Show that an alternating series diverges for $a=0$

Question

Let $a \geq 0$ be a given nonnegative number and consider the series $$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{ \sqrt {n^a}}.$$ Show that this series is divergent for $a = 0$.

What I did was:

$$\lim_{n\to\infty} \frac{(-1)^{n-1}}{ \sqrt {n^a}} = \frac {1}{\sqrt{n^0}} = 1.$$

So by failing the condition of the limit not being equal to $0$ by the alternating series test, is this enough to show that the series is divergent when $a = 0$? Also, how would you go about showing for which values of $a$ the series is convergent?

Answer 1

For $0\lt a\le 2$, the series is conditional convergent, using the fact that the terms are alternating and decreasing in magnitude. For $a\gt 2$, the series is absolutely convergent. The easiest test is compare with $\int_1^\infty \frac {dx}{x^{\frac{a}{2}}}$.

Your idea for $a=0$ is essentially correct. Else you can show the series alternates between $1$ and $0$, so never converges.

Answer 2

This has nothing to do with the alternating series test. Since $\left(\dfrac{(-1)^{n+1}}1\right)_{n\in\mathbb N}$ does not converge to $0$, the series $\displaystyle\sum_{n=1}^\infty\dfrac{(-1)^{n+1}}1$ diverges.

If $a>0$, then, yes, the series converges by the alternating series test.