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.

About these ads

3 responses to this post.

  1. Combining both cases s=4m+3 and s=4m+5 using (-1)^s confirms the connection between the Bernoulli numbers and your V1(s),V2(s) as follows:

    Bernoulli[n_] := 4n ((-1)^n V1[1-2n]- 2^n V2[1-2n])/((-1)^n-2^n)

    BernoulliB[2n] == Bernoulli[n] for n>1

    Reply

  2. The rational F(s) relationship between V1,V2 and the Odd Zeta values can also be combined as follows

    F[s_] := (2^s V1[2 s + 1] – (-1)^s V2[2 s + 1] +
    1/2 (2^s – (-1)^s) Zeta[1 + 2 s])/(Sqrt[2] Pi^(2 s + 1))

    giving the sequence (first 10 values):

    {1/24, 1/270, 41/37800, 19/103950, 29/714420, 5017/638512875, 707339/434188755000, 1069039/3275571048750,
    30021059/451339210822500, 13097369/974482387003125}

    Reply

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

Follow

Get every new post delivered to your Inbox.

%d bloggers like this: