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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

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Your writing on Problem 1 seems a bit scrambled to me, however for Problem 2 you can apply the root test. –  process91 Jan 11 '12 at 22:55
    
Problem 1 uses $k$ and $n$, presumably meaning the same thing. –  Henry Jan 12 '12 at 0:00

1 Answer 1

Problem 1: the series converges absolutely if and only if $\alpha+\frac12\gt1$ and it converges if and only if $\alpha+\frac12\gt0$.

There are some errors with the signs, and a beautifully false step: the sequence of general term $\sin\frac1{\sqrt{n}}$ is decreasing, yes, but certainly not because $\sin\frac1{\sqrt{n}}\leqslant\frac1{\sqrt{n}}$ and the sequence of general term $\frac1{\sqrt{n}}$ is decreasing.

Problem 2: since $1\leqslant 2+\cos n\leqslant3$ and $\beta=\dfrac{\sqrt{1+\alpha}}{|1-\alpha|}$ is nonnegative, the series converges absolutely if and only if it converges if and only if $\beta\lt1$. Maybe you can finish this...

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