Answer
Converges
Work Step by Step
For alternating series $\Sigma_{k=1}^\infty(-1)^k a_k$ (let us consider), to be convergent , we must follow the two conditions such as: a) The magnitude of terms must form a non-increasing sequence.
b) $\lim\limits_{k \to \infty} a_k=0$
We are given that $a_k=\dfrac{1}{\sqrt k}$
1) In the given sequence, $a_k=\dfrac{1}{\sqrt k}$ , and $a_{k+1}=\dfrac{1}{\sqrt k} \lt \dfrac{1}{\sqrt k}$.
This implies that $a_{k+1} \lt a_k$ and $a_{k+1} \gt 0$ , so the terms are a non-increasing sequence.
2) $\lim\limits_{k \to \infty} a_k=\lim\limits_{k \to \infty} \dfrac{1}{\sqrt k}= 0$
This means that the given sequence converges.
