Convergence of $\sum(u_n)$ if $\frac{u_{n+1}}{u_n} \leq \left ( \frac{n}{n+1} \right )^{\alpha}$

I got this question:let $\alpha >1$, and let $u_{n}$ be a sequence of positive numbers such that for all n:

$$ \frac{u_{n+1}}{u_n}\leq \left ( \frac{n}{n+1} \right )^{\alpha}. $$

I have to prove that $\sum(u_n)$ converges.

I did the following: since $u_{n}, u_{n+1}$ are positive, make the comparison test:$$ \sum \frac{u_n}{u_{n+1} }\leq \sum \left ( \frac{n}{n+1} \right ) ^{\alpha}. $$I proved that$\sum \left ( \frac{n}{n+1} \right ) ^{\alpha}$ converges, and then it means that
$\displaystyle \lim_{n \to \infty}\frac{u_n}{u_{n+1}}=0$and then by D'Alembert's Ratio Test $\sum(u_n)$ converges.Is it legit?

Thank you in advance.