## Posts Tagged ‘constants’

### A new formula for Apery’s constant and other zeta(2n+1)?

I. Introduction

In Identities Inspired from Ramanujan’s Notebooks, Simon Plouffe recounts how, based on Ramanujan’s,

\begin{aligned}\sum_{k=1}^\infty \frac{\coth(\pi k)}{k^3} = \frac{7}{180}\pi^3\end{aligned}

he found,

\begin{aligned}\zeta(3) &= \frac{7\pi^3}{180}-2\sum_{k=1}^\infty \frac{1}{k^3(e^{2\pi k}-1)}\\ \zeta(7) &= \frac{19\pi^7}{56700}-2\sum_{k=1}^\infty\frac{1}{k^7(e^{2\pi k}-1)} \end{aligned}

and similar ones for other s = 4m+3.  On a hunch, and using Mathematica’s LatticeReduce function, I found that,

\begin{aligned}\frac{3}{2}\,\zeta(3) &= \frac{\pi^3}{24}\sqrt{2}-2\sum_{k=1}^\infty \frac{1}{k^3(e^{\pi k\sqrt{2}}-1)}-\sum_{k=1}^\infty\frac{1}{k^3(e^{2\pi k\sqrt{2}}-1)}\\ \frac{3}{2}\,\zeta(5) &= \frac{\pi^5}{270}\sqrt{2}-4\sum_{k=1}^\infty \frac{1}{k^5(e^{\pi k\sqrt{2}}-1)}+\sum_{k=1}^\infty \frac{1}{k^5(e^{2\pi k\sqrt{2}}-1)}\\ \frac{9}{2}\,\zeta(7) &= \frac{41\pi^7}{37800}\sqrt{2}-8\sum_{k=1}^\infty\frac{1}{k^7(e^{\pi k\sqrt{2}}-1)}-\sum_{k=1}^\infty\frac{1}{k^7(e^{2\pi k\sqrt{2}}-1)} \end{aligned}

etc.

II. Functions

If we define,

\begin{aligned} &U_a(s) = \sum_{k=1}^\infty \frac{1}{k^s(e^{a\pi k}-1)}\end{aligned}

then Plouffe discovered integer relations between,

$\zeta(s), \pi^s, U_1(s),\, U_2(s),\, U_4(s)$

for odd s, with s = 3 being,

\begin{aligned} &\zeta(3) = 28U_1(3)-37U_2(3)+7U_4(3)\\ &\text{and,}\\ &\pi^3 =720U_1(3)-900U_2(3)+180U_4(3)\end{aligned}

Eliminating $U_1(s),\, U_4(s)$ leads to the 3-term equalities in the Introduction.  See Chamberland’s and Lopatto’s Formulas for Odd Zeta Values.  On the other hand, by defining the function,

\begin{aligned} &V_b(s) = \sum_{k=1}^\infty \frac{1}{k^s(e^{b\pi k\sqrt{2}}-1)}\end{aligned}

I observed integer relations between,

$\zeta(s), \pi^s \sqrt{2},\, V_1(s),\, V_2(s),\, V_3(s),\, V_6(s)$

also for odd s, with s = 3 as,

\begin{aligned} &\zeta(3)-102V_1(3)+99V_2(3)+10V_3(3)-5V_6(3) = 0\\ &\text{and,}\\ &\pi^3\,\sqrt{2}-3720V_1(3)+3540V_2(3)+360V_3(3)-180V_6(3) = 0\end{aligned}

and so on.  Eliminating $V_3(s),\, V_6(s)$ leads to the 4-term equalities in the Introduction.

III. Conjecture

The 4-term equalities have coefficients that are simple except for one term. Recall that,

\begin{aligned} V_1(s) &= \sum_{k=1}^\infty \frac{1}{k^s(e^{\pi k \sqrt{2}}-1)}\\ V_2(s) &= \sum_{k=1}^\infty \frac{1}{k^s(e^{2\pi k \sqrt{2}}-1)}\end{aligned}

Conjecture:

“Using the positive case of $\pm 1$ for s = 4m+3, and the negative for s = 4m+5, then in the equation,

\begin{aligned}&F(s)\zeta(s-1)\pi\sqrt{2} = \big(2^{(s-1)/2}\pm 1\big)\frac{\zeta(s)}{2}+2^{(s-1)/2}\,V_1(s) \pm V_2(s) \end{aligned}

$F(s)$ is a rational number.”

The first few for s = {3, 7, 11,…} are $F(s) = \frac{1}{4}, \frac{41}{40}, \frac{319}{84}, \dots$ while for s = {5, 9, 13,…} are $F(s) = \frac{1}{3}, \frac{19}{11}, \frac{5017}{691},\dots$  These rationals may have a closed-form expression in terms of Bernoulli numbers, but I do not yet know the exact formulation.

### The zeta function and roots of unity

In Mathworld’s entry on the Riemann zeta function, one finds in eq. 119-121 the curious evaluations,

\begin{aligned}\sum_{n=1}^\infty [\zeta(2n)-1] &= \frac{3}{4}\\ \sum_{n=1}^\infty [\zeta(3n)-1] &= \frac{1}{3}\left[-(-1)^{2/3}H_{(3-\sqrt{-3})/2}+(-1)^{1/3}H_{(3+\sqrt{-3})/2} \right]\\ \sum_{n=1}^\infty [\zeta(4n)-1] &= \frac{1}{8}\,(7-2\pi\coth(\pi))\end{aligned}

However, using the Inverse Symbolic Calculator, the first and the third, plus another one, can also be expressed as,

\begin{aligned}\sum_{n=1}^\infty [\zeta(2n)-1] &= \frac{5}{4}-\sum_{n=1}^\infty \frac{1}{2n^2+2n} = \frac{3}{4}\\ \sum_{n=1}^\infty [\zeta(4n)-1] &= \frac{5}{8}-\sum_{n=1}^\infty \frac{1}{2n^2+2}=\frac{7}{8}-\frac{1}{4}\,\pi i \cot(\pi w_4)\\ \sum_{n=1}^\infty [\zeta(6n)-1] &= \frac{5}{12}-\sum_{n=1}^\infty \frac{1}{2n^2+2n+2}=\frac{11}{12}-\frac{1}{6}\sqrt{3}\pi i\cot(\pi w_6)\end{aligned}

where $w_p = e^{2\pi i/p}$.  Interesting similar forms, isn’t it?

Unfortunately, it doesn’t seem to generalize to $\zeta(pn)$ for p = 8.  However, there is still p = 3 and, based on the even case, I assumed perhaps roots of unity are also involved.  First, given the Euler-Mascheroni constant $\gamma$, and the digamma function,

$\psi_0(z) = \psi[z]$

where we suppress the subscript for ease of notation.  Define,

$u_p = e^{\pi i/p } = (-1)^{1/p}$

and the pth root chosen such that $(-1)^{1/p} \not = -1$, then I found that p = 3 generalizes as,

\begin{aligned} 3\sum_{n=1}^\infty [\zeta(3n)-1] &= 3 + \gamma + u_3^{-1}\, \psi[u_3^{-1}]+u_3\,\psi[u_3]\\&= 0.66506\dots\\ 5\sum_{n=1}^\infty [\zeta(5n)-1] &= 6 + \gamma + \sum_{k=0}^1 \Big(u_5^{-(2k+1)}\, \psi[u_5^{-(2k+1)}]+u_5^{(2k+1)}\,\psi[u_5^{(2k+1)}]\Big)\\&=0.18976\dots\\ 7\sum_{n=1}^\infty [\zeta(7n)-1] &= 9 + \gamma + \sum_{k=0}^2 \Big(u_7^{-(2k+1)}\, \psi[u_7^{-(2k+1)}]+u_7^{(2k+1)}\,\psi[u_7^{(2k+1)}]\Big)\\&=0.05887\dots\end{aligned}

and so on, though a rigorous proof is needed that it holds true for all odd numbers p.

P.S. Going back to even p, note that p = 2, 4, 6 can also be expressed by the digamma function since,

\begin{aligned} &\sum_{n=1}^\infty \frac{1}{an^2+bn+c} = \frac{1}{\sqrt{b^2-4ac}}\Big(\psi[\tfrac{2a+b+\sqrt{b^2-4ac}}{2a}]-\psi[\tfrac{2a+b-\sqrt{b^2-4ac}}{2a}]\Big)\end{aligned}

for $a \not=0$.

### A missing binomial sum identity?

D.Bailey, J. Borwein, and D.Bradley found the beautiful pair involving binomial sums.  In Theorem 1 of this paper (2008), let x $\not=$ non-zero integer, then,

\begin{aligned} \sum_{k=1}^\infty \frac{1}{k^2-x^2} &= 3\,\sum_{k=1}^\infty \frac{1}{k^2 \binom{2k}k (1-x^2/k^2)} \prod_{m=1}^{k-1} \left(\frac{1-4x^2/m^2}{1-x^2/m^2}\right)\\[2.5mm] \sum_{k=1}^\infty \frac{1}{k^3(1-x^4/k^4)} &= \frac{5}{2}\,\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^3 \binom{2k}k (1-x^4/k^4)} \prod_{m=1}^{k-1} \left(\frac{1+4x^4/m^4}{1-x^4/m^4}\right)\end{aligned}

When x = 0, they reduce into,

\begin{aligned} \zeta(2) &= 3\,\sum_{k=1}^\infty \frac{1}{k^2\binom{2k}k }\\ \zeta(3) &= \frac{5}{2}\,\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^3\binom{2k}k }\end{aligned}

However, there is a third single-term equality,

\begin{aligned} \frac{17}{36}\,\zeta(4) &= \sum_{k=1}^\infty \frac{1}{k^4\binom{2k}k }\end{aligned}

so there might be a third identity that reduces to this as the special case x = 0.

To compare, there are three identities such that as $x \to 0$, then those zeta values are the respective limit.  For x $\not=$ integer, then,

\begin{aligned} \sum_{k=1}^\infty \frac{1}{k^2\,\binom{2k}k}\,\frac{3k^2+x^2}{k^2-x^2}\prod_{m=1}^{k-1}\left(1-\frac{x^2}{m^2}\right) &=\frac{\pi x \csc(\pi x)-1}{x^2}\\[2.5mm] \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k^3\,\binom{2k}k}\,\frac{5k^2-x^2}{2(k^2-x^2)}\prod_{m=1}^{k-1}\left(1-\frac{x^2}{m^2}\right) &=\frac{-\psi^{(0)}(1-x)-\psi^{(0)}(1+x)-2\gamma}{2x^2}\\[2.5mm] \sum_{k=1}^\infty \frac{1}{k^2\,\binom{2k}k}\,\frac{1}{k^2-x^2}\prod_{m=1}^{k-1}\left(1-\frac{x^2}{m^2}\right) &=\frac{\pi x \csc(\pi x)+3\cos(\pi x/3)-4}{4x^4}\end{aligned}

The first two were found by Leshchiner and Koecher, respectively, while the third is Theorem 2 in the same paper by Bailey, Borwein, and Bradley. The function $\psi^{(0)}$ is given in Mathematica as,

$\psi^{(0)}(z) = \text{PolyGamma[0,z]}$

while $\gamma$ is the Euler-Mascheroni constant.  So are the Bailey-Borwein-Bradley pair of binomial sum identities in fact a triplet?

### Sequences 2, Padovan and Perrin numbers

Just like the golden ratio and tribonacci constant, powers of the plastic constant P can also be expressed in terms of sequences associated with it. P is a root of the equation,

$P^3=P+1$

or,

$P = \frac{1}{3}\left(\frac{27+3\sqrt{69}}{2}\right)^{1/3}+\frac{1}{3}\left(\frac{27-3\sqrt{69}}{2}\right)^{1/3}$

Define,

\begin{aligned} a & = \left(\tfrac{27+3\sqrt{69}}{2}\right)^{1/3}\\b&=\left(\tfrac{27-3\sqrt{69}}{2}\right)^{1/3}\end{aligned}

then powers of P  are,

$P^{n} = \frac{1}{9}(a^2+b^2)U_{n+1}+\frac{1}{3}(a+b)U_{n+2}+\frac{1}{3}V_n$

where U and V are the Padovan and Perrin sequences, respectively,

\begin{aligned} U_n &= 1,0,0,1,0,1,1,1,2,2,3,4,5,7,9,12,16\dots\\ V_n &=3,0,2,3,2,5,5,7,10,12,17,22,29,\dots\end{aligned}

$P = \frac{0}{9}(a^2+b^2)+\frac{1}{3}(a+b)+\frac{0}{3}$

$P^2 =\frac{1}{9}(a^2+b^2)+\frac{0}{3}(a+b)+\frac{2}{3}$

$P^3 =\frac{0}{9}(a^2+b^2)+\frac{1}{3}(a+b)+\frac{3}{3}$

and so on.  These sequences obey,

$W_n = W_{n-2} + W_{n-3}$

and their limiting ratio, of course, is P.  While the Fibonacci sequence has a nice representation as a square spiral, the Padovan is a spiral of equilateral triangles,

The Perrin sequence also has a notable feature regarding primality testing.  Let $x_1, x_2, x_3$ be the roots of,

$P^3=P+1$

then, starting with n = 0,

$V_n=x_1^n+x_2^n+x_3^n = 3,0,2,3,2,5,5,7,10,12,17,22,29,\dots$

Indexed in this manner, if n is prime, then n divides $V_n$.  For example $V_{11} = 22$.  However, there are Perrin pseudoprimes, composite numbers that pass this test, with the smallest being n = 521^2.

Lastly, like all the four limiting ratios of this family of recurrences, the plastic constant P  can be expressed in terms of the Dedekind eta function as,

\begin{aligned} P &=\frac{e^{\pi i/24}\,\eta(\tau) }{\sqrt{2}\,\eta(2\tau)}\end{aligned}

where,

$\tau=\frac{1+\sqrt{-23}}{2}$

### Ramanujan’s continued fraction for Catalan’s constant

Ramanujan was a goldmine when it came to continued fractions (and many others).  In this post, two families will be given: they involve pi and Catalan’s constant as special cases.  However, versions will be given that roughly double the rate of convergence.

Recall that Catalan’s constant C is given by,

$C = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2} = 0.915965\dots$

Ramanujan gave the beautiful pair of continued fractions.  Let $|x| > 1$, then,

$f(x) = 2\sum_{n=0}^\infty \frac{(-1)^n}{(x+2n+1)^2} = \cfrac{1}{x^2-1 + \cfrac{2^2}{1 + \cfrac{2^2}{x^2-1 + \cfrac{4^2}{1 + \cfrac{4^2}{x^2-1 +\ddots}}}}}$

$g(x) \; = \;\frac{\Gamma^4 (\frac{x+1}{4})}{\Gamma^4 (\frac{x+3}{4})}\; =\; \cfrac{16}{x^2-1 + \cfrac{1^2}{1 + \cfrac{1^2}{x^2-1 + \cfrac{3^2}{1 + \cfrac{3^2}{x^2-1 +\ddots}}}}}$

One can just admire Ramanujan’s artistry — one continued fraction uses even numerators, the other, odd numerators.  For even x > 0, it is easily seen the first one involves Catalan’s constant.  For example,

$f(2) = 2(1-C)$

On the other hand, for odd x > 1, both involve pi,

$f(3) = \frac{1}{24}(12-\pi^2)$

$g(3) = \pi^2$

Notice though that numerators are repeated. Thanks to the insight of J.M. from a mathstackexchange post about Apery’s constant, we can speed up the rate of convergence of this particular form by getting rid of every other level and extracting even convergents.  After some slightly tedious algebraic manipulation, given,

$y = \cfrac{a_1}{b_1 + \cfrac{a_2}{1 + \cfrac{a_3}{b_2 + \cfrac{a_4}{1 +\ddots}}}}$

then,

$y_{even} = \cfrac{a_1}{b_1+a_2 - \cfrac{a_2\, a_3}{b_2+a_3+a_4 - \cfrac{a_4\, a_5}{b_3+a_5+a_6 - \cfrac{a_6\, a_7}{b_4+a_7+a_8 -\ddots}}}}$

So Ramanujan’s continued fractions are now the more compact,

$f(x) = 2\sum_{n=0}^\infty \frac{(-1)^n}{(x+2n+1)^2} = \cfrac{1}{u_1 - \cfrac{2^4}{u_2 - \cfrac{4^4}{u_3 - \cfrac{6^4}{u_4 -\ddots}}}}$

where, starting with n = 1,

$u_n = (2n-2)^2 + (2n)^2 + x^2-1$

and,

$g(x) \; = \; \frac{\Gamma^4 (\frac{x+1}{4})}{\Gamma^4(\frac{x+3}{4})}\; =\; \cfrac{16}{-1+v_1 - \cfrac{1^4}{v_2 - \cfrac{3^4}{v_3 - \cfrac{5^4}{v_4 -\ddots}}}}$

$v_n = (2n-3)^2 + (2n-1)^2 + x^2-1$

Thus, we have the slightly faster continued fraction for Catalan’s constant C,

$f(2) = 2(1-C) = \cfrac{1}{u_1 - \cfrac{2^4}{u_2 - \cfrac{4^4}{u_3 - \cfrac{6^4}{u_4 -\ddots}}}}$

where,

$u_n = 8n^2-8n+7$

(There is an even faster one by Zudilin given in An Apery-like difference equation for Catalan’s constant though he states this still does not prove C is irrational.)