Asymptotics of recursion

suppose we have the following two sequences

$$\alpha_k = (k-1)\left(1-\frac {1}{1+(k+1)l}\right) \quad , k \geq 2$$ $$\beta_k = (k-1)\left(1+\frac {1}{1+(k-1)l}\right) \quad , k \geq 2$$

where $l$ is a positive constant and define the sequence $c_k$ recursively by: $$c_2 = - 1/\beta_2$$ $$c_3 = 0$$ $$c_{k+1} = \frac{\alpha_{k-1}}{\beta_{k+1}}c_{k-1} \quad , k \geq 3$$

it is not hard to see that this would give $$c_2 = - 1/\beta_2$$ $$c_{2k} = -\frac{\alpha_2}{\beta_2 }\cdot\frac{\alpha_4}{\beta_4 }\cdot\cdot\cdot \frac{\alpha_{2k-2}}{\beta_{2k-2} }\cdot \frac{1} {\beta_{2k}} \quad , k \geq 2$$

$$c_{2k+1} = 0 \quad , k \geq 1$$

apparently we must have $d_k = c_{2k} \sim k^{-(1+1/l)}$ but I have no idea how to show this. can anyone shed some light on this?

also a similar question for the recursion $$c_2 = - 1/\beta_2$$ $$c_3 = \frac{\gamma_2}{\beta_3}c_2$$ $$c_{k+1} = \frac{\alpha_{k-1}}{\beta_{k+1}}c_{k-1} + \frac{\gamma_k}{\beta_{k+1}}c_k \quad , k \geq 3$$

where $$\gamma_k = \sigma \frac{l(k^3-k)}{1+kl} \quad , k \geq 2$$ $\sigma$ being also a positive constant

How would the asymptotics look like in this case?