Answer
$\dfrac{1}{\sqrt {2}}+\dfrac{1}{\sqrt {3}}$
Work Step by Step
Here, we have
$S_n= \Sigma_{k=1}^{n} (\dfrac{1}{\sqrt {k+1}}-\dfrac{1}{\sqrt {k+3}})$
or, $= (\dfrac{1}{\sqrt {2}}-\dfrac{1}{\sqrt {4}})+ (\dfrac{1}{\sqrt {3}}-\dfrac{1}{\sqrt {5}})+....... +(\dfrac{1}{\sqrt {n}}-\dfrac{1}{\sqrt {n+2}})+ (\dfrac{1}{\sqrt {n+1}}-\dfrac{1}{\sqrt {n+3}})$
or, $= \dfrac{1}{\sqrt {2}}+\dfrac{1}{\sqrt {3}}-\dfrac{1}{\sqrt {n+2}}-\dfrac{1}{\sqrt {n+3}} $
Now, $S=\lim\limits_{n \to \infty} \dfrac{1}{\sqrt {2}}+\lim\limits_{n \to \infty}\dfrac{1}{\sqrt {3}}-\lim\limits_{n \to \infty}\dfrac{1}{\sqrt {n+2}}-\lim\limits_{n \to \infty}\dfrac{1}{\sqrt {n+3}} $
or, $S= \dfrac{1}{\sqrt {2}}+\dfrac{1}{\sqrt {3}}-0-0$
or, $= \dfrac{1}{\sqrt {2}}+\dfrac{1}{\sqrt {3}}$
