相关论文: $q$-Analogues of some series for powers of $\pi$

We obtain a new $q$-analogue of the classical Leibniz series $\sum_{k=0}^\infty(-1)^k/(2k+1)=\pi/4$, namely \begin{equation*}…

We give $q$-analogues of the following two Ramanujan-type formulas for $1/\pi$: \begin{align*} \sum_{k=0}^\infty (6k+1)\frac{(\frac{1}{2})_k^3}{k!^3 4^k} =\frac{4}{\pi} \quad\text{and}\quad \sum_{k=0}^\infty…

Motivated by Ramanujan-type series and Zeilberger-type series, in this paper we investigate two new types of series for powers of $\pi$. For example, we prove that…

Via symbolic computation we deduce 97 new type series for powers of $\pi$ related to Ramanujan-type series. Here are three typical examples: $$\sum_{k=0}^\infty \frac{P(k) \binom{2k}k\binom{3k}k…

Let $H_n^{(2)}$ denote the second-order harmonic number $\sum_{0<k\le n}1/k^2$ for $n=0,1,2,\ldots$. In this paper we obtain the following identity: $$\sum_{k=1}^\infty\frac{2^kH_{k-1}^{(2)}}{k\binom{2k}k}=\frac{\pi^3}{48}.$$ We explain how…

In this paper, we evaluate some series of the form $$\sum_{k=1}^\infty\frac{ak^2+bk+c}{k(3k-1)(3k-2)m^k\binom{4k}k}.$$ For example, we prove that $$\sum_{k=1}^\infty\frac{(5k^2-4k+1)8^{k}}{k(3k-1)(3k-2)\binom{4k}k}=\frac{3}2\pi$$ and…

We give the q-analogue of the sums of the n-th powers of positive integers up to k-1.

Let $$ S_{m,n}(q):=\sum_{k=1}^{n}\frac{1-q^{2k}}{1-q^2} (\frac{1-q^k}{1-q})^{m-1}q^{\frac{m+1}{2}(n-k)}. $$ Generalizing the formulas of Warnaar and Schlosser, we prove that there exist polynomials $P_{m,k}(q)\in\mathbb{Z}[q]$ such that $$…

$q$-analogs of sum equals integral relations $\sum_{n\in\mathbb{Z}}f(n)=\int_{-\infty}^\infty f(x)dx$ for sinc functions and binomial coefficients are studied. Such analogs are already known in the context of $q$-hypergeometric series. This…

By applying multiplicate forms of the Carlitz inverse series relations to the $q$-Pfaff-Saalsch{\"u}tz summation theorem, we establish twenty five nonterminating $q$-series identities with several of them serving as $q$-analogues of…

In this paper, by constructing some identities, we prove some $q$-analogues of some congruences. For example, for any odd integer $n>1$, we show that \begin{gather*} \sum_{k=0}^{n-1} \frac{(q^{-1};q^2)_k}{(q;q)_k} q^k \equiv (-1)^{(n+1)/2}…

For any odd prime p we obtain q-analogues of Van Hamme's supercongruence: $$ \sum_{k=0}^{\frac{p-1}{2}}{2k\choose k}^3\frac{1}{64^k} \equiv 0 \pmod{p^2} \quad\text{for}\quad p\equiv 3\pmod 4, $$ and Rodriguez-Villegas' Beukers-like…

In this paper we construct the $q$-analogue of Barnes's Bernoulli numbers and polynomials of degree 2, for positive even integers, which is an answer to a part of Schlosser's question. For positive odd integers, Schlosser's question is…

In this paper we construct the q-analogue of Barnes' Bernoulli numbers and plynomials of degree 2, which is an answer to a part of Schlosser's question. Finally, we treat the q-analogue of the sums of powers of consecutive integrs.

We give a $q$-congruence whose specializations $q=-1$ and $q=1$ correspond to supercongruences (B.2) and (H.2) on Van Hamme's 1997 list: $$ \sum_{k=0}^{(p-1)/2}(-1)^k(4k+1)A_k\equiv p(-1)^{(p-1)/2}\pmod{p^3} \quad\text{and}\quad…

We prove that $\sum_{k=0}^{q-1}\binom{2k}{k}\equiv q^2\pmod{3q^2}$ if q>1 is a power of 3, as recently conjectured by Z.W. Sun and R. Tauraso. Our more precise result actually implies that the value of $(1/q^2)\sum_{k=0}^{q-1}\binom{2k}{k}$…

In this paper we apply a formula of the very-well poised $_{2k+4}\phi_{2k+3}$ to write a $k$-tuple sum of $q$-series as a linear combination of terms wherein each term is a product of expressions of the form $\frac{1}{(qy,…

Following an idea due to J. Bernoulli, we explore the q-analogue of the sums of powers of consecutive integers.

In this paper, we derive eight basic identities of symmetry in three variables related to $q$-Bernoulli polynomials and the $q$-analogue of power sums. These and most of their corollaries are new, since there have been results only about…

It is known that $\sum_{k=0}^\infty\binom{2k}{k}/((2k+1)4^k)=\pi/2$ and $\sum_{k=0}^\infty\binom{2k}{k}/((2k+1)16^k)=\pi/3$. In this paper we obtain their p-adic analogues such as…