q-series and Finite Continued Fraction

For $n \in \mathbb{N}$,

$$\begin{align*} \mu &:= \mu _n (a,q) := \sum _{k=0} ^{ [(n+1)/2] } \frac { (q)_{n-k+1} a^k q^{k^2} }{ (q)_k (q)_{n-2k+1} } \\ \nu &:= \nu _n (a,q) := \sum _{k=0} ^{[n/2]} \frac { (q)_{n-k} a^k q^{k(k+1)} }{(q)_k (q)_{n-2k} } \end{align*}$$

Then

$$\newcommand{\surplus}{\mathbin{\genfrac{}{}{0pt}{}{}{+}}} \frac \mu \nu = 1+ \frac {aq} 1 \surplus \frac {aq^2} 1 \surplus \cdots \surplus \frac {aq^n } 1 .$$

To prove this, we define a new function to handle $\mu$ and $\nu$.

$$F_r := F_r (a,q ) := \sum _{k=0} ^{ [(n-r+1)/2] } \frac { (q)_{n-r-k+1} a^k q^{k (r+k)} } {(q)_k (q)_{n-r-2k+1} }$$

Then $F_0 = \mu, \quad F_1 = \nu$. Also we can observe

$$F_n = 1, \quad F_{n-1} = \sum _{k=0} ^1 \frac { (q)_{2-k} a^k q^{n+k-1} }{ (q)_k (q)_{2-2k} } = 1 + aq^n$$  

If we find the recurrence relation of $F_r$, we can expand known value $F_n , F_{n-1}$ to $F_1 ,F_0$. Using the fact that $\frac {1}{(q)_{-1}} = 0$,

$$\begin{align*}  F_r - F_{r-1}  &=  \sum _{k=0} ^{ [(n-r+1)/2] } \frac { (q)_{n-r-k+1} a^k q^{k (r+k)} } {(q)_k (q)_{n-r-2k+1} } - \sum_{k=0} ^{[(n-r=2)/2]} \frac { (q)_{n-r-k} a^k q^{k (r+k+1}} {(q)_k (q)_{n-r-2k} } \\ &= \sum _{k=1} ^{[(n-r+1)/2]} \frac { (q)_{n-r-k} a^k q^{k(r+k)} }{(q)_k (q)_{n-r-2k} } \Bigg( \frac {1- q^{n-r-k+1} }{1-q^{n-r-2k+1} } -q^k \Bigg) \\ &= \sum _{k=1} ^{[(n-r+1)/2]} \frac { (q)_{n-r-k} a^k q^{k(r+k)} }{(q)_k (q)_{n-r-2k} } \frac {1-q^k} {1-q^{n-r-2k+1}} \\ &= \sum _{k=1} ^{[(n-r+1)/2]} \frac { (q)_{n-r-k} a^k q^{k(r+k)} }{(q)_{k-1} (q)_{n-r-2k+1} } \\&= aq^{r+1} F_{r+2} \end{align*}$$

Using this result, we can easily get

$$\newcommand{\surplus}{\mathbin{\genfrac{}{}{0pt}{}{}{+}}} \frac \mu \nu = 1+ \frac {aq} 1 \surplus \frac {aq^2} 1 \surplus \cdots \surplus \frac {aq^n } 1 .$$