Absolute and Uniform Convergence of a Series

Question

$$\sum_{k=1}^\infty \frac{(-1)^k}{\sqrt{k-1}+(-1)^k}$$ I put it on wolfram alpha and it shows that converges, but I think the analysis has to be different, because I need to know if the series converge absolutely or uniform or conditional and wolfram doesn't show it. Any help you can give will be greatly appreciated

Answer 1

hint for absolute

$$|u_k|\ge \frac {1}{\sqrt {k-1}+1} $$

$$\frac {1}{\sqrt {k-1}+1}\sim \frac {1}{\sqrt {k}} $$

$\sum \frac {1}{\sqrt {k}} $ is divergent thus

$\sum u_k $ is not absolutely conv.

Answer 2

The series fails to converge even conditionally since

$$\frac{(-1)^k}{\sqrt{k-1} + (-1)^k} = \frac{(-1)^k(\sqrt{k-1} - (-1)^k)}{(\sqrt{k-1} + (-1)^k)(\sqrt{k-1} - (-1)^k)} = \frac{(-1)^k \sqrt{k-1}}{k-2} - \frac{1}{k-2}$$

and we have divergence of

$$\sum_{k=3}^\infty \frac{1}{k-2}$$

and convergence by the Dirichlet test of

$$\sum_{k=3}^\infty \frac{(-1)^k\sqrt{k-1}}{k-2}$$