Convergence of $a_k=\sum_{k=5}^{+\infty}(-1)^k({3\over2})^{-k}(k^2+5)\sin(k+5)$

Question

Firstly, $\lim_{k \to +\infty}a_k=0$,so necessary condition for convergence is satisfied.If we start to study absolute convergence we have : $$a_k=({3\over2})^{-k}(k^2+5)|\sin(k+5)|\leq({3\over2})^{-k}(k^2+5)\sim({3\over2})^{-k}(k^2)=\frac{k^2}{({3\over2})^{k}}$$
So if we apply ratio test : $$\lim_{k\to+\infty}\frac{(k+1)^2({3\over2})^k}{({3\over2})^k({3\over2})(k^2)}={2\over3}\lt1$$ And $a_k$ is absolutely and simply convergent.Is this approach correct?