13
votes
Accepted
Searching for a proof for a series identity
Follow the comments of Lucia and note that
$$\sum_{n\ge 1}\frac{n^2q^n}{(1-q^n)^2}=q\frac{\,d}{\,dq}\sum_{n\ge 1}\frac{nq^n}{1-q^n}.$$
I believe the identity actually is the well known
$$q\frac{\,d}{\...
9
votes
Searching for a proof for a series identity
$$ \sum_{k=1}^\infty \frac{k^2 q^k}{(1-q^k)^2} = \sum_{n=1}^\infty \sigma(n) n q^n$$
$$ \sum_{k=1}^\infty \frac{k q^k}{1-q^k} = \sum_{n=1}^\infty \sigma(n) q^n$$
$$ \left(\sum_{k=1}^\infty \frac{k q^k}...
9
votes
Accepted
A curious $q$-series identity on a truncated Euler function
Let ${n\choose k}_q=\frac{(q)_n}{(q)_k(q)_{n-k}}$ denote a $q$-binomial coefficient. We start with the following version of $q$-Vandermonde convolution identity:
$$
(x-y)(x-qy)\ldots(x-q^{n-1}y)\\=\...
6
votes
Accepted
(Conceptual) proof and/or interpretation of a $q$-binomial identity
Note that it will be enough to check the identity when $q$ is a prime power, in which case we can choose a field $F$ with $|F|=q$, and vector spaces $A$ and $B$ of dimension $a$ and $b$ over $F$. In ...
6
votes
Accepted
A $q$-series identity: proof request
As darij grinberg says, the result is equivalent to the main result in the linked paper. On the right-hand side we have
$$\sum_{k=1}^\infty \frac{q^k}{1-q^k} = \sum_{k=1}^\infty (q^k + q^{2k} + \...
5
votes
(Conceptual) proof and/or interpretation of a $q$-binomial identity
Denoting $q^a=x$ (and fixing $b$), this is an expansion of the polynomial $x^b$ in the basis of $f_k(x):=(1-x)(1-x/q)\ldots (1-x/q^{k-1})$:
$$
x^b=\sum_{k=0}^b (-1)^k q^{{k\choose 2}}{b\choose k}_qf_k(...
5
votes
On Ramanujan's beautiful cubic identity
I do not know how Ramanujan proved this, but to prove such a thing is not really a big deal. Say, $a_n$ is a linear combination of $p_i^n$, where $(1-p_1x)(1-p_2x)(1-p_3x)=1-82x-82x^2+x^3$. Thus $a_n^...
4
votes
Accepted
Sum of $q$-binomial coefficients
Let's look at the ratio of two adjacent $q$-binomials as we move away from the center, for simplicity I'll do the even case.
$\binom{2n}{n-a}_q / \binom{2n}{n-a-1}_q = \frac{[n+a+1]_q}{[n-a]_q} > q^...
4
votes
Accepted
notation for $(a-b)(a-qb)\dots (a-q^{n-1}b)$
There are some different notations in the literature:
B.A. Kupershmidt used $ {(a\dot - b)^n} $,
Victor Kac and Pokman Cheung in “Quantum Calculus” used $(a-b)_q^n$.
In my lecture notes I used $...
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