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..
 A: 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 $x$.
A: For the first series it would be desirable to replace $n^{2/3}+(-1)^n$ by $n^{2/3}.$ To this end observe that $${\cos n\over  n^{2/3}+(-1)^n}={\cos n\over  n^{2/3}}+{(-1)^{n+1}\cos n\over [n^{2/3}+(-1)^n]n^{2/3}}$$ The first terms are summable by the Dirichlet test. The second terms are absolutely summable as the denominator is greater than  $n^{4/3}/2$ for $n\ge 3.$
For the second series we have $$e^{\cos n/n}-\cos(1/n)\\ = \left [e^{\cos n/n}-1-{\cos n\over n}\right ]\\ +[1-\cos(1/n)]+{\cos n\over n}$$ The absolute value of the term in the first square brackets is less than $1/n^2$ because $$|e^x-1-x|\le x^2$$ The term in the second square bracket is equal $2\sin^2(1/n).$ The last terms are summable by the Dirichlet test.
A: 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.
