Do 1.$\sum_{n=2}^{\infty} \frac{\cos{n}}{n^\frac{2}{3} + (-1)^n}$ 2. $\sum_{n=1}^\infty(e^{\cos n/n}−\cos\frac{1}n)$ converge? Does
$$ 1. \hspace{8mm} \sum_{n=2}^{\infty} \frac{\cos{n}}{n^\frac{2}{3} + (-1)^n} $$
and
$$ 2. \hspace{8mm} \sum_{n=1}^{\infty} \left(e^\frac{\cos{n}}{n} - \cos\frac{1}{n}\right) $$
converge? In $\sum_{n=2}^{\infty} \frac{\cos{n}}{n^\frac{2}{3} + (-1)^n}$,  I allocate two sequences of 1. cos n is bounded and  $\frac{1}{n^\frac{2}{3} + (-1)^n}$, where $(-1)^n$ irrelevant next to $n^\frac{2}{3}$ for large and, so I get sequence $\frac{1}{n^\frac{2}{3}}$, this sequence is monotonically decreases to $0$.
On the basis of Dirichlet, the series converges. Is it so?
But I don't have any ideas about the second number..

For the first series, even though $(-1)^n$ is neglible compared with $n^{2/3}$ as $n$ gets large, the monotone hypothesis of the Dirichlet test is not met and it cannot be applied directly.
However, we have
$$\sum_{n=2}^m \frac{\cos{n}}{n^\frac{2}{3} + (-1)^n}= \sum_{n=2}^m \frac{\cos{n}}{n^\frac{2}{3} + (-1)^n}\frac{n^\frac{2}{3} - (-1)^n}{n^\frac{2}{3} - (-1)^n} = \underbrace{\sum_{n=2}^m \frac{n^{\frac{2}{3}}\cos{n}}{n^\frac{4}{3} -1}}_{A_m}- \underbrace{\sum_{n=2}^m \frac{(-1)^n\cos{n}}{n^\frac{4}{3} -1}}_{B_m}$$
Now we can apply the Dirichlet test to show that the sequences of partial sums $A_m$ and $B_m$ on the RHS converges.
To handle $B_m$, note that $(-1)^n \cos n = \cos n\pi\cos n = \frac{1}{2}[\cos (n(\pi+1)) + \cos (n(\pi-1))]$, and use the fact that $\sum_{n=1}^m \cos nx$ is bounded for all $m$ and fixed $x$.
