I have some difficulty with these two problems, the second of which I do not know how to set
$${\bf Problem\,\,1}$$
Consider the series:
$$ \sum_{k=1}^\infty \frac{(-1)^n}{n^{\alpha}}\sin{\frac{1}{\sqrt{n}}}e^{\frac{1}{\sqrt{n}}}$$
and say for which values of $ \alpha \in \mathbb {R^+} $ is absolute convergence and simply convergent
$${\bf Solution.}$$ begin by studying the absolute convergence, ie we study the convergence of the series with positive terms: $$ \sum_{k=1}^\infty \frac{1}{n^{\alpha}}\sin{\frac{1}{\sqrt{n}}}e^{\frac{1}{\sqrt{n}}}\quad \sim \quad\sum_{k=1}^\infty \frac{1}{n^{\alpha+\frac{1}{2}}} $$ when $n\to\infty$,\, $\frac{1}{n^{\alpha}}\sim\frac{1}{n^{\alpha}}$, \, $\sin{\frac{1}{\sqrt{n}}}\sim\frac{1}{\sqrt{n}}$, \,$e^{\frac{1}{\sqrt{n}}}\sim 1.$ Then the series we know to be a generalized harmonic series, which converges if the exponent is less than one, then the criterion of asymptotic comparison we can establish that the series converges absolutely (and thus simply) if: $$\alpha+\frac{1}{2}<1 \quad\Rightarrow \quad \alpha<\frac{1}{2}.$$ Now consider the case where $ 0 <\alpha \le 1/2. $ The series is alternate signs, and then see if the Leibniz's criterion is applicable:
we have that: $$\lim_{n\to+\infty}\,\,\frac{1}{n^{\alpha}}\sin{\frac{1}{\sqrt{n}}}e^{\frac{1}{\sqrt{n}}}=0\cdot1=0$$ and the the sequence $$a_n=\frac{1}{n^{\alpha}}\sin{\frac{1}{\sqrt{n}}}e^{\frac{1}{\sqrt{n}}}$$ is decreasing, in fact: $$\frac{1}{n^{\alpha}},\alpha > 0 \quad\text{decreases},$$ $$\sin{\frac{1}{\sqrt{n}}} $$ is decreasing because it is less a function decrecente: $$\sin{\frac{1}{\sqrt{n}}}\le \frac{1}{\sqrt{n}} \quad \text{and the reciprocal function of the root is decrecente }$$ finally $e^{\frac{1}{\sqrt{n}}}$ is decreasing: $$e^{\frac{1}{\sqrt{n}}}>e^{\frac{1}{\sqrt{n+1}}}\quad\Rightarrow \quad \frac{1}{\sqrt{n}}>\frac{1}{\sqrt{n+1}}\quad\Rightarrow \quad \sqrt{n+1}>\sqrt{n}$$ Then the series assigned is monotone decreasing as the product of monotone sequences decreasing, and being satisfied the hypotheses of Leibniz criterion we can conclude that, when $ 0 <\alpha \le 1 /2,$the series is simply convergent.
$${\bf Problem\,\,2}$$ Consider the series:
$$ \sum_{k=1}^{+\infty} \,\ (\cos n +2)\left(\frac{(\sqrt{1+\alpha)}}{|1-\alpha|}\right)^n$$
and say for which values of $ \alpha \in \mathbb {R} $ is absolute convergence and simply convergent