Search Results

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 …

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 …

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 …

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 …

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 …

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 …

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 …

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} …

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 …

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- …

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

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 …

Exhaustive analysis of lattice sums can be found in the book Lattice sums then and now. This book can be found for free on the web. OP's sum is given by formula 1.3.14 on page 33: L. Lorenz. Bidrag …

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 …

$\bf{Step~1}.$ $B_{n,a}(q)=B_{n,n-a}(q)$. $\it{Proof}$. 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 …