Does the series $\sum_{n=1}^\infty\frac{\sin n}{\ln n+\cos n}$ converge? $$\sum_{n=1}^\infty\frac{\sin n}{\ln n+\cos n}$$
My guess is "yes", but I can't prove it.
 A: Why not just write 
$$\frac{\sin n}{\log n + \cos n} = - \frac{\sin n}{\log n } \left(1- \frac{1}{1 + \frac{\cos n}{\log n }} \right) + \frac{\sin n}{\log n } $$ 
The first term gives an absolutely convergent series, since 
$$ \left| \frac{\sin n}{\log n } \left(1- \frac{1}{1 + \frac{\cos n}{\log n }} \right) \right| \sim \left| \frac{\sin 2n}{2 \log^2 n} \right| \leq \frac{1}{2 \log^2 n}  $$
And for the second term, the Dirichlet criterion applies because $\frac{1}{\log n}$ is decreasing to $0$. 
Thus the series converges. 
A: This method was inspired by the OP's heuristic argument in a comment to the question. We approximate the summand by Taylor polynomials, but with more and more terms as $n$ grows.
We need the fact that for every integer $k\ge1$,
$$
\bigg| \sum_{B\le n< C} \sin n \cdot \cos^k n \bigg| = O(k^7)
$$
uniformly for integers $B<C$. To see this, verify that $\sin n \cdot \cos^k n = q_k(e^{in})$, where $q_k(t)=\sum_{j=-k-1}^{k+1} c_{k,j}t^j$ is a Laurent polynomial with no constant term that satisfies $\sum_{j=-k-1}^{k+1} |c_{k,j}| = 2^{-k}\binom{k}{\lfloor k/2\rfloor} \le 1$. Then
\begin{align*}
\sum_{B\le n< C} \sin n \cdot \cos^k n &= \sum_{B\le n< C} q_k(e^{in}) \\
&= \sum_{j=-k-1}^{k+1} \sum_{B\le n< C} c_{k,j}e^{ijn} = \sum_{j=-k-1}^{k+1} c_{k,j} \frac{e^{ijC}-e^{ijB}}{e^{ij}-1},
\end{align*}
so that
$$
\bigg| \sum_{B\le n< C} \sin n \cdot \cos^k n \bigg| \le \sum_{j=-k-1}^{k+1} |c_{k,j}| \frac{2}{|e^{ij}-1|}.
$$
The fact that $\pi$ has an irrationality measure of less than $8$ (Salikhov, 2012) means that $|e^{ij}-1|\gg j^{-7}$. Therefore
$$
\bigg| \sum_{B\le n< C} \sin n \cdot \cos^k n \bigg| \ll k^7 \sum_{j=-k-1}^{k+1} |c_{k,j}| \ll k^7.
$$
Also note that for any positive integer $A$ and any real number $x\in[-\frac12,\frac12]$,
$$
\frac1{1+x} = \sum_{k=0}^{A-1} (-x)^k + r(x)
$$
where $|r(x)| \le |2x|^A$ (a consequence of Taylor's theorem).
Now we establish the convergence of the series in question by showing that it is Cauchy, i.e., that
$$
\sum_{n=M}^N \frac{\sin n}{\log n+\cos n} \to 0
$$
as $M\to\infty$ (and $N>M>e^{2e^2}$, say). For each $n$, we choose $x=\frac{\cos n}{\log n}$ and $A=\lceil{\log n}\rceil$, giving
$$
\sum_{n=M}^N \frac{\sin n}{\log n} \bigg( \sum_{k<\log n} \bigg( -\frac{\cos n}{\log n} \bigg)^k + s(n) \bigg)
$$
where $|s(n)| \le |2\frac{\cos n}{\log n}|^A \le (\frac2{\log n})^{\log n} \le \frac1{n^2}$ for $n>e^{2e^2}$. Therefore the contribution from $\sum_{n=M}^N \frac{\sin n}{\log n}s(n)$ is $O(\frac1M)$. As for the other terms, we can write
\begin{multline*}
\sum_{n=M}^N \frac{\sin n}{\log n} \sum_{k<\log n} \bigg( -\frac{\cos n}{\log n} \bigg)^k = \sum_{k<\log M} (-1)^k \sum_{n=M}^N \frac{\sin n \cdot \cos^k n}{\log^{k+1} n} \\
+ \sum_{\log M<k<\log N} (-1)^k \sum_{e^k < n \le N} \frac{\sin n \cdot \cos^k n}{\log^{k+1} n}.
\end{multline*}
Let $S_k(t) = \sum_{n\le t} \sin n \cdot \cos^k n$, which is $O(k^7)$ as noted above. Then
\begin{align*}
\sum_{B<n\le C} \frac{\sin n \cdot \cos^k n}{\log^{k+1} n} = \int_B^C \frac{dS_k(t)}{\log^{k+1} t} &= \frac{S_k(t)}{\log^{k+1} t} \bigg|_B^C + (k+1) \int_B^C \frac{S_k(t)}{t\log^{k+2} t}\,dt \\
&\ll k^7 \bigg(\frac1{\log^{k+1} B} + (k+1) \int_B^C \frac1{t\log^{k+2} t}\,dt \bigg) \\
&\ll k^7 \frac1{\log^{k+1} B}.
\end{align*}
In particular,
\begin{align*}
\sum_{n=M}^N \frac{\sin n}{\log n} \sum_{k<\log n} \bigg( -\frac{\cos n}{\log n} \bigg)^k &\ll \sum_{k<\log M} \frac{k^7}{\log^{k+1} M} + \sum_{\log M<k<\log N} \frac{k^7}{\log^{k+1} e^k} \\
&\ll \sum_{k=1}^\infty \frac{k^7}{\log^{k+1} M} + \sum_{k=2}^\infty \frac1{k^{\log M-6}},
\end{align*}
and both these series tend to $0$ as $M\to\infty$ by the dominated convergence theorem.
A: This is not an answer, but some computational "evidence". Below is a plot of the values of the sum for up to $20.000$  summations. As can be seen in the picture, the sum oscillates for a very long time, but one might guess that the oscillations die away because of the $\ln n$-term, after a very long time.  What can be concluded however, is that if the answer is "no", then it is not because the sum tends to infinity, but because of the oscillations. 

A: The sum $ | \sum_{n=3}^{\infty} \frac{\sin(n)}{\ln(n)-1}|$ converges by the Dirichlet-Criterion since $ | \sum_{n=3}^{\infty} sin(n)|$ is bounded and $a_n = \frac{1}{\ln(n)-1}$ is a decreasing positive sequence converging to zero. Then it follows that
$$  \sum_{n=3}^{\infty} \frac{\sin(n)}{\ln(n)+\cos(n)} <  \sum_{n=3}^{\infty} \frac{\sin(n)}{\ln(n)-1} < \infty$$
Surely that would be too easy though, but where is my mistake? :)
