Asymptotics of a 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?
 A: For the first sequence, $$d_k=\frac{\alpha_{2k-2}}{\beta_{2k}}d_{k-1}=\frac{k-3/2}{k+1/2-1/\ell}d_{k-1}$$
so one has 
$$d_k=C\frac{\Gamma(k-1/2)}{\Gamma(k+3/2-1/\ell)}\ ,$$
the constant $C$ being determined by the initial condition $d_1$, namely
$$C=d_1\frac{\Gamma(5/2- 1/\ell)}{\Gamma(1/2)}\ . $$ Recall that $\Gamma(x+a)=x^a\Gamma(x)(1+o(1))$ as $x\to+\infty$, so
$$d_k=Ck^{-1-1/\ell}(1+o(1)). $$
For the second sequence, $$c_{k+1}=\frac{(k-1)(k+1)}{k+2/\ell }c_k+\frac{k-2}{k+2/\ell}c_{k-1}$$
which implies $c_{k-1}/c_k=O(1/k)$; if we plug this in the recursion again we have
$$c_{k+1}=\frac{(k-1)(k+1)}{k+2/\ell }(1+O(1/k^2))\ , $$
whence $$c_k= A\frac{\Gamma(k+1)\Gamma(k-1)}{\Gamma(k+2/\ell) }(1+o(1))\ ,$$  because the infinite product of $1+O(1/k^2)$ is convergent. By the Stirling formula  $$c_k= B k^{k-1/2+2/\ell}e^{-k} (1+o(1))$$ for a certain constant $B$. 
A: The first factor telescopes completely, and the second becomes easier by combining $\alpha_{2\kappa}$ with $\beta_{2\kappa+2}$, so
$$
c_{2k} = \frac{1+\ell}{(2k-1)(2+\ell)}\prod_{\kappa=1}^k \left(1-\frac{2}{2+(2\kappa+1)\ell}\right).
$$
Now take logarithms, and apply Euler-MacLaurin summation.
