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}{\...
Zhou's user avatar
  • 967
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}...
Robert Israel's user avatar
9 votes
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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)\\=\...
Fedor Petrov's user avatar
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 ...
Neil Strickland's user avatar
6 votes
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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} + \...
Mark Wildon's user avatar
  • 10.8k
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(...
Fedor Petrov's user avatar
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^...
Fedor Petrov's user avatar
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^...
Nate's user avatar
  • 1,992
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 $...
Johann Cigler's user avatar

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