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Results tagged with sequences-and-series
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user 82588
for questions about sequences and series, e.g. convergence, closed form expressions, etc. Note that there is a different tag for spectral sequences, and also note that MathOverflow is not for homework. Please consider consulting the online encyclopedia for integer sequences, if you are trying to identify a given sequence that you have found in your research.
6
votes
Double series problems
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 …
- 5,624
1
vote
Accepted
Does $\sum_{n=-\infty}^\infty (bq^n,p/aq^n;p)_\infty z^n q^{n(n-1)/2}$ have a closed form?
It turns out that the answer is quite easy.
Take $p=q^2$,$~z=1$,$~b=q^2a$, then
$$
\sum_{n=-\infty}^\infty (bq^n,p/aq^n;p)_\infty
z^n q^{n(n-1)/2}=2 \left(q a,q/a;q^2\right){}_{\infty }\sum _{n=- …
- 5,624
8
votes
Accepted
An infinite series involving the mod-parity of Euler's totient function
The only odd values of $\phi(n)$ are $\phi(1)=\phi(2)=1$.
$\phi(n)$ is even but not divisible by $4$ when:
$n=4$
$n=2^{\left\{0,1\right\}}p^m$, where $p=4k+3$ is prime, $m=1,2,3,...$
We have
$$
\ …
- 5,624
3
votes
1
answer
1k
views
Does $\sum_{n=-\infty}^\infty (bq^n,p/aq^n;p)_\infty z^n q^{n(n-1)/2}$ have a closed form?
The formula
$$
\small\sum_{n=-\infty}^\infty (bq^n,p/aq^n;p)_\infty
z^n q^{n(n-1)/2}=\frac{(-z,-q/z;q)_\infty}{\ln\frac{1}{q}}\int\limits_0^\infty\frac{\left(bt/z,pz/at;p\right)_\infty}{\left(-t,- …
- 5,624
8
votes
Accepted
Partition numbers and Gaussian binomial coefficient
Consider Rogers-Szego polynomials defined by
$$
H_n(t)=\sum\limits_{m=0}^{n} \binom{n}{m}_qt^m.
$$
In Andrews, "Theory of partitions", exercise 6 in chapter 3 gives three term recurrence satisfied by …
- 5,624
4
votes
Accepted
Approximating a finite sum with an integral
First, we rewrite the sum as a sum over the full period
$$
S(a,N)=\frac{2}{N+1}\sum_{j=1}^{N+1} \sin^2\left( \frac{2\pi j}{N+1} \right)\sin \left( 2 a \cos \left( \frac{2\pi j}{N+1} \right) \right).
…
- 5,624
10
votes
Identity with Pochhammer and harmonic numbers
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
…
- 5,624
12
votes
Accepted
Is it true that $\sum_{k=1}^\infty\frac{\binom{2k}k^2}{k16^k}(H_{2k}-H_k)=\frac23\sum_{k=1}^...
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 …
- 5,624
3
votes
Accepted
$q$-Eulerian type B enjoy symmetry
$\bf{Step~1}.$ $B_{n,a}(q)=B_{n,n-a}(q)$.
$\it{Proof}$. Write
$$
\sum_{n\geq1}\dfrac{B_n(t,q)}{t^{n/2}}\frac{z^n}{(q;q)_n}=\frac{e(z/\sqrt{t};q)-e(z\sqrt{t};q)}{\dfrac{e(2z\sqrt{t};q)}{\sqrt{t}}-\sqr …
- 5,624
18
votes
Accepted
A mystery sequence
The conjectured identity
$$
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}
$$
using Euler's pentagona …
- 5,624
13
votes
Ramanujan's Lost Notebook page 1 first equation and OEIS sequence A260195
This conjecture is equivalent to the following
$$\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} …
- 5,624
4
votes
Accepted
Double Series involving Gamma function
This problem can be reduced at least formally to a compact double integral, which might be easier to solve.
Starting with the integral representation for the Gamma function, we write the double sum a …
- 5,624
10
votes
Accepted
generating $q$-Catalan numbers
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- …
- 5,624
25
votes
Bernoulli sum meets golden number
Using the integral representation of Bernoulli numbers I obtain formally the integral representation of the double summation
$$
\sum_{k=1}^{\infty}\sum_{j=0}^{k}\binom{k}{j}\frac{B_{j+k+1}}{j+k+1}=2\c …
- 5,624
6
votes
3 divides coefficents of this $q$-series
Trivially
$$
\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 …
- 5,624