Answer
Diverges
Work Step by Step
Let us consider an alternating series $\Sigma_{k=1}^\infty(-1)^k a_k$ that can be convergent , when it satisfy the following two conditions such as: 1) The magnitude of terms must form a non-increasing sequence.
2) $\lim\limits_{k \to \infty} a_k=0$
Here, in the problem we have $a_k=k^{1/k}$
a) In the given sequence, $a_k=k^{1/k}$ , and $a_{k+1}=(k+1)^{1/k}$ and $(k+1)^{1/k} \leq k^{1/k}$
This implies that $a_{k+1} \lt a_k$ and $a_{k+1} \gt 0$ , so the terms are a non-increasing sequence.
b) $\lim\limits_{k \to \infty} a_k=\lim\limits_{k \to \infty} k^{1/k}$
$\lim\limits_{k \to \infty} \ln a_k=\dfrac{1}{k} \lim\limits_{k \to \infty} \ln k=\dfrac{\ln k}{k}$
Apply L-Hospital's Rule to obtain:
$\lim\limits_{k \to \infty} \ln a_k =$\lim\limits_{k \to \infty} (\dfrac{1}{k})=0$
and $\lim\limits_{k \to \infty} a_k=e^0=1 \ne 0$
This implies that the given series diverges.
