Does $\sum{\frac{\sin (n+1/n)}{ \ln n}} \sim \sum{\frac{\sin n}{\ln n}}$? As was mentioned in the title, can we say that both the series
$$
\sum{\frac{\sin (n+1/n)}{\ln n}} \quad \text{ and } \quad \sum{\frac{\sin n}{\ln n}}
$$
converges or diverges simultaneously? I mean, it looks very intuitive since
$$
\sin(n+1/n) \approx \sin(n)
$$
for sufficiently large $n$. But I didn't find any rigorous Test or Rule to show that.
 A: Since $\sin(a+b) = \sin a \cos b + \cos a \sin b$, by a Taylor expansion around $0$ we get
$$\begin{align*}
\sin\!\left(n+\frac{1}{n}\right) &= \sin n \cos \frac{1}{n} + \cos n \sin \frac{1}{n} \\
&= \sin n\left(1-\frac{1}{n^2} + o\!\left(\frac{1}{n^2}\right) \right)
+ \cos n\left(\frac{1}{n} + o\!\left(\frac{1}{n^2}\right) \right)
\end{align*}$$
which means that
$$\begin{align*}
\frac{\sin\!\left(n+\frac{1}{n}\right)}{\ln n}
&= \frac{\sin n}{\ln n} 
+ \frac{\cos n}{n\ln n}+ o\!\left(\frac{1}{n^2}\right) 
\end{align*}$$
since $|\sin|,|\cos|\leq 1$. This shows that, if you take for granted that
$$
\sum_{n=2}^\infty \frac{\cos n}{n\ln n}
$$
converges, then indeed
$$
\sum_{n=2}^\infty \frac{\sin\!\left(n+\frac{1}{n}\right)}{\ln n},\quad\sum_{n=2}^\infty \frac{\sin n}{\ln n}
$$
have same nature (since the remainder, "$\sum_n o\!\left(\frac{1}{n^2}\right)$", is absolutely convergent by comparison). The question then becomes: can one easily show that
$$
\sum_{n=2}^\infty \frac{\cos n}{n\ln n}
$$
converges? (One way: Dirichlet's test.)
