Answer
$-\dfrac{1}{2}$
Work Step by Step
Here, we have: $ s_n=(\dfrac{1}{2^{\frac{1}{1}}}-\dfrac{1}{2^{\frac{1}{2}}})+(\dfrac{1}{2^{\frac{1}{2}}}-\dfrac{1}{2^{\frac{1}{3}}})+.....(\dfrac{1}{2^{\frac{1}{n}}}-\dfrac{1}{2^{\frac{1}{n+1}}})=\dfrac{1}{2}-\dfrac{1}{2^{\frac{1}{n+1}}}$
Thus, we have the sum:
$\lim\limits_{n \to \infty} s_n=\lim\limits_{n \to \infty} [\dfrac{1}{2}-\dfrac{1}{2^{\frac{1}{n+1}}}]=\lim\limits_{n \to \infty} \dfrac{1}{2}- \lim\limits_{n \to \infty}\dfrac{1}{2^{\frac{1}{n+1}}}=\dfrac{1}{2}-1=-\dfrac{1}{2}$
