| context | $p\in Q$ |
| context | $u:{\mathbb N}\to{}Q\to{\mathbb A}$ |
| definition | $[n]_u(q) := \sum_{k=1}^n \dfrac{u_k(q)}{u_k(p)}$ |
And clearly the denominator must be nonzero.
E.g., for another sequence $a_n$ consider $u(n,q):=q^{a_n}$.
In particular, consider $a_n:=n*x+d$ for some $d$.
a[k_] = k x + d;
Sum[q^a[k], {k, 1, n}]
Limit[%, q -> 1]
In particular, consider $x:=1, d:=0$ for quantum_integers.
$\lim_{q\to p}[n]_u(q) = \sum_{k=1}^n 1 = n$
Also, for any sequence $(a_k)$,
$\sum_{k=1}^n a_k = \lim_{q\to 1}\dfrac{\partial}{\partial{}q} \sum_{k=1}^n q^{a_k}$
Wikipedia: q-analog