Answer
$$-1$$
Work Step by Step
Let us consider that $S_n=\Sigma_{k=2}^{n+1} [\dfrac{1}{\sqrt k}-\dfrac{1}{\sqrt {k-1}}]\\=(\dfrac{1}{\sqrt 2}-\dfrac{1}{\sqrt 1})+(\dfrac{1}{\sqrt 3}-\dfrac{1}{\sqrt 2})+(\dfrac{1}{\sqrt 4}-\dfrac{1}{\sqrt 3})+(\dfrac{1}{\sqrt {n+1}}-\dfrac{1}{\sqrt n})\\=\dfrac{1}{\sqrt {n+1}}-1$
Now, $\lim\limits_{n \to \infty} S_n=\lim\limits_{n \to \infty}\dfrac{1}{\sqrt {n+1}}-\lim\limits_{n \to \infty}(1)\\=0-1\\=-1$
