Pith. sign in

REVIEW

Two q-analogues of Euler's formula zeta(2)=π²/6

Not yet reviewed by Pith; the record is open.

This paper has not been read by Pith yet. Machine review is queued; the pith claim, tier, and objections will appear here once it completes.

SPECIMEN: schema-true, not a live event

T0 review · schema-true

One-sentence machine reading of the paper's core claim.

pith:XXXXXXXX · record.json · timestamp

arxiv 1802.01473 v6 pith:5ETSQMVE submitted 2018-02-05 math.NT math.CO

Two q-analogues of Euler's formula zeta(2)=π²/6

classification math.NT math.CO
keywords fracinftyzetaanalogueseulerformulaprodanalogue
verification ladder T0 review T1 audit T2 compute T3 formal T4 reserved
0 comments
read the original 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 $\zeta(2m)\ (m=3,4,\ldots)$.

discussion (0)

Sign in with ORCID, Apple, or X to comment. Anyone can read and Pith papers without signing in.