Convergence of $\sum_{n = 10}^\infty \frac{\sin n}{n + 10 \sin n}$

Convergence of $\sum_{n = 10}^\infty \frac{\sin n}{n + 10 \sin n}$

It is easy to see that $$\sum_{n = 10}^\infty \frac{\sin n}{n + 10 \sin n}
= K + \sum_{n = 11}^\infty \frac{\sin n}{n + 10 \sin n} \le K +
\sum_{n=11}^\infty \frac{\sin n}{n - 10},$$
where $$K = \sum_{n = 10}^{10} \frac{\sin n}{n + 10\sin n} = \frac{\sin
10}{10 + 10\sin10} \in \mathbb{R}.$$
(This is to avoid the undefined term $\frac{\sin 10}{10 - 10}$ in the
upper bound.)
Let's write $a_n := \sin n$ and $b_n := \frac1{n - 10}$. The sequence of
partial sums of $a_n$ is bounded and $b_n$ is a monotonic sequence that
tends towards $0$ as $n$ tends to infinity.
According to Dirichlet test, the series $\sum_{n=11}^\infty \frac{\sin
n}{n - 10}$ converges. Accoding to the comparison test, the series
$\sum_{n = 11}^\infty \frac{\sin n}{n + 10 \sin n}$ converges as well.
Then the given series $\sum_{n = 10}^\infty \frac{\sin n}{n + 10 \sin n}$
is convergent, too.
Is my argumentation correct?
Thank you!