Determining the convergence of $\sum_{n=2}^\infty \frac{\sin(n + 1/n)}{\log(\log n)}$

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.

• $\sum_{A}^{B}\sin k$ is bounded so any series $\Sigma{a_n\sin n}, a_n$ decreasing to zero, converges by partial summation; $\cos(\frac{1}{n})=1 + O(\frac{1}{n^2})$ so $\Sigma{a_n\sin n \cos(\frac{1}{n}})=\Sigma{a_n\sin n}$ plus an absolute convergent series when $a_n$ is just bounded – Conrad Aug 16 at 0:57
• Or, it appears that $\frac{\cos(1/n)}{\log(\log n)}$ is eventually decreasing and of course its limit is 0, so Dirichlet's test would apply directly to that part of the sum. – Daniel Schepler Aug 16 at 15:53

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 $$\left\lvert \sum_{n = p}^{p+v} \frac{\sin(t_n)}{\log(\log n)} \right\rvert \le \left\lvert \sum_{k=m}^{m+u} b_k \right\rvert + \left\lvert \sum_{n \in M'} \frac{\sin(t_n)}{\log(\log n)} \right\rvert + \left\lvert \sum_{n \in M''} \frac{\sin(t_n)}{\log(\log n)} \right\rvert < \varepsilon .$$