I am trying to determine convergence of the series
$$\sum_{n=2}^\infty \frac{\sin(n + 1/n)}{\log(\log n)}$$
I used the expansion $\sin(n+1/n) = \sin (n) \cos(1/n) + \cos(n) \sin(1/n)$.
I can see that the series $\sum \frac{\cos (n) \sin(1/n)}{\log(\log n)}$ converges by the Dirichlet test, but I'm not sure how to determine if $\sum \frac{\sin (n) \cos(1/n)}{\log(\log n)}$ converges.
Op has reduced the question to asking if
\begin{align} \sum\limits_{n=2}^{\infty} \frac{\sin(n)\cos(1/n)}{\log(\log(n))} \end{align}
converges. We know that $\sum\limits_{n=2}^{N} \sin(n)$ is bounded by a single $M$ for all $N$ (see here for example). Then by the Dirichlet test, we only need to check when $\frac{\cos(1/n)}{\log(\log(n))}$ is decreasing.
Compute the derivative:
\begin{align} \frac{\mathrm d}{\mathrm dx} \left(\frac{\cos(1/x)}{\log(\log(x))}\right) &= \frac{\log (x) \log (\log (x)) \sin \left(\frac{1}{x}\right)-x \cos \left(\frac{1}{x}\right)}{x^2 \log (x)( \log(\log (x)))^2} \end{align}
The denominator is positive as long as $\log(\log (x)) > 0$ or equivalently, $x > e$. So let's restrict to $x > e$.
The numerator is negative for $x > e$, because $ \cos(1/x) > \sin(1/x)$ and $x > \log (x) \log(\log (x))$ (because $x > (\log(x))^2 > \log (x) \log(\log (x))$).
Then for all $n > e$, $\frac{\cos(1/n)}{\log(\log(n))} $ is decreasing and so by the Dirichlet test the original series converges.
For $n \in \mathbb N$ let $t_n = n + 1/n$. We have $t_{n+1} - t_n = 1 + 1/(n+1) - 1/n < 1$. For $k \in \mathbb N$ define $$n_k = \min \{ n \ge 2 \mid t_n \ge (k-1)\pi \} .$$ Then $n_1 = 2$. It is also clear that $n_k \le n_{k+1}$ and $t_{n_{k+1}-1} < k\pi$. Thus we have $$(*) \quad M_k = \{ t_{n_k},\dots,t_{n_{k+1}-1} \} \subset [(k-1)\pi,k\pi) .$$ This shows in particular that $n_k < n_{k+1}$. Moreover, $n_{k+1} \le n_k + 4$. To see this, note that we have $$t_{n_k+4} - t_{n_k} = (n_k+4) + 1/(n_k+4) - (n_k + 1/n_k) = 4 - 1/n_k +1/(n_k+4) > 4 - 1/n_k > 4 -1/2 > \pi ,$$ i.e. $t_{n_k+4} > t_{n_k} + \pi \ge k\pi$. This means that $M_k$ has at most $4$ elements.
Define $$b_k = \sum_{n \in M_k}\frac{\sin(t_n)}{\log(\log n)} = \sum_{n =n_k}^{n_{k+1}-1}\frac{\sin(t_n)}{\log(\log n)} .$$ By $(*)$ the series $\sum_{k=1}^\infty b_k$ is alternating, hence it converges since $$\lvert b_k \rvert \le \sum_{n \in M_k}\frac{\lvert \sin(t_n) \rvert}{\log(\log n)} \le \frac{4}{\log(\log n_k)} .$$ Note that also for any subset $M' \subset M_k$ we have $\lvert \sum_{n \in M'}\frac{\sin(t_n)}{\log(\log n)} \rvert \le \frac{4}{\log(\log n_k)}$.
Let $\varepsilon > 0$. We find $r \in \mathbb N$ such that for all $m \ge r$ and $u \ge 0$ we have $\lvert \sum_{k=m}^{m+u} b_k \rvert < \varepsilon/3$. W.l.o.g. we may assume that $\frac{4}{\log(\log n_r)} < \varepsilon/3$.
Let $p \ge n_r$ and $v \ge 0$. Let $m$ be the maximal integer such that $n_m \le p$ and $u$ be the minimal integer such that $p+v < n_{m+u} $. Since $v \ge 0$, we have $u > 0$. Hence $$\sum_{n = p}^{p+v} \frac{\sin(t_n)}{\log(\log n)} = \sum_{n = n_m}^{n_{m+u}-1} \frac{\sin(t_n)}{\log(\log n)} - \sum_{n = n_m}^{p-1} \frac{\sin(t_n)}{\log(\log n)} - \sum_{n = p+v+1}^{n_{m+u}-1} \frac{\sin(t_n)}{\log(\log n)} \\ = \sum_{k=m}^{m+u} b_k - \sum_{n = n_m}^{p-1} \frac{\sin(t_n)}{\log(\log n)} - \sum_{n = p+v+1}^{n_{m+u}-1} \frac{\sin(t_n)}{\log(\log n)} \\ = \sum_{k=m}^{m+u} b_k - \sum_{n \in M'} \frac{\sin(t_n)}{\log(\log n)} - \sum_{n \in M''} \frac{\sin(t_n)}{\log(\log n)} $$ with suitable $M' \subset M_{k_m}$ and $M'' \subset M_{k_{m+u}}$. Thus $$\left\lvert \sum_{n = p}^{p+v} \frac{\sin(t_n)}{\log(\log n)} \right\rvert \le \left\lvert \sum_{k=m}^{m+u} b_k \right\rvert + \left\lvert \sum_{n \in M'} \frac{\sin(t_n)}{\log(\log n)} \right\rvert + \left\lvert \sum_{n \in M''} \frac{\sin(t_n)}{\log(\log n)} \right\rvert < \varepsilon .$$
