\frac{u}{v}\right) ^{\alpha }\left( 1-q\right) ^{\alpha -1}\Gamma _{q}\left(
\alpha \right) \sum\limits_{n\geq 0}A_{n}\frac{\left( q^{\alpha };q\right)
_{n}}{k\left( \dfrac{u}{v};\alpha +n\right) }\left( \frac{u}{v}\right) ^{n}.
\tag{37}$$<span style="font-variant:small-caps;">Proof</span> Let the hypothesis of the theorem be satisfied for some $\alpha >0.$ Then, on account of $\left( 22\right) ,$ we declare that$$\begin{aligned}
_{q}N\left( x^{\alpha -1}f\left( x\right) \right) \left( u;v\right) &=&\frac{1}{\left( -\frac{v}{u};q\right) _{\infty }}\sum_{k\in 
\mathbb{Z}
}q^{\alpha k}\sum\limits_{n\geq 0}A_{n}q^{kn}\left( -\frac{v}{u};q\right)
_{k}  \notag \\
&=&\frac{1}{\left( -\frac{v}{u};q\right) _{\infty }}\sum\limits_{n\geq
0}A_{n}\left( \frac{u}{v}\right) ^{\alpha +n}\sum_{k\in 
\mathbb{Z}
}\left( \frac{q^{k}}{\frac{u}{v}}\right) ^{\alpha +n}\left( \frac{-v}{u};q\right) _{k}\text{ }.  \TCItag{38}\end{aligned}$$Hence, by taking into account the fact that$$\Gamma _{q}\left( \alpha \right) =\frac{K\left( A;\alpha \right) }{\left(
1-q\right) ^{\alpha -1}\left( -\left( 1/A\right) ;q\right) _{\infty }}\sum_{k\in 
\mathbb{Z}
}\left( \frac{q^{k}}{A}\right) ^{\alpha }\left( \frac{-1}{A};q\right) _{k}$$where $K\left( A;\alpha \right) =A^{\alpha -1}\dfrac{\left( -q/\alpha
;q\right) _{\infty }}{\left( -q^{t}/\alpha ;q\right) _{\infty }}\dfrac{\left( -\alpha ;q\right) _{\infty }}{\left( -\alpha q^{1-t}/;q\right)
_{\infty }}$ \[25, Equ.$\left( 24\right) ]$ (for $A=\dfrac{u}{v}$ and $\alpha
=\alpha +n),$ we get$$\begin{aligned}
_{q}N\left( x^{\alpha -1}f\left( x\right) \right) \left( u;v\right)
&=&\sum_{k\in 
\mathbb{Z}
}A_{n}\left( \frac{u}{v}\right) ^{\alpha +n}\frac{\Gamma _{q}\left( \alpha
+n\right) \left( 1-q\right) ^{\alpha +n-1}}{K\left( \frac{u}{v};\alpha
+n\right) } \\
&=&\left( \frac{u}{v}\right) ^{\alpha }\left( 1-q\right) ^{\alpha -1}\Gamma
_{q}\left( \alpha \right) \sum\limits_{n\geq 0}A_{n}\frac{\left( q^{\alpha
};q\right) _{n}}{K\left( \frac{u}{v};\alpha +n\right) }\left( \frac{u}{v}\right) ^{n}.\end{aligned}$$This completes the proof of the theorem.

As a straightforward corrollary of Theorem 9, the previous 