Answer
$-\dfrac{1}{2}$
Work Step by Step
The nth partial sums are: $ 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}}}]$
Now, we have $\lim\limits_{n \to \infty} s_n=\lim\limits_{n \to \infty} [\dfrac{1}{2}-\dfrac{1}{2^{\frac{1}{n+1}}}]$
or, $=\lim\limits_{n \to \infty} \dfrac{1}{2}- \lim\limits_{n \to \infty}\dfrac{1}{2^{\frac{1}{n+1}}}$
or, $=\dfrac{1}{2}-1$
or, $=-\dfrac{1}{2}$