$\underset{n\to \infty }{\lim }\left\{\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right)....\left(2^{\frac{1}{2}}-2^{\frac{1}{2n+1}}\right)\right\}$ is equal to

Let,

$L=\underset{n\to \infty }{\lim }\left\{\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right)....\left(2^{\frac{1}{2}}-2^{\frac{1}{2n+1}}\right)\right\}$

Now,

 $2^{\frac{1}{2}}-2^{\frac{1}{3}}<1$

$2^{\frac{1}{2}}-2^{\frac{1}{5}}<1$

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$2^{\frac{1}{2}}-2^{\frac{1}{2n+1}}<1 \forall n\in N$

And ${\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)}^n<\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right)....\left(2^{\frac{1}{2}}-2^{\frac{1}{2n+1}}\right)<{\left(2^{\frac{1}{2}}-2^{\frac{1}{2n+1}}\right)}^n$

$\Rightarrow \underset{n\to \infty }{\lim }{\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)}^n<\underset{n\to \infty }{\lim }\left\{\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right)....\left(2^{\frac{1}{2}}-2^{\frac{1}{2n+1}}\right)\right\}<\underset{n\to \infty }{\lim }{\left(2^{\frac{1}{2}}-2^{\frac{1}{2n+1}}\right)}^n$

$\Rightarrow \underset{n\to \infty }{\lim }{\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)}^n<L<\underset{n\to \infty }{\lim }{\left(2^{\frac{1}{2}}-2^{\frac{1}{2n+1}}\right)}^n$

And $\underset{n\to \infty }{\lim }{\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)}^n=0 \& \underset{n\to \infty }{\lim }{\left(2^{\frac{1}{2}}-2^{\frac{1}{2n+1}}\right)}^n=0$

$\left\{\text{as} 2^{\frac{1}{2}}-2^{\frac{1}{3}}<1 \& 2^{\frac{1}{2}}-2^{\frac{1}{2n+1}}<1 \right\}$

Hence, $\underset{n\to \infty }{\lim }\left\{\left(2^{\frac{1}{2}}-2^{\frac{1}{3}}\right)\left(2^{\frac{1}{2}}-2^{\frac{1}{5}}\right)....\left(2^{\frac{1}{2}}-2^{\frac{1}{2n+1}}\right)\right\}=0$