1 Introduction

Recently, Sun [3] obtained a very nice q-analogue of Euler’s formula \(\zeta (2)=\pi ^2/6\).

Theorem 1.1

(Sun [3]) For a complex q with \(|q|<1\), we have:

$$\begin{aligned} \sum _{k=0}^\infty \dfrac{q^k(1+q^{2k+1})}{(1-q^{2k+1})^2}=\prod _{n=1}^\infty \dfrac{(1-q^{2n})^4}{(1-q^{2n-1})^4}. \end{aligned}$$
(1.1)

Motivated by Theorem 1.1, the present author obtained the q-analogue of \(\zeta (4)=\pi ^4/90\) and noted that it was simultaneously and independently obtained by Sun in his subsequent revised paper.

Theorem 1.2

(Sun [3]) For a complex q with \(|q|<1\), we have:

$$\begin{aligned} \sum _{k=0}^\infty \dfrac{q^{2k}(1+4q^{2k+1}+q^{4k+2})}{(1-q^{2k+1})^4}=\prod _{n=1}^\infty \dfrac{(1-q^{2n})^8}{(1-q^{2n-1})^8}. \end{aligned}$$
(1.2)

Furthermore, Sun commented that one does not know how to find q-analogues of Euler’s formula for \(\zeta (6)\) and beyond, similar to Theorems 1.1 and 1.2. This further motivated the author to consider the problem, and indeed, we obtained the q-analogue of \(\zeta (6)\). As we shall see shortly, the q-analogue formulation of \(\zeta (6)\) is more difficult as compared to \(\zeta (2)\) and \(\zeta (4)\) due to an extra term that shows up in the identity; however, in the limit as \(q\uparrow 1\) (where \(q\uparrow 1\) means q is approaching 1 from inside the unit disk), this term vanishes. We also state the q-analogue of \(\zeta (4)=\pi ^4/90\), since we found it independently of Sun’s result; however, we skip the proof of this, since it essentially uses the same idea as Sun.

We emphasize here that the q-analogue of \(\zeta (6)=\pi ^6/945\) is the first non-trivial case where we notice the occurrence of an interesting extra term which essentially is the twelfth power of a well-known function of Euler (see Theorem 2.2). After obtaining this result, we obtained q-analogues of Euler’s general formula for \(\zeta (2k), k=4, 5,\ldots \) (see [1]). Each of these q-analogues has an extra term that arises from the general theory of modular forms all of which approach zero in the limit \(q\uparrow 1\). The case \(k=3\) or the q-analogue of \(\zeta (6)\) is special, since the extra term that we obtain in this case has a beautiful product representation, and has connections to well-known identities of Euler (see below).

2 Main Theorems

Theorem 2.1

For a complex q with \(|q|<1\), we have:

$$\begin{aligned} \sum _{k=0}^\infty \dfrac{q^{2k}\;P_2(q^{2k+1})}{(1-q^{2k+1})^4}=\prod _{n=1}^\infty \dfrac{(1-q^{2n})^8}{(1-q^{2n-1})^8}, \end{aligned}$$
(2.1)

where \(P_2(x) = x^2+4x+1\). In other words, (2.1) gives a q-analogue of \(\zeta (4)=\pi ^4/90\).

Theorem 2.2

For a complex q with \(|q|<1\), we have:

$$\begin{aligned} \sum _{k=0}^\infty \dfrac{q^{k}(1+q^{2k+1})\;P_4(q^{2k+1})}{(1-q^{2k+1})^6}-\phi ^{12}(q)=256q\prod _{n=1}^\infty \dfrac{(1-q^{2n})^{12}}{(1-q^{2n-1})^{12}}, \end{aligned}$$
(2.2)

where \(P_4(x) = x^4+236x^3+1446x^2+236x+1\) and \(\phi (q)=\displaystyle \prod \nolimits _{n=1}^\infty (1-q^n)\) is Euler’s function. In other words, (2.2) gives a q-analogue of \(\zeta (6)=\pi ^6/945\).

Remark 2.3

We note that \(\phi ^{12}(q)\) has a beautiful product representation and is uniquely determined by:

$$\begin{aligned} \phi ^{12}(q)=\sum _{k=0}^\infty \dfrac{q^{k}(1+q^{2k+1})\;P_4(q^{2k+1})}{(1-q^{2k+1})^6}-256q\prod _{n=1}^\infty \dfrac{(1-q^{2n})^{12}}{(1-q^{2n-1})^{12}}. \end{aligned}$$
(2.3)

In the general q-analogue formulation (see [1]), we do not have very elegant representations of these functions, although we obtain expressions for them similar to (2.3).

Remark 2.4

Since the coefficients in the q-series expansion of \(\phi ^{12}(q)\) are related to the pentagonal numbers by Euler’s pentagonal number theorem, and the coefficients of the product in the right-hand side of (2.2) are related to the triangular numbers, it will be worthwhile to understand the relationships of these coefficients via identity (2.2).

3 Some Useful Lemmas

Let \(q=e^{2\pi i\tau }\), \(\tau \in {\mathcal {H}}\) where \({\mathcal {H}}=\{\tau \in {\mathbb {C}} : \text{ Im }(\tau )>0\)}. Then, the Dedekind \(\eta \)-function defined by:

$$\begin{aligned} \eta (\tau ) = q^{1/24}\prod _{n=1}^\infty (1-q^n), \end{aligned}$$
(3.1)

is a modular form of weight 1/2. Also, let us denote by \(\psi (q)\) the following sum:

$$\begin{aligned} \psi (q)=\sum _{n=0}^\infty q^{T_n}, \end{aligned}$$
(3.2)

where \(T_n=\dfrac{n(n+1)}{2}\) (for \(n=0,1,2,\ldots \)) are triangular numbers. Then, we have the following well-known result due to Gauss:

Lemma 3.1

$$\begin{aligned} \psi (q)=\prod _{n=1}^\infty \dfrac{(1-q^{2n})}{(1-q^{2n-1})}. \end{aligned}$$
(3.3)

Thus, we have from Lemma 3.1 that:

$$\begin{aligned} \prod _{n=1}^\infty \dfrac{(1-q^{2n})^{12}}{(1-q^{2n-1})^{12}} = \psi ^{12}(q) = \sum _{n=1}^\infty t_{12}(n) q^n, \end{aligned}$$
(3.4)

where \(t_{12}(n)\) is the number of ways of representing a positive integer n as a sum of 12 triangular numbers. Next, we have the following well-known result of Ono, Robins and Wahl [2].

Theorem 3.2

Let \(\eta ^{12}(2\tau )=\sum \nolimits _{k=0}^\infty a(2k+1)q^{2k+1}\). Then, for a positive integer n, we have:

$$\begin{aligned} t_{12}(n) = \dfrac{\sigma _5(2n+3)-a(2n+3)}{256}, \end{aligned}$$
(3.5)

where

$$\begin{aligned} \sigma _5(n) = \sum _{d|n}d^5. \end{aligned}$$
(3.6)

4 Proof of Theorem 2.2

Since \(\zeta (6)=\dfrac{\pi ^6}{945}\) has the following equivalent form:

$$\begin{aligned} \sum _{k=0}^\infty \dfrac{1}{(2k+1)^6} = \dfrac{63}{64}\zeta (6) = \dfrac{\pi ^6}{960}, \end{aligned}$$
(4.1)

it will be sufficient to get the q-analogue of (4.1). Now, from q-analogue of Euler’s Gamma function, we know that:

$$\begin{aligned} \lim _{q\uparrow 1} \;(1-q)\prod _{n=1}^\infty \dfrac{(1-q^{2n})^2}{(1-q^{2n-1})^2}=\dfrac{\pi }{2}, \end{aligned}$$
(4.2)

so that from (4.2), we have:

$$\begin{aligned} \lim _{q\uparrow 1} \;(1-q)^6\prod _{n=1}^\infty \dfrac{(1-q^{2n})^{12}}{(1-q^{2n-1})^{12}}=\dfrac{\pi ^6}{64}. \end{aligned}$$
(4.3)

Next, we consider the following infinite series

$$\begin{aligned} S_6(q) := \sum _{k=0}^\infty \dfrac{q^{k}(1+q^{2k+1})\;P_4(q^{2k+1})}{(1-q^{2k+1})^6}, \end{aligned}$$
(4.4)

where \(P_4(x)=x^4+236x^3+1446x^2+236x+1\).

By partial fractions, we have:

$$\begin{aligned} S_6(q) = \sum _{k=0}^\infty q^k\left\{ \dfrac{3840}{(1-q^{2k+1})^6}-\dfrac{9600}{(1-q^{2k+1})^5}+\dfrac{8160}{(1-q^{2k+1})^4}\right. \nonumber \\\left. -\dfrac{2640}{(1-q^{2k+1})^3}+\dfrac{242}{(1-q^{2k+1})^2}-\dfrac{1}{(1-q^{2k+1})}\right\} . \end{aligned}$$
(4.5)

Lemma 4.1

With \(S_6(q)\) represented by (4.5), we have:

$$\begin{aligned} S_6(q) = 256q\sum _{n=0}^\infty t_{12}(n)q^n + \phi ^{12}(q). \end{aligned}$$
(4.6)

Proof

From (4.5), we have:

$$\begin{aligned} S_6(q)&=\sum _{k=0}^\infty \sum _{j=0}^\infty q^{k}\left\{ 3840\left( {\begin{array}{c}-6\\ j\end{array}}\right) -9600\left( {\begin{array}{c}-5\\ j\end{array}}\right) +8160\left( {\begin{array}{c}-4\\ j\end{array}}\right) \right. \\&\quad \left. -2640\left( {\begin{array}{c}-3\\ j\end{array}}\right) +242\left( {\begin{array}{c}-2\\ j\end{array}}\right) -\left( {\begin{array}{c}-1\\ j\end{array}}\right) \right\} (-q)^{j(2k+1)}\\&=\sum _{k=0}^\infty \sum _{j=0}^\infty \left\{ 32(j+1)(j+2)(j+3)(j+4)(j+5)\right. \\&\quad \left. -400(j+1)(j+2)(j+3)(j+4)+1360(j+1)(j+2)(j+3)\right. \\&\quad \left. -1320(j+1)(j+2)+242(j+1)-1\right\} q^{k+j(2k+1)}\\&=\sum _{k=0}^\infty \sum _{j=0}^\infty (2j+1)^5 q^{\frac{(2j+1)(2k+1)-1}{2}}\\&=\sum _{n=0}^\infty \sigma _5(2n+1)q^n\\&=1 + \sum _{n=1}^\infty \sigma _5(2n+1)q^n\\&=1 + q\sum _{n=0}^\infty \sigma _5(2n+3)q^{n}. \end{aligned}$$

Also from (3.1), we have:

$$\begin{aligned} \phi ^{12}(q)&=\dfrac{\eta ^{12}(\tau )}{q^{\frac{1}{2}}}\\&=\sum _{n=0}^\infty a(2n+1)q^n\\&=1+\sum _{n=1}^\infty a(2n+1)q^n\\&=1+q\sum _{n=0}^\infty a(2n+3)q^n. \end{aligned}$$

Thus, from above, we have:

$$\begin{aligned} S_6(q)-\phi ^{12}(q)&=q\sum _{n=0}^\infty \left\{ \sigma _5(2n+3)-a(2n+3)\right\} q^n\\&=256\;q\sum _{n=0}^\infty t_{12}(n) q^n, \end{aligned}$$

where the last step follows from Theorem 3.2. This completes the proof of Theorem 2.2. \(\square \)

We also note that

$$\begin{aligned} \lim _{q\uparrow 1}\;(1-q)^6 (S_6(q)-\phi ^{12}(q))&=\lim _{q\uparrow 1}\;(1-q)^6 S_6(q)-\lim _{q\uparrow 1}\;(1-q)^6\phi ^{12}(q)\nonumber \\&=\sum _{k=0}^\infty \dfrac{3840}{(2k+1)^6}, \end{aligned}$$
(4.7)

where \(\lim _{q\uparrow 1}\;(1-q)^6\phi ^{12}(q)=0\) and \(q\uparrow 1\) indicates \(q\rightarrow 1\) from within the unit disk. Hence, combining Eqs. (4.1), (4.3), (4.7), and Lemma 4.1, Theorem 2.2 follows.