How to prove this series converges (possibly by modifying the Dirichlet test)? I am trying to prove convergence of the series
$(1) \sum_{n=1}^\infty \frac{\sin(n)}{\sqrt{n}+\sin(n)}$. 
If this series were $(2) \sum_{n=1}^\infty \frac{\sin(n)}{\sqrt{n}}$, then it converges by the Dirichlet test because partial sums of $\sin(n)$ are bounded and $1/\sqrt{n}$ decreases monotonically to $0$.
The difficulty I have is that $\sqrt{n}+ \sin(n)$ in the denominator is not monotonic and $\sin(n)$ in the numerator changes sign. But the denominator is bracketed by monotonic sequences $\sqrt{n} - 1 \leq \sqrt{n} + \sin(n) \leq \sqrt{n} + 1$ and it seems possible to adapt the Dirichlet test. 
Can this be done to prove convergence of (1) or is there a more general approach that I am not aware of?
 A: Let me help you with this one.
We can rewrite the following expression as follows: 
$$
\sum_{n=1}^\infty {\sin(n)}\cdot(\sin(n)+n^{1/2})^{-1} \tag{1}
$$ then we can take $n^{1/2}$ out and write it like 
$$ \sum_{n=1}^\infty {\frac{\sin(n)}{n^{1/2}}}\cdot\left (1+\frac{\sin(n)}{n^{1/2}}\right)^{-1}\tag{2}
$$ 
now we can apply the taylor series like
$$ \sum_{n=1}^\infty {\frac{\sin(n)}{n^{1/2}}}\cdot\left(1-\frac{\sin(n)}{n^{1/2}} + O\left(\frac{1}{n}\right)\right)\tag{3}$$
from here you can solve it, first one converges by Dirichlet, second one diverges, third one converges. 
A: $a(m)
=\sum_{n=1}^m \dfrac{\sin(n)}{\sqrt{n}+\sin(n)},
b(m)
=\sum_{n=1}^m \dfrac{\sin(n)}{\sqrt{n}}
$.
$\begin{array}\\
b(m)-a(m)
&=\sum_{n=1}^m (\dfrac{\sin(n)}{\sqrt{n}}-\dfrac{\sin(n)}{\sqrt{n}+\sin(n)})\\
&=\sum_{n=1}^m \sin(n)(\dfrac{\sin(n)}{(\sqrt{n}+\sin(n))(\sqrt{n})})\\
&=\sum_{n=1}^m \dfrac{\sin^2(n)}{n+\sqrt{n}\sin(n)}\\
\end{array}
$
This last sum
looks like it diverges.
Numerical experimentation
suggests it grows like
$\ln(m)/2+.3
\approx(\ln(m)+\gamma)/2
$.
Since
$b(m)$ converges
(to about 1, it seems),
this would imply that
$a(m)$ diverges
like about
$-\ln(m)/2$.
