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Two $q$-analogues of Euler’s formula $\zeta (2)=\pi ^2/6$

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It is well known that $\zeta(2)=\pi^2/6$ as discovered by Euler. In this paper we present the following two $q$-analogues of this celebrated formula: $$\sum_{k=0}^\infty\frac{q^k(1+q^{2k+1})}{(1-q^{2k+1})^2}=\prod_{n=1}^\infty\frac{(1-q^{2n})^4}{(1-q^{2n-1})^4}$$ and $$\sum_{k=0}^\infty\frac{q^{2k-\lfloor(-1)^kk/2\rfloor}}{(1-q^{2k+1})^2} =\prod_{n=1}^\infty\frac{(1-q^{2n})^2(1-q^{4n})^2}{(1-q^{2n-1})^2(1-q^{4n-2})^2},$$ where $q$ is… Click to show full abstract

It is well known that $\zeta(2)=\pi^2/6$ as discovered by Euler. In this paper we present the following two $q$-analogues of this celebrated formula: $$\sum_{k=0}^\infty\frac{q^k(1+q^{2k+1})}{(1-q^{2k+1})^2}=\prod_{n=1}^\infty\frac{(1-q^{2n})^4}{(1-q^{2n-1})^4}$$ and $$\sum_{k=0}^\infty\frac{q^{2k-\lfloor(-1)^kk/2\rfloor}}{(1-q^{2k+1})^2} =\prod_{n=1}^\infty\frac{(1-q^{2n})^2(1-q^{4n})^2}{(1-q^{2n-1})^2(1-q^{4n-2})^2},$$ where $q$ is any complex number with $|q|<1$. We also give a $q$-analogue of the identity $\zeta(4)=\pi^4/90$, and pose a problem on $q$-analogues of Euler's formula for $\zeta(2m)\ (m=3,4,\ldots)$.

Keywords: formula zeta; euler formula; two analogues; infty frac; analogues euler

Journal Title: Colloquium Mathematicum
Year Published: 2019

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