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 one you have
$$\frac{\cos(n)}{n^{2/3}+(-1)^n} = \frac{\cos(n)}{n^{4/3}-1}\,\left(n^{2/3}-(-1)^n\right) = \frac{\cos(n)}{n^{2/3}-n^{-2/3}} - \frac{(-1)^n}{n^{4/3}-1}$$
and so the first term converges by the Dirichlet-test, since $\left(n^{2/3}-n^{-2/3}\right)^{-1}$ is decreasing. The second term converges absolutely.
For the second you could argue as follows:
$$\sum_{n=1}^{\infty} \left(e^\frac{\cos{n}}{n} - \cos\frac{1}{n}\right) = \sum_{n=1}^\infty \left(1 + \frac{\cos(n)}{n} + O(1/n^2) - 1 - O(1/n^2) \right) \\= \sum_{n=1}^\infty \left( \frac{\cos(n)}{n} + O(1/n^2) \right)$$
The O-term converges clearly, while the first term is a convergent Fourier-series.
