Prove that sum is convergent How to prove that the following sum is convergent? $$\sum_1^\infty\frac{\sin(n + \ln{n})}{n}$$
I tried to use formula $$\sin(n+ \ln{n}) = \sin{n}\cos \ln{n} + \sin \ln{n}\cos{n}$$ and $$\sum_1^N \sin{n} \leq \frac{1}{\sin{1/2}}$$
But I can't make same estimates for $\sin{n}\cos \ln{n} $ and $\sin \ln{n} \cos{n}$.
 A: Here give another method for the convergence of
$$\sum_{n=1}^\infty\frac{\sin(n+\ln{n})}{n}.$$
Use the OP's formula:
$$\sin(n+\ln{n})=\sin{n}\cos\ln{n}+\sin\ln{n}\cos{n},$$
We will prove the two series
$$\sum_{n=1}^\infty\frac{\sin{n}\cos\ln{n}}{n},\quad
\sum_{n=1}^\infty\frac{\sin\ln{n}\cos{n}}{n}$$
are convergent.
For the convergence, we use the following $\bf{Dirichlet-Abel\ test}$:Dirichlet-Abel


If the series $\sum\limits_{n=1}^\infty a_n,\sum\limits_{n=1}^\infty b_n$ satisfy the following conditions:
(1) $\left\{\sum_{k=1}^n a_k\right\}$ is bounded;
(2) $\lim\limits_{n\to\infty}b_n=0;$
(3) $\sum\limits_{n=1}^\infty|b_n-b_{n+1}|$ is convergent,
then the series $\sum\limits_{n=1}^\infty a_nb_n$ is convergent.


For the first series $\sum_{n=1}^\infty\frac{\sin{n}\cos\ln{n}}{n}$,
take $a_n=\sin n,b_n=\frac{\cos\ln n}{n}$, it is easy to see conditions $(1),(2)$
are satisfied, as for condition $(3)$, just notice that:
$$\left(\frac{\cos\ln x}{x}\right)'=-\frac{\sin\ln x+\cos\ln x}{x^2},$$
and (Lagrange's mean value theorem)
$$\left|\frac{\cos\ln n}{n}-\frac{\cos\ln(n+1)}{n+1}\right|<\frac{2}{n^2}.$$
By the Dirichlet-Abel test, we know the convergence of
$$\sum_{n=1}^\infty\frac{\sin{n}\cos\ln{n}}{n}.$$
In the same way, we can prove the second is convergent.
So the series $$\sum_{n=1}^\infty\frac{\sin(n+\ln{n})}{n}$$
is convergent.
A: Look separately at the intervals $N^2\le n < (N+1)^2$, with $N\gg 1$. For these $n$, we have that
$$
|\sin (n+\log n) - \sin (n+\log N^2)| \le \log (n/N^2) \lesssim 1/N \lesssim 1/n^{1/2} ,
$$
and since $\sum n^{-3/2}<\infty$, we may replace $\sin (n+\log n)$ by $\sin (n+\alpha_N)$, for the $n$ currently under consideration.
However, as you already indicated, it's easy to control $\sum \sin (n+\alpha)/n$. Using summation by parts and the fact that $\sum_{n=N_1}^{N_2}\sin (n+\alpha)=O(1)$, we obtain that
$$
\sum_{N^2\le n<(N+1)^2} \frac{\sin (n+\alpha)}{n} = O(1/N^2) + \sum_{N^2\le n<(N+1)^2} O(1)/n^2 = O(1/N^2) ,
$$
and this is summable over $N$.
