Does $\sum (-1)^k 2^{1/k}$ converge or diverge?

Question

How am I supposed to determine the convergence of this series if I only know about the alternating series test and the divergence test?

Answer 1

As $k\to\infty$, $\frac 1k\to 0$, $2^{1/k}\to 1$, and $|(-1)^k2^{1/k}|\to 1$.

So if these are the terms of the series, since the terms of the series do not tend to zero, the series is divergent.

If these are the partial sums of the series, the even sums tend to $1$ and the odd ones tend to $-1$, so this also does not converge.

Answer 2

Assuming you're talking about the convergence of the sequence $\{(-1)^k2^{1/k}\}_{k\in \mathbb N}$, you can prove it diverges since it has a subsequence that converges to $1$ and another one that converges to $-1$. Can you identify those subsequences?

HINT: $2^{1/k}$ converges to $1$ as $k$ goes to infinity and $(-1)^k$ is either $-1$ or $1$ depending on the parity of $k$.

If you're talking about the convergence of the series

$$\sum_{k=1}^{\infty} (-1)^k2^{1/k}$$

the same argumet shows that it doesn't converges since the its terms doesn't converge (in particular, doesn't converge to zero).