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!
 A: Your approach is a bit awkward.  The simplest approach would be to compare your $n$th term to $\frac{\sin n}{n}$ by showing that the difference $\frac{\sin n}{n + 10 \sin n}-\frac{\sin n}{n}$ is dominated by $\frac{1}{n^2}$ which is absolutely convergent.  The convergence of your series then follows from the convergence of the classical series $\sum\frac{\sin n}{n}$.
A: That is an interesting proof.
I am going to generalize it
by showing that,
if $f(n)$
is a bounded function
such that
$\sum_{n=1}^{\infty} \frac{f(n)}{n}$ converges
and $a$ is a real number
such that
$n+a f(n) \ne 0$
for all $n$,
then
$\sum_{n=1}^{m} \frac{f(n)}{n+a f(n)}$
converges as $m \to \infty$.
$a=10$
and
$f(n) = \sin(n)$
in the original problem.
Consider
$\sum_{n=1}^{m} \frac{f(n)}{n+a f(n)}$.
$\begin{align}
\frac{f(n)}{n+a f(n)}
-\frac{f(n)}{n}
&=f(n)(\frac1{n+a f(n)}-\frac1{n})\\
&=f(n)\frac{n-(n+af(n))}{n(n+af(n))}\\
&=f(n)\frac{-af(n))}{n(n+af(n))}\\
&=\frac{-af^2(n))}{n(n+af(n))}\\
\end{align}
$
so that
$\sum_{n=1}^m \frac{f(n)}{n+a f(n)}
=\sum_{n=1}^m \frac{f(n)}{n}
+\sum_{n=1}^m \frac{-af^2(n))}{n(n+af(n))}
$.
Since,
as $m \to \infty$,
$\sum_{n=1}^m \frac{f(n)}{n}$ converges
(by assumption)
and
$\sum_{n=1}^m \frac{-af^2(n))}{n(n+af(n))}$
converges
(by comparison with
$\sum_{n=1}^m \frac1{n^2}$),
$\sum_{n=1}^m \frac{f(n)}{n+a f(n)}$
also converges.
If $n+a f(n) = 0$
for some $n$,
the result still holds
if we start the sums 
at an $k$ such that
$k > |a| M$,
where $M> |f(n)|$
for all $n$
($M$ exists since $f$ is bounded).
