Answer
Converges
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 this problem we have $a_k=\dfrac{1}{\sqrt {k^2+4}}$
a) In the given sequence, $a_k=\dfrac{1}{\sqrt {k^2+4}}$ , and $a_{k+1}=\dfrac{1}{\sqrt {(k+1)^2+4}}$ and $\dfrac{1}{\sqrt {(k+1)^2+4}} \leq \dfrac{1}{\sqrt {k^2+4}}$
This implies that $a_{k+1} \lt a_k$ and $a_{k+1} \gt 0$ , so the terms are a non-increasing sequence.
b) $\lim\limits_{k \to \infty} a_k=\lim\limits_{k \to \infty} \dfrac{1}{\sqrt {k^2+4}}=\dfrac{1}{\infty}=0$
This implies that the given series converges.
