# An alternating series convergence

I was wondering whether the following series converges or diverges,

$$\sum_{n=1}^\infty (-1)^n \sqrt[n]{a}$$ $$\forall a>0, a\ne1$$

The divergence test cannot be applied, since the sequence does not have a limit.

• If $a\gt 1$, all the terms have absolute value $\gt 1$. If $0\lt a\lt 1$, all the terms have absolute value $\ge a$. In particular, the absolute value of the terms does not have limit $0$, so we have divergence. – André Nicolas Mar 20 '16 at 1:13
• $\sum_{n=1}^\infty (-1)^n (1- \sqrt[n]{a})$ is convergent, right? And $\sum_{n=1}^\infty (-1)^n 1$ is divergent, So$\sum_{n=1}^\infty (-1)^n \sqrt[n]{a}$ looks like it should be divergent, But I do not know the order in which you sum the terms. – user480281 Oct 15 '17 at 4:14

Since $$\lim_{n \to \infty} (-1)^n \sqrt[n]{a} \neq 0$$ the series is not convergent.
Note that $\sqrt[n]a \to 1$, hence $(-1)^n \sqrt[n]a \not\to 0$, so the series diverges.