ k k l (1) ( 1) ( 1) alSk (l) = − l=1 − . x(x + 1)(Px + 2) (x + k) Xk=1 ···

5 Now it is time for our summation formulas. For every single finite sum there are infinitely many different such formulas, therefore we give here just the canonical ones. It is not hard to prove, using the asymptotic expansions of the Bernoulli numbers and the Stirling numbers of the first kind as well as that of the Euler numbers, that all series below converges very rapidly. Theorem 11. (Summation formulas for the harmonic series) For every natural number n N, we have that ∈ l n ∞ k+1 k (−1) (1) 1 1 ( 1) l l Bl+1Sk (l) = log(n)+ γ + + − =1 +1 k 2n n(n + 1)(Pn + 2) (n + k) Xk=1 Xk=1 ··· 1 1 1 19 = log(n)+ γ + 2n − 12n(n + 1) − 12n(n + 1)(n + 2) − 120n(n + 1)(n + 2)(n + 3) 9 .... − 20n(n + 1)(n + 2)(n + 3)(n + 4) − We also have that l n ∞ k+1 k (−1) (1) 1 ( 1) l BlSk (l) = log(n)+ γ + − =1 l k (n + 1)(Pn + 2) (n + k) Xk=1 Xk=1 ··· 1 5 3 = log(n)+ γ + + + 2(n + 1) 12(n + 1)(n + 2) 4(n + 1)(n + 2)(n + 3) 251 + + .... 120(n + 1)(n + 2)(n + 3)(n + 4)

1 Proof. Applying the Euler-Maclaurin summation formula to the function f(x) := x , we get that n ∞ 1 Bk log(n)+ γ . k ∼ − knk Xk=1 Xk=1 Applying now the Weniger transformation to the equivalent series n ∞ 1 1 Bk log(n)+ γ + k ∼ 2n − knk Xk=1 Xk=2 ∞ 1 Bk log(n)+ γ + +1 , ∼ 2n − (k + 1)nk+1 Xk=1 we get the ﬁrst claimed formula. To obtain the second expression, we apply the Weniger transformation to the identity n ∞ 1 Bk log(n)+ γ n . k ∼ − knk+1 Xk=1 Xk=1

6 All summation formulas in this article, which we list below, are obtained using the same universal technique: For each finite sum, we used first the Euler-Maclaurin summation formula and then applied to it the Weniger transformation to get convergent formulas. Therefore, we omit full proofs and list here just our final results.

1.) Summation formulas for the harmonic series: For every natural number n N, we have ∈ l n ∞ k+1 k (−1) (1) 1 1 ( 1) l l Bl+1Sk (l) = log(n)+ γ + + − =1 +1 k 2n n(n + 1)(Pn + 2) (n + k) Xk=1 Xk=1 ··· 1 1 1 19 = log(n)+ γ + 2n − 12n(n + 1) − 12n(n + 1)(n + 2) − 120n(n + 1)(n + 2)(n + 3) 9 ... − 20n(n + 1)(n + 2)(n + 3)(n + 4) −

and

l n ∞ k+1 k (−1) (1) 1 ( 1) l BlSk (l) = log(n)+ γ + − =1 l k (n + 1)(Pn + 2) (n + k) Xk=1 Xk=1 ··· 1 5 3 = log(n)+ γ + + + 2(n + 1) 12(n + 1)(n + 2) 4(n + 1)(n + 2)(n + 3) 251 + + .... 120(n + 1)(n + 2)(n + 3)(n + 4)

2.) Summation formulas for the n-th partial sum of ζ(2): For every natural number n N, we have ∈ n ∞ k+1 k l (1) 1 1 1 1 ( 1) ( 1) Bl Sk (l) = ζ(2) + + − l=1 − +1 k2 − n 2n2 n n(n + 1)(Pn + 2) (n + k) Xk=1 Xk=1 ··· 1 1 1 1 3 = ζ(2) + − n 2n2 − 6n2(n + 1) − 6n2(n + 1)(n + 2) − 10n2(n + 1)(n + 2)(n + 3) 4 ... − 5n2(n + 1)(n + 2)(n + 3)(n + 4) −

7 and n ∞ k+1 k l (1) 1 1 ( 1) ( 1) BlSk (l) = ζ(2) + − l=1 − k2 − n n(n + 1)(Pn + 2) (n + k) Xk=1 Xk=1 ··· ∞ 1 1 (k 1)! = ζ(2) + − − n k +1 · n(n + 1)(n + 2) (n + k) Xk=1 ··· 1 1 1 1 = ζ(2) + + + − n 2n(n + 1) 3n(n + 1)(n + 2) 2n(n + 1)(n + 2)(n + 3) 6 + + .... 5n(n + 1)(n + 2)(n + 3)(n + 4)

3.) Summation formulas for the n-th partial sum of ζ(3): For every natural number n N, we have ∈ n ∞ k+1 k l (1) 1 1 1 1 ( 1) ( 1) (l + 2)Bl Sk (l) = ζ(3) + + − l=1 − +1 k3 − 2n2 2n3 2n2 n(nP+ 1)(n + 2) (n + k) Xk=1 Xk=1 ··· 1 1 1 1 5 = ζ(3) + − 2n2 2n3 − 4n3(n + 1) − 4n3(n + 1)(n + 2) − 12n3(n + 1)(n + 2)(n + 3) 1 ... − n3(n + 1)(n + 2)(n + 3)(n + 4) − and n ∞ k+1 k l (1) 1 1 1 ( 1) ( 1) (l + 1)BlSk (l) = ζ(3) + − l=1 − k3 − 2n2 2 n2(n P+ 1)(n + 2) (n + k) Xk=1 Xk=1 ··· 1 1 1 1 = ζ(3) + + + − 2n2 2n2(n + 1) 4n2(n + 1)(n + 2) 4n2(n + 1)(n + 2)(n + 3) 1 + + .... 3n2(n + 1)(n + 2)(n + 3)(n + 4)

4.) Summation formulas for the sum of the square roots of the ﬁrst n natural numbers: For every natural number n N, we have ∈ n ∞ k l (2l−3)!! (1) l l 2 3 1 1 3 k l=1( 1) 2 (l+1)! B +1Sk (l) √k = n 2 + √n ζ + √n ( 1) − 3 2 − 4π 2 − P(n + 1)(n + 2) (n + k) Xk=0 Xk=1 ··· 2 3 1 1 3 √n √n 53√n = n 2 + √n ζ + + + 3 2 − 4π 2 24(n + 1) 24(n + 1)(n + 2) 640(n + 1)(n + 2)(n + 3) 79√n + + ..., 320(n + 1)(n + 2)(n + 3)(n + 4)

8 as well as

n ∞ k l (2l−1)!! (1) l 2 3 1 1 3 1 k l=1( 1) 2l+1(l+2)! B +2Sk (l) √k = n 2 + √n ζ + + ( 1) − 3 2 − 4π 2 24√n − P√n(n + 1)(n + 2) (n + k) Xk=0 Xk=1 ··· 2 3 1 1 3 1 1 = n 2 + √n ζ + 3 2 − 4π 2 24√n − 1920√n(n + 1)(n + 2) 1 259 − 640√n(n + 1)(n + 2)(n + 3) − 46080√n(n + 1)(n + 2)(n + 3)(n + 4) 115 ... − 4608√n(n + 1)(n + 2)(n + 3)(n + 4)(n + 5) −

and

n ∞ k l (2l−5)!! (1) 2 3 1 3 k l=1( 1) l−1l BlSk (l) √k = n 2 ζ + n√n ( 1) − 2 ! 3 − 4π 2 − P(n + 1)(n + 2) (n + k) Xk=0 Xk=1 ··· 2 3 1 3 n√n 13n√n 9n√n = n 2 ζ + + + 3 − 4π 2 2(n + 1) 24(n + 1)(n + 2) 8(n + 1)(n + 2)(n + 3) 2213n√n + + .... 640(n + 1)(n + 2)(n + 3)(n + 4)

5.) Summation formulas for the n-th partial sum of ζ( 3/2): For every natural number n N, we have − ∈ n ∞ k l (2l−5)!! (1) l 2 5 1 3 3 5 3 3 k l=1( 1) 2l−1(l+1)! B +1Sk (l) k√k = n 2 + n 2 ζ + n 2 ( 1) +1 − 5 2 − 16π2 2 2 − P(n + 1)(n + 2) (n + k) Xk=0 Xk=1 ··· 2 5 1 3 3 5 n√n n√n 481n√n = n 2 + n 2 ζ + + + 5 2 − 16π2 2 8(n + 1) 8(n + 1)(n + 2) 1920(n + 1)(n + 2)(n + 3) 241n√n + + ..., 320(n + 1)(n + 2)(n + 3)(n + 4)

as well as

n ∞ k l (2l−1)!! (1) l 2 5 1 3 1 3 5 3 k l=1( 1) 2l+1(l+3)! B +3Sk (l) k√k = n 2 + n 2 + √n ζ + ( 1) +1 − 5 2 8 − 16π2 2 2 − P√n(n + 1)(n + 2) (n + k) Xk=0 Xk=1 ··· 2 5 1 3 1 3 5 1 1 = n 2 + n 2 + √n ζ + + 5 2 8 − 16π2 2 1920√n(n + 1) 1920√n(n + 1)(n + 2) 107 51 + + + ... 107520√n(n + 1)(n + 2)(n + 3) 17920√n(n + 1)(n + 2)(n + 3)(n + 4)

9 and

n ∞ k l (2l−7)!! (1) 2 5 3 5 3 5 k l=1( 1) l−2l BlSk (l) k√k = n 2 ζ + n 2 ( 1) +1 − 2 ! 5 − 16π2 2 2 − P(n + 1)(n + 2) (n + k) Xk=0 Xk=1 ··· 2 2 2 2 5 3 5 n √n 5n √n 11n √n = n 2 ζ + + + 5 − 16π2 2 2(n + 1) 8(n + 1)(n + 2) 8(n + 1)(n + 2)(n + 3) 8401n2√n + + .... 1920(n + 1)(n + 2)(n + 3)(n + 4)

6.) Summation formulas for the n-th partial sum of ζ( 5/2): For every natural number n N, we have − ∈ n ∞ k l (2l−7)!! (1) l− l 2 7 1 5 15 7 15 5 k l=1( 1) 2 2(l+1)! B +1Sk (l) k2√k = n 2 + n 2 + ζ + n 2 ( 1) − 7 2 64π3 2 4 − P(n + 1)(n + 2) (n + k) Xk=0 Xk=1 ··· 2 2 2 7 1 5 15 7 5n √n 5n √n = n 2 + n 2 + ζ + + 7 2 64π3 2 24(n + 1) 24(n + 1)(n + 2) 53n2√n 79n2√n + + 128(n + 1)(n + 2)(n + 3) 64(n + 1)(n + 2)(n + 3)(n + 4) 35187n2√n + + ..., 7168(n + 1)(n + 2)(n + 3)(n + 4)(n + 5)

as well as n 2 7 1 5 5 3 15 7 1 k2√k = n 2 + n 2 + n 2 + ζ 7 2 24 64π3 2 − 384√n Xk=0 ∞ k l (2l−1)!! (1) 15 l=1( 1) l+1 l Bl+4Sk (l) + ( 1)k − 2 ( +4)! 4 − P√n(n + 1)(n + 2) (n + k) Xk=1 ··· 2 7 1 5 5 3 15 7 1 1 = n 2 + n 2 + n 2 + ζ + 7 2 24 64π3 2 − 384√n 21504√n(n + 1)(n + 2) 1 115 + + 7168√n(n + 1)(n + 2)(n + 3) 229376√n(n + 1)(n + 2)(n + 3)(n + 4) 255 + + ... 114688√n(n + 1)(n + 2)(n + 3)(n + 4)(n + 5)

10 and

n ∞ k l (2l−9)!! (1) 2 7 15 7 15 7 k l=1( 1) l−3l BlSk (l) k2√k = n 2 + ζ + n 2 ( 1) − 2 ! 7 64π3 2 4 − P(n + 1)(n + 2) (n + k) Xk=0 Xk=1 ··· 3 3 3 2 7 15 7 n √n 17n √n 13n √n = n 2 + ζ + + + 7 64π3 2 2(n + 1) 24(n + 1)(n + 2) 8(n + 1)(n + 2)(n + 3) 677n3√n + + .... 128(n + 1)(n + 2)(n + 3)(n + 4)

7.) Summation formulas for the sum of the inverses of the square roots of the ﬁrst n natural numbers: For every natural number n N, we have ∈ n ∞ k l (2l−1)!! (1) 1 1 1 l=1( 1) l l Bl+1Sk (l) =2√n + ζ + + ( 1)k+1 − 2 ( +1)! √k 2 2√n − √Pn(n + 1)(n + 2) (n + k) Xk=1 Xk=1 ··· 1 1 1 1 =2√n + ζ + 2 2√n − 24√n(n + 1) − 24√n(n + 1)(n + 2) 31 15 ... − 384√n(n + 1)(n + 2)(n + 3) − 64√n(n + 1)(n + 2)(n + 3)(n + 4) − and

n ∞ k l (2l−3)!! (1) 1 1 l ( 1) l−1 BlSk (l) =2√n + ζ + √n ( 1)k+1 =1 − 2 l! √k 2 − P(n + 1)(n + 2) (n + k) Xk=1 Xk=1 ··· 1 √n 11√n 7√n =2√n + ζ + + + 2 2(n + 1) 24(n + 1)(n + 2) 8(n + 1)(n + 2)(n + 3) 977√n + + .... 384(n + 1)(n + 2)(n + 3)(n + 4)

8.) Summation formulas for the n-th partial sum of ζ(3/2): For every natural number n N, we have ∈ n ∞ k+1 k l (2l+1)!! (1) 1 3 2 1 2 ( 1) l=1( 1) l+1 l Bl+1Sk (l) = ζ + + − − 2 ( +1)! k√k 2 − √n 2n√n √n n(nP+ 1)(n + 2) (n + k) Xk=1 Xk=1 ··· 3 2 1 1 1 = ζ + 2 − √n 2n√n − 8n√n(n + 1) − 8n√n(n + 1)(n + 2) 89 41 ... − 384n√n(n + 1)(n + 2)(n + 3) − 64n√n(n + 1)(n + 2)(n + 3)(n + 4) −

11 and

n ∞ k+1 k l (2l−1)!! (1) 1 3 2 ( 1) l ( 1) l BlSk (l) = ζ +2 − =1 − 2 l! k√k 2 − √n √n(nP+ 1)(n + 2) (n + k) Xk=1 Xk=1 ··· 3 2 1 3 5 = ζ + + + 2 − √n 2√n(n + 1) 8√n(n + 1)(n + 2) 8√n(n + 1)(n + 2)(n + 3) 631 + .... 384√n(n + 1)(n + 2)(n + 3)(n + 4) −

9.) Summation formulas for the n-th partial sum of ζ(5/2): For every natural number n N, we have that ∈ n ∞ k+1 k l (2l+3)!! (1) l l 1 5 2 1 4 ( 1) l=1( 1) 2 +2(l+1)! B +1Sk (l) = ζ 3 + + − − 2 2 P k √k 2 − 3n 2 2n √n 3n√n n(n + 1)(n + 2) (n + k) Xk=1 Xk=1 ··· 5 2 1 5 5 = ζ 3 + 2 2 2 2 − 3n 2 2n √n − 24n √n(n + 1) − 24n √n(n + 1)(n + 2) 139 59 ... − 384n2√n(n + 1)(n + 2)(n + 3) − 64n2√n(n + 1)(n + 2)(n + 3)(n + 4) −

and that

n ∞ k+1 k l (2l+1)!! (1) l 1 5 2 4 ( 1) l=1( 1) 2l+1l! B Sk (l) = ζ 3 + − − k2√k 2 − 3n 2 3√n n(n +P 1)(n + 2) (n + k) Xk=1 Xk=1 ··· 5 2 1 7 3 = ζ 3 + + + 2 − 3n 2 2n√n(n + 1) 24n√n(n + 1)(n + 2) 8n√n(n + 1)(n + 2)(n + 3) 293 + + .... 384n√n(n + 1)(n + 2)(n + 3)(n + 4)

10.) Convergent versions of Stirling’s formula: For every natural number n N, we have that ∈ n ∞ k (−1)l (1) 1 1 l=1 l l Bl+1Sk (l) log(k)= n log(n) n + log (2π)+ log(n)+ ( 1)k ( +1) − 2 2 − (nP+ 1)(n + 2) (n + k) Xk=1 Xk=1 ··· 1 1 1 1 = n log(n) n + log (2π)+ log(n)+ + − 2 2 12(n + 1) 12(n + 1)(n + 2) 59 29 + + + ..., 360(n + 1)(n + 2)(n + 3) 60(n + 1)(n + 2)(n + 3)(n + 4)

12 as well as

n ∞ k k (−1)l (1) 1 1 1 ( 1) l=1 l l Bl+2Sk (l) log(k)= n log(n) n + log (2π)+ log(n)+ + − ( +1)( +2) − 2 2 12n n(n P+ 1)(n + 2) (n + k) Xk=1 Xk=1 ··· 1 1 1 1 = n log(n) n + log (2π)+ log(n)+ − 2 2 12n − 360n(n + 1)(n + 2) 1 5 − 120n(n + 1)(n + 2)(n + 3) − 168n(n + 1)(n + 2)(n + 3)(n + 4) 11 ... − 84n(n + 1)(n + 2)(n + 3)(n + 4)(n + 5) −

and

n ∞ k k (−1)l (1) 1 1 ( 1) l=2 l l− BlSk (l) log(k)= n log(n) n + log (2π)+ log(n)+ n − ( 1) − 2 2 (n + 1)(Pn + 2) (n + k) Xk=1 Xk=1 ··· 1 1 n n = n log(n) n + log (2π)+ log(n)+ + − 2 2 12(n + 1)(n + 2) 4(n + 1)(n + 2)(n + 3) 329n 149n + + + .... 360(n + 1)(n + 2)(n + 3)(n + 4) 36(n + 1)(n + 2)(n + 3)(n + 4)(n + 5)

11.) First logarithmic summation formulas: For every natural number n N, we have ∈ n 1 1 1 1 1 ′ k log(k)= n2 log(n) n2 + n log(n)+ log(n)+ ζ ( 1) 2 − 4 2 12 12 − − Xk=0 ∞ k (−1)l (1) l=1 l l l Bl+2Sk (l) + ( 1)k+1 ( +1)( +2) − P(n + 1)(n + 2) (n + k) Xk=1 ··· 1 1 1 1 1 ′ 1 = n2 log(n) n2 + n log(n)+ log(n)+ ζ ( 1) + 2 − 4 2 12 12 − − 720(n + 1)(n + 2) 1 19 + + 240(n + 1)(n + 2)(n + 3) 1260(n + 1)(n + 2)(n + 3)(n + 4) 17 + + ... 252(n + 1)(n + 2)(n + 3)(n + 4)(n + 5)

13 and n 1 1 1 1 1 ′ k log(k)= n2 log(n) n2 + n log(n)+ log(n)+ ζ ( 1) 2 − 4 2 12 12 − − Xk=0 ∞ k (−1)l (1) l=1 l l l Bl+3Sk (l) + ( 1)k+1 ( +1)( +2)( +3) − Pn(n + 1)(n + 2) (n + k) Xk=1 ··· 1 1 1 1 1 ′ 1 = n2 log(n) n2 + n log(n)+ log(n)+ ζ ( 1) + 2 − 4 2 12 12 − − 720n(n + 1) 1 13 + + 720n(n + 1)(n + 2) 5040n(n + 1)(n + 2)(n + 3) 1 + + .... 140n(n + 1)(n + 2)(n + 3)(n + 4)

12.) Second logarithmic summation formula: For every natural number n N, we have ∈ n ∞ k (−1)l (1) (1) log(k) 1 log(n) l=1 l Bl+1Sl+1(2)Sk (l) = log(n)2 + + γ + ( 1)k ( +1)! k 2 2n 1 − Pn(n + 1)(n + 2) (n + k) Xk=1 Xk=1 ··· ∞ k k 1 (1) ( 1) l l Bl+1Sk (l) + log(n) − =1 +1 n(n + 1)(P n + 2) (n + k) Xk=1 ··· 1 log(n) 1 1 = log(n)2 + + γ + + 2 2n 1 12n(n + 1) 12n(n + 1)(n + 2) 109 49 + + + ... 720n(n + 1)(n + 2)(n + 3) 120n(n + 1)(n + 2)(n + 3)(n + 4) log(n) log(n) 19 log(n) − 12n(n + 1) − 12n(n + 1)(n + 2) − 120n(n + 1)(n + 2)(n + 3) 9 log(n) .... − 20n(n + 1)(n + 2)(n + 3)(n + 4) −

14 13.) Third logarithmic summation formula (Summation formula for the n-th partial sum of ζ′(2)): For every natural number n N, we have ∈ n k 1 l−1 m+1 (1) ∞ Bl S (l) log(k) ′ log(n) 1 log(n) 1 l=1 l+1 m=0 l−m +1 k = ζ (2) + + ( 1)k+1 k2 − − n − n 2n2 n − P n(n + 1)(Pn + 1) (n + k) Xk=1 Xk=1 ··· ∞ k k (1) log(n) ( 1) Bl Sk (l) + − l=1 +1 n n(n + 1)(Pn + 2) (n + k) Xk=1 ··· ′ log(n) 1 log(n) 1 1 = ζ (2) + + + − − n − n 2n2 12n2(n + 1) 12n2(n + 1)(n + 2) 47 17 + + + ... 360n2(n + 1)(n + 2)(n + 3) 60n2(n + 1)(n + 2)(n + 3)(n + 4) log(n) log(n) 3 log(n) − 6n2(n + 1) − 6n2(n + 1)(n + 2) − 10n2(n + 1)(n + 2)(n + 3) 4 log(n) .... − 5n2(n + 1)(n + 2)(n + 3)(n + 4) −

14.) Fourth logarithmic summation formula:

15 For every natural number n N, we have ∈ n 1 log(n) γ2 π2 log(2)2 log(k)2 = n log(n)2 2n log(n)+2n + log(n)2 + + − 2 6n 2 − 24 − 2 Xk=1 k k − l (1) (1) 2 ∞ ( 1) log(π) ( 1) l=1 l Bl+2Sl+1(2)Sk (l) log(2) log(π) + γ +2 − ( +2)! − − 2 1 n(Pn + 1)(n + 2) (n + k) Xk=1 ··· ∞ k k 1 (1) ( 1) l=1 l l Bl+2Sk (l) + 2 log(n) − ( +1)( +2) n(n P+ 1)(n + 2) (n + k) Xk=1 ··· 1 log(n) γ2 π2 log(2)2 = n log(n)2 2n log(n)+2n + log(n)2 + + − 2 6n 2 − 24 − 2 log(π)2 1 log(2) log(π) + γ + − − 2 1 120n(n + 1)(n + 2) 1 167 + + 40n(n + 1)(n + 2)(n + 3) 1890n(n + 1)(n + 2)(n + 3)(n + 4) 145 + + ... 378n(n + 1)(n + 2)(n + 3)(n + 4)(n + 5) log(n) log(n) − 180n(n + 1)(n + 2) − 60n(n + 1)(n + 2)(n + 3) 5 log(n) 11 log(n) .... − 84n(n + 1)(n + 2)(n + 3)(n + 4) − 42n(n + 1)(n + 2)(n + 3)(n + 4)(n + 5) −

15.) Summation formulas for the n-th partial sum of the Gregory-Leibniz series: For every natural number n N , we have that ∈ 0 n ∞ k l k n+1 n+1 k+1 (−1) (1) ( 1) π ( 1) ( 1) ( 1) l l ElSk (l) − = − + − − =1 2 2k +1 4 − 4(n + 1) 4 (n + 1)(nP+ 2) (n + k + 1) Xk=0 Xk=1 ··· π ( 1)n+1 ( 1)n+1 3( 1)n+1 = − + − + − 4 − 4(n + 1) 16(n + 1)(n + 2)(n + 3) 16(n + 1)(n + 2)(n + 3)(n + 4) 39( 1)n+1 + − 64(n + 1)(n + 2)(n + 3)(n + 4)(n + 5) 75( 1)n+1 + − 32(n + 1)(n + 2)(n + 3)(n + 4)(n + 5)(n + 6) + ...

16 and for all n N that n ∈ ∞ k l k+1 n n k+1 (−1) (1) ( 1) π ( 1) ( 1) ( 1) l l ElSk (l) − = − + − − =1 2 2k 1 4 − 4n 4 n(n + 1)(Pn + 2) (n + k) Xk=1 − Xk=1 ··· π ( 1)n ( 1)n 3( 1)n = − + − + − 4 − 4n 16n(n + 1)(n + 2) 16n(n + 1)(n + 2)(n + 3) 39( 1)n 75( 1)n + − + − + .... 64n(n + 1)(n + 2)(n + 3)(n + 4) 32n(n + 1)(n + 2)(n + 3)(n + 4)(n + 5)

16.) Summation formula for the n-th partial sum of the alternating harmonic series: For every natural number n N, we have ∈ 2 l+1 n ∞ k l +l (2 −1) (1) k+1 n 2 ( 1) ( 1) n k l ( 1) l Bl+1 Sk (l) − = log(2) − +( 1) ( 1) =1 − +1 | | k − 2n − − P n(n + 1)(n + 2) (n + k) Xk=1 Xk=1 ··· ( 1)n ( 1)n ( 1)n 3( 1)n = log(2) − + − + − + − − 2n 4n(n + 1) 4n(n + 1)(n + 2) 8n(n + 1)(n + 2)(n + 3) 3( 1)n + − + .... 4n(n + 1)(n + 2)(n + 3)(n + 4) To obtain the formulas 15.) and 16.), we have used the references [13, 18, 19, 20] and the Boole summation formula [13, 18], which is the analog of the Euler-Maclaurin summation formula for alternating sums.

4 Conclusion

We have proved many rapidly convergent summation formulas for different sums of functions. These formulas are highly useful, because of their fast convergence and their connections to many different special functions, like for example the polygamma functions. Perhaps one can implement these formulas in a future version of Mathematica to compute for example the digamma function via the formula l ∞ k+1 k (−1) (1) 1 ( 1) l l Bl+1Sk (l) ψ(x) = log(x) + − =1 +1 . − 2x x(x + 1)(Px + 2) (x + k) Xk=1 ··· Computing for example ψ(1010) up to 400 correct decimal digits with the current version of Mathematica 10 takes more than 10 minutes on our home computer, but with the above formula, we can do it in less than 3 seconds. With our technique, one can obtain much more such summation formulas for finite sums of n the form f(k), where n N and f is any function, showing again that this universal k=1 ∈ techniqueP for obtaining such formulas is highly suitable for computer implementations.

17 References

[1] Ernst Joachim Weniger, Summation of divergent power series by means of factorial series, arXiv:1005.0466v1 [math.NA], 4 May 2010.

[2] A. C. Aitken, A note on inverse central factorial series, Proceedings of The Edinburgh Mathematical Society - PROC EDINBURGH MATH SOC, Vol. 7, No. 03, 1946.

[3] Donatella Merlini, Renzo Sprugnoli, and M. Cecilia Verri, The Cauchy numbers, Elsevier, Science Direct, Discrete Mathematics 306 (2006), 1906–1920.

[4] https://en.wikipedia.org/wiki/EulerMascheroni_constant, Relations with the reciprocal logarithm.

[5] Feng Qi, Explicit formulas for computing Bernoulli numbers of the second kind and Stirling numbers of the ﬁrst kind, Filomat 28:2 (2014), DOI: 10.2298/FIL14023190, 319– 327.

[6] https://oeis.org/A002206.

[7] https://oeis.org/A002207.

[8] https://en.wikipedia.org/wiki/Stirling%27s_approximation, A convergent version of Stirling’s formula.

[9] https://oeis.org/A122252.

[10] https://oeis.org/A122253.

[11] K. Goldberg, M. Newman, and E. Haynsworth (1972), Stirling numbers of the ﬁrst kind, Stirling numbers of the second kind, in Milton Abramowitz, Irene A. Stegun, Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, 10th printing, New York: Dover, 824–825.

[12] Feng Qi and Bai-Ni Guo, Alternative proofs of a formula for Bernoulli numbers in terms of Stirling numbers, De Gruyter, Analysis 34 (2014), DOI: 10.1515/anly-2014-0003, 187– 193.

[13] Jonathan M. Borwein, Neil J. Calkin, and Dante Manna, Euler-Boole summation revisited, The American Mathematical Monthly, Vol. 116, No. 5 (May, 2009), DOI: 10.4169/193009709X470290, 387–412.

[14] http://mathworld.wolfram.com/TangentNumber.html.

[15] http://oeis.org/A000182.

[16] Victor Kac and Pokman Cheung, Quantum Calculus, Springer, 2002.

18 [17] A. V. Ustinov, A discrete analog of Euler’s summation formula, Mathematical Notes, Vol. 71, No. 6, 2002, 851–856.

[18] J. Borwein and D. Bailey, A curious anomaly in the Gregory series, Mathematics by Experiment: Plausible Reasoning in the 21st Century (2003), Wellesley, MA: A K Peters, 48–50.

[19] http://mathworld.wolfram.com/GregorySeries.html.

[20] http://mathworld.wolfram.com/NaturalLogarithmof2.html.

2010 Mathematics Subject Classification: Primary 40G99; Secondary 40A05. Keywords: Rapidly convergent summation formulas for finite sums of functions involving Stirling series; Inverse factorial series; Stirling series; Stirling summation formulas; Conver- gent version of the Euler-Maclaurin summation formula; Convergent version of the Boole summation formula; Weniger transformation; Stirling numbers of the first kind; Stirling numbers of the second kind; Bernoulli numbers; Euler numbers; tangent numbers; Gregory numbers; Convergent versions of Stirling’s formula; Convergent formulas for the harmonic series; Gregory-Leibniz series; alternating harmonic series.