Asymptotics of a recursion

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 $B$.

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.