What is that function? What is the set of the convergence in the reals of the series
$$f(x):= \sum\limits_{n=1}^{\infty} \frac 1 n \sin \left (n+\frac x n\right)?$$ Is the function $f(x)$ bounded?
Edit: more exact title.
 A: For any fixed $x$; and any $N\geq 1$
\begin{align*}
\sum_{n=1}^{N} \frac{1}{n} \sin \left (n+\frac x n\right) &= 
\sum_{n=1}^{N} \frac{1}{n} \sin n \cos \frac{x}{n} + \sum_{n=1}^{N} \frac{1}{n} \cos n \sin \frac{x}{n}
\end{align*}
Now, 


*

*$\frac{\sin n}{n} \cos \frac{x}{n} = \frac{\sin n}{n} - \frac{\sin n}{2n^3}x^2 + o(\frac{1}{n^3})$; the first series converges iff $\sum \frac{\sin n}{n}$ does. It does.

*$\lvert \frac{1}{n} \cos n \sin \frac{x}{n} \rvert \leq \frac{\lvert x\rvert}{n^2}$ (as $\vert \sin t\rvert \leq \lvert t\rvert$): the second series converges as well.

A: Let us break that $\sin$.
$$\sin(n+\frac{x}{n})=\sin(n)\cos(\frac{x}{n})+\cos(n)\sin(\frac{x}{n}).$$
You already know the convergence of the series $\sum \frac{1}{n}\sin(\frac{x}{n})$, therefore the same happens for $\sum \frac{1}{n}\cos(n)\sin(\frac{x}{n})$, since $\cos(n)$ is bounded.
We only need to study
$$\sum \frac{1}{n}\sin(n)\cos(\frac{x}{n})$$
The convergen of this is equivalent to $\sum \frac{\sin(n)}{n}$, because $\cos(\frac{x}{n})\rightarrow1$ as $n\rightarrow\infty$.
For $\sum \frac{\sin(n)}{n}$ we can use summation by parts. Notice that the summation of $\sin(n)$ has bounded partial sums and the finite differences of $1/n$ give us a $1/n^2$ factor.
