For which values of $\alpha$ the serie $$\sum_{n=2}^{\infty}\frac{e^{in}}{n [\ln (n)]^{\alpha}}$$ converges absolutely? and for which values is conditional convergent?
I know that $\sum e^{in}$ diverges (I proved it before), and now I think I need to consider cases for $\alpha$: if $\alpha=0$ or $\alpha\neq0$.
If $\alpha=0$, $\sum_{n=2}^{\infty}\frac{e^{in}}{n [\ln (n)]^{\alpha}}$ converges by Dirichlet test.
Now I just need to analyze the convergence of $\sum_{n=2}^{\infty}|\frac{e^{in}}{n [\ln (n)]^{\alpha}}|$, but I think it would be the same result because all the terms involved in the series would be positive, in conclusion the whole serie is Absolutely convergent.
But if $\alpha\neq0$, should I consider the cases where $\alpha<0$ and $\alpha>0$ or would be more simple considering the general case?
Also can someone give a hint/help to proof the rest ?
Answer
It is conditionally convergent (by the Dirichlet test) for any $\alpha$ since $\frac{1}{n \ln(n)^\alpha}$ decreases to zero as $n \to \infty$ for any $\alpha$ (I'm cheating a bit here because $\frac{1}{n\ln(n)^\alpha}$ is not a monotone decreasing sequence; however, it is decreasing for large enough $n$ and the limit is zero which is sufficient). To test absolute convergence, you need $$\sum_{n=2}^\infty \frac{1}{n \ln(n)^\alpha}$$ to converge (since $\lvert e^{in} \rvert = 1$ for all $n$). For this we use the integral test. Using $u = \ln (x)$, we see $$\int_2^\infty \frac{dx}{x \ln(x)^\alpha} = \int^{\infty}_{\ln(2)} \frac{du}{u^\alpha}.$$ This converges iff $\alpha > 1$; thus the sum also converges absolutely for $\alpha > 1$.
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