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The paper Two-dimensional series evaluations via the elliptic functions of Ramanujan and Jacobi deals exactly with double sums of this kind and shows how to evaluate them in terms of elliptic function …

The identity $$ \begin{align} \sum_{s=1}^{m}{2s\choose s}{s\choose m-s}\frac{(-1)^s }{s+1}H_{s}^{(2)}=\frac{2(-1)^m}{m+1}\sum_{s=1}^m H_{s}^{(2)}. \tag{1} \end{align} $$ is equivalent to the following …

This proves the case $\delta=0$, namely for $z^{mn-1}=1$ $$ S=\sum_{k=1}^{n-1}\left(\frac{z^k}{1+z^{km}}-(-1)^{n-k}\frac{z^k}{1-z^{km}}\right),\quad \text{Re} \,S=(-1)^{n-1}\left\lfloor \frac n2\right …

Consider Rogers-Szego polynomials defined by $$ H_n(t)=\sum\limits_{m=0}^{n} \binom{n}{m}_qt^m. $$ In Andrews, "Theory of partitions", exercise 6 in chapter 3 gives three term recurrence satisfied by …

There are such examples in Ramanujan's Notebooks, Part 2, page 116

The identity under question can be found in the paper S. Boettner, V.H. Moll The integrals in Gradshteyn and Ryzhik. Part 16: Complete elliptic integrals, pages 11-12 https://arxiv.org/abs/1005.2941 …

Let $a_n=\frac{1}{16^n}\binom{2n}n^2$. We have $$ \sum_{n\ge 1} a_n(2H_{2n}-H_n)k^{2n}=-\frac{1}{\pi}K(k)\log(1-k^2). $$ Here $K(x)=\frac{\pi}{2}\sum_{n\ge 0}a_nx^{2n}$ is complete elliptic integral o …

$\bf{Step~1}.$ $B_{n,a}(q)=B_{n,n-a}(q)$. $\it{Proof}$. Write $$ \sum_{n\geq1}\dfrac{B_n(t,q)}{t^{n/2}}\frac{z^n}{(q;q)_n}=\frac{e(z/\sqrt{t};q)-e(z\sqrt{t};q)}{\dfrac{e(2z\sqrt{t};q)}{\sqrt{t}}-\sqr …

The conjectured identity $$ f(q)=(q;q)_\infty\left(1+\sum_{k=1}^\infty q^k(-q;q)^2_{k-1}\right)=\sum_{\substack{m,n\geqslant0\\n\ne1}}(-1)^mq^{\frac{(m+n)(3m+n+1)}2},\tag{1} $$ using Euler's pentagona …

This conjecture is equivalent to the following $$\frac{q}{(1-q)^2}\sum_{n=0}^\infty(-q)^n \frac{(q;q^2){}_n(-q^2;q^2){}_n}{(q^3;q^2){}_n^2}=\sum_{1\le r,s\le t}q^{t^2-\frac{1}{2}(r^2-r+s^2-s)},\tag{1} …

The functions $$ C_n(q)=\sum_{P\in\square_n}q^{area(P)} $$ satisfy the following recurrence relation $$ C_n(q)=\sum_{k=1}^nq^{k-1}C_{k-1}(q)C_{n-k}(q).\tag{1} $$ Proof. (taken from the book "The q, t- …

Using the integral representation of Bernoulli numbers I obtain formally the integral representation of the double summation $$ \sum_{k=1}^{\infty}\sum_{j=0}^{k}\binom{k}{j}\frac{B_{j+k+1}}{j+k+1}=2\c …

Not a complete answer, more like an extended comment. It is possible to remove the inequalities by introducing new variables $y_i\ge 0$ according to \begin{align} x_n&=y_n\\ x_{n-1}-x_n&=y_{n-1}\\ \ld …

Trivially $$ \prod_{n\geq1} f_n(q)=\frac{\left(q^3;q^3\right)^4_{\infty }}{(q;q)_{\infty } \left(q^9;q^9\right){}_{\infty }}. $$ Denoting $A(q)=(q;q)_{\infty }(q^2;q^2)_{\infty }$, one can see that in …

We will use the following well known fact (e.g., see sections 1.1 and 1.2 in this article): Given $f(x)$ with period $1$, its Fourier series $$ f(x)=\sum_{j=0}^\infty a_j\cos(2\pi jx), $$ and a positi …